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Question: <p>Adding acids or bases to water, so that either pH or pOH decreases independently of the other, will that affect the auto-ionization of water? For example, pure water + auto-ionized state, with some base added to remove some protons, will it auto-ionize a bit more (create more H+ and OH-) or a bit less (remove some H+ and OH-)?</p>
Answer: <p>It depends how you write the reaction.</p>
<blockquote>
<p>For example, pure water + auto-ionized state, with some base added to remove some protons, will it auto-ionize a bit more (create more H+ and OH-) or a bit less (remove some H+ and OH-)?</p>
</blockquote>
<p>If we call the base <span class="math-container">$\ce{B}$</span> and the conjugate acid <span class="math-container">$\ce{BH+}$</span>, you could write:</p>
<p><span class="math-container">$$\ce{B + H3O+ <=> BH+ + H2O}$$</span></p>
<p>This would mean that you just lost some hydronium ions, and water will ionize a bit more to attain equilibrium with its ions.</p>
<p>Or you could write it like this:</p>
<p><span class="math-container">$$\ce{B + H2O <=> BH+ + OH-}$$</span></p>
<p>This would mean that you just gained some hydroxide ions, and water will ionize a bit less to attain equilibrium with its ions.</p>
<p><strong>What about adding acid?</strong></p>
<p>Same story:</p>
<p><span class="math-container">$$\ce{AH + H2O <=> A- + H3O+}$$</span></p>
<p>or</p>
<p><span class="math-container">$$\ce{AH + OH- <=> A- + H2O}$$</span></p>
<p><strong>Which one is correct?</strong></p>
<p>Either one is fine. You will arrive at the same equilibrium concentrations if you consider one of the acid/base reactions and water auto-dissociation. Depending on what is the major species and what is the minor species, you can choose one or the other. Or if you can't decide, you can be creative and write:</p>
<p><span class="math-container">$$\ce{AH + 1/2 OH- <=> A- + 1/2 H3O+}$$</span></p>
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https://chemistry.stackexchange.com/questions/129275/auto-ionization-equilibrium-of-water-shifted-with-acid-base-addition
|
Question: <p>I learned that </p>
<p>a) the conjugate base to a weak acid is a strong base (and vice versa).</p>
<p>b) a buffer consists of a weak acid (base) and its conjugate base (acid). </p>
<p>However, <a href="http://chemcollective.org/activities/tutorials/buffers/buffers3" rel="nofollow">this</a> explanation of buffers says the following:</p>
<p>"[...] a weak acid is one that only rarely dissociates in water [...]. Likewise, since the conjugate base is a weak base, [...]"</p>
<p>which seems to stand in conflict with my assumption a) above. So, what is correct?</p>
<p>Furthermore, if b) is correct, isn't any solution of a weak acid solution a buffer, since any weak acid in water makes an equilibrium of the form</p>
<p>$HA \text{ (weak acid)} \leftrightarrow H^+ + A^- \text{ (conjugate base)}$</p>
<p>and is thus a solution of a weak acid and its base?</p>
Answer: <p>A strong acid (or base) forms a weak conjugate base (or acid). This is correct. By saying that $\ce{HA}$ is a weak acid and $\ce{A-}$ is a weak conjugate base they mean:</p>
<ul>
<li>Acid shouldn't be too strong - don't use $\ce{NaCl + HCl }$ </li>
<li>Conjugated base shouldn't be too strong - don't use $\ce{EtOH/NaOEt }$ </li>
<li>Use a "not-strong" acid (weak acid) whose conjugate base is also "not-strong" (weak conjugate base). For example, $\ce{H3CCOOH}$ is a relatively weak acid. But $\ce{H3CCOONa}$ is also a relatively weak base. Choose such acids to make a buffer.</li>
</ul>
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https://chemistry.stackexchange.com/questions/55059/confused-about-weak-strong-acids-conjugated-acid-base-pairs-and-buffers
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Question: <p>This question came up in figuring out what the fundamental differences are between redox chemistry and acid/base chemistry (if any).</p>
<p>In redox chemistry, strong oxidants and strong reductants can co-exist in solution because they react slowly with each other (or not at all). An example from biology is a low NADH/NAD+ ratio in the same compartment as a high NADPH/NADP+ ratio.</p>
<p>Are there substances that act as acid or base that are far from equilibrium in aqueous solution for a long (minute, hours) time period? Or are all acid/base reactions in aqueous solution fast and reach equilibrium within seconds?</p>
Answer: <p>Since water is a protic solvent, the predominant form of aqueous acid-base reactions involve proton transfer and that mechanism tends to be quite fast.</p>
<p>But not always.</p>
<p>Most readers are familiar with the use of phenolphthalein as an indicator for titrating acid solutions with a strong base such as sodium hydroxide. When the acid is neutralized, the next increment of sodium hydroxide deprotonates the phenolphthalein molecules and they rapidly turn pink/fuchsia. We proceed to record the data, unaware that with too much additional sodium hydroxide the phenolphthalein color may fade.</p>
<p>That is the slow reaction, for instead of a relatively rapid proton transfer this excess stage is a much slower nucleophilic addition of hydroxide ions to the deprotonated phenolphthalein. From <a href="https://en.wikipedia.org/wiki/Phenolphthalein" rel="nofollow noreferrer">Wikipedia</a>:</p>
<blockquote>
<p>The lactone form (H2In) is colorless between strongly acidic and slightly basic conditions. The doubly deprotonated (In2-) phenolate form (the anion form of phenol) gives the familiar pink color. In strongly basic solutions, phenolphthalein is converted to its In(OH)3− form, and its pink color undergoes a rather slow fading reaction[<a href="https://doi.org/10.1016/S1386-1425(00)00371-1" rel="nofollow noreferrer">1</a>] and becomes completely colorless when pH is greater than 13.</p>
</blockquote>
<p>How slow is "rather slow" is indicated in the educational article by <a href="https://www.researchgate.net/publication/336362530_Primary_Kinetic_Salt_Effect_on_Fading_of_Phenolphthalein_in_Strong_Alkaline_Media_Experimental_Design_for_a_Single_Lab_Session" rel="nofollow noreferrer">González-Arjona et al.[2]</a>. The table below, taken from this reference, indicates that the second-order rate constant for the nucleophilic reaction is about <span class="math-container">$0.02\text{ mol}^{-1}\text{ s}^{-1}$</span>.</p>
<p><a href="https://i.sstatic.net/vN37t.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/vN37t.jpg" alt="enter image description here" /></a></p>
<p>At pH <span class="math-container">$13$</span> we multiply by <span class="math-container">$0.1\text{ M}$</span> hydroxide ion concentration to get a pseudo-first order constant of <span class="math-container">$0.002\text{ s}^{-1}$</span>, so the time constant here is about <span class="math-container">$500$</span> seconds — certainly long enough for kinetic control and convenient for laboratory kinetic measurements.</p>
<p>So if you want to use phenolphthalein indicator for a chemically inspired "It's a girl!" reveal, sodium hydroxide is not the best base to set off the color reveal. A small amount of milk of magnesia in water has the optimal pH to make the pink color but not set off the slow fading reaction.</p>
<p><strong>Reference</strong></p>
<ol>
<li><p>Kunimoto, Ko-Ki (February 2001). "Molecular structure and vibrational spectra of phenolphthalein and its dianion". <em>Spectrochimica Acta Part A</em>. <strong>57</strong> (2): 265–271. Bibcode:2001AcSpA..57..265K. doi:10.1016/S1386-1425(00)00371-1. PMID 11206560.</p>
</li>
<li><p>González-Arjona, D. & Domínguez, Manuel & López-Pérez, German & Mulder, Willem. (2019). "Primary Kinetic Salt Effect on Fading of Phenolphthalein in Strong Alkaline Media: Experimental Design for a Single Lab Session". <em>The Chemical Educator</em>. <strong>24</strong>. 126-132.</p>
</li>
</ol>
|
https://chemistry.stackexchange.com/questions/133203/are-there-acid-base-processes-in-aqueous-solution-that-are-so-slow-that-they-are
|
Question: <p>I was looking at acid-base equilibria of weak acids and weak bases. To determine the equilibrium constant of a reaction of those, I found a formula on the libretexts page:</p>
<p><a href="https://i.sstatic.net/ndfSk.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ndfSk.png" alt="enter image description here" /></a></p>
<p>Where ΔpKa is the difference of the product acid's pKa minus the reactant acid's pKa. This formula makes sense to me, but how can one arrive at this formula, starting with the general definitions of pKa and law of mass action?</p>
<p>An example:</p>
<p><a href="https://i.sstatic.net/UInfb.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/UInfb.png" alt="enter image description here" /></a>
<a href="https://i.sstatic.net/wdrOv.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/wdrOv.png" alt="enter image description here" /></a></p>
<p>Now if we add the two reactions together (using NaOH, hydronium and hydroxide cancel):</p>
<p><a href="https://i.sstatic.net/wKfqp.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/wKfqp.png" alt="enter image description here" /></a></p>
<p>But how can we get the equilibrium constant for this reaction, using only the K values listed above, and thus arriving at the supposed result of K=0.01 (pKa of water is 14)? I am kind of stuck, any help is appreciated.</p>
Answer: <p>Consider a generic acid dissociation reaction of the form:</p>
<p><span class="math-container">$$\ce{HE <=> H^+ + E^-}$$</span></p>
<p>Let:</p>
<p><span class="math-container">$A$</span> represent <span class="math-container">$HE$</span></p>
<p><span class="math-container">$B$</span> represent <span class="math-container">$H^+$</span></p>
<p><span class="math-container">$C$</span> represent <span class="math-container">$E^-$</span></p>
<p>So that the reaction becomes:</p>
<p><span class="math-container">$$\ce{A <=> B + C}$$</span></p>
<p>We're mixing 2 different weak acids together, so each acid will have a different <span class="math-container">$K_a$</span> value. Assuming (1) represents the stronger acid (higher <span class="math-container">$K_a$</span>), and (2) represents the weaker acid (lower <span class="math-container">$K_a$</span>), we have:</p>
<p><span class="math-container">$$\ce{A_1 <=> B_1 + C_1, K_a_1}$$</span></p>
<p><span class="math-container">$$\ce{A_2 <=> B_2 + C_2, K_a_2}$$</span></p>
<p>Now, because (1) is stronger than (2), it will work as an acid and force (2) to work as a base.</p>
<p>This means we need to flip reaction (2) so that the products are the reactants and vice versa.</p>
<p>When we flip a reaction, the new <span class="math-container">$K$</span> value is the inverse of the old <span class="math-container">$K$</span> value, and when we add two reactions together, the resulting <span class="math-container">$K$</span> value is the product of each constant.</p>
<p>So, we get the following system:</p>
<p><span class="math-container">$$\ce{A_1 <=> B_1 + C_1, K_a_1}$$</span></p>
<p><span class="math-container">$$\ce{B_2 + C_2 <=> A_2, \frac{1}{K_a_2}}$$</span></p>
<p><span class="math-container">$$------------------$$</span></p>
<p><span class="math-container">$$\ce{A_1 + C_2 <=> A_2 + C_1 + B_1 - B_2, \frac{K_a_1}{K_a_2}}$$</span></p>
<p>Where: <span class="math-container">$$K_{eq}=\frac{K_{a1}}{K_{a2}}$$</span></p>
<p>If we take the base 10 logarithm on both sides:</p>
<p><span class="math-container">$$log(K_{eq}) = log\left(\frac{K_{a1}}{K_{a2}}\right) = log(K_{a1})-log(K_{a2})=pK_{a2} - pK_{a1}$$</span></p>
<p>We then define:</p>
<p><span class="math-container">$$\Delta P_{ka}=pK_{a2} - pK_{a1}$$</span></p>
<p>So then:</p>
<p><span class="math-container">$$log(K_{eq})=\Delta P_{ka}$$</span></p>
<p>Finally:</p>
<p><span class="math-container">$$K_{eq}=10^{\Delta P_{ka}}$$</span></p>
|
https://chemistry.stackexchange.com/questions/168215/predicting-acid-base-equilibria
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Question: <p>I'm in high school chemistry and just learning about buffer solutions. For example, given a buffered solution of <span class="math-container">$\ce{CH3COOH}$</span> and <span class="math-container">$\ce{NaCH3COO}$</span> with <span class="math-container">$K_a = 1.8 \times 10^{-5}$</span>, we might want to find what the pH will become if <span class="math-container">$\ce{0.010 M KOH}$</span> is added. Given the dissociation of acetic acid:</p>
<p><span class="math-container">$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$</span></p>
<p>My question is, which acid (<span class="math-container">$\ce{CH3COOH}$</span> or the conjugate <span class="math-container">$\ce{H3O+}$</span>) will the base <span class="math-container">$\ce{KOH}$</span> react with? My teacher said that the base will react with <span class="math-container">$\ce{H3O+}$</span> to shift equilibrium in the forward direction, but why wouldn't it react with <span class="math-container">$\ce{CH3COOH}$</span> since it also is an acid?</p>
<p>Is there a specific rule that an added acid/base will react only with the base/acid in the product side of the reaction?</p>
Answer: <p><span class="math-container">$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$</span></p>
<blockquote>
<p>My question is, which acid (CH3COOH or the conjugate H3O+) will the base KOH react with?</p>
</blockquote>
<p>You have to satisfy two equilibria, the one you wrote and the auto-dissociation of water.</p>
<p><span class="math-container">$$\ce{H3O+ + OH- <=> 2 H2O}$$</span></p>
<p>If you add up the two reaction, you get a third one:</p>
<p><span class="math-container">$$\ce{CH3COOH + OH- <=> H2O + CH3COO-}$$</span></p>
<p>You can go either way (use the second or the third reaction) to explain that <span class="math-container">$\ce{OH-}$</span> is part of a neutralization reaction. As a consequence, the concentration of <span class="math-container">$\ce{CH3COO-}$</span> increases and that of <span class="math-container">$\ce{CH3COOH}$</span> and hydronium decreases.</p>
<p><strong>Hydroxide reacts with hydronium</strong></p>
<p>In this case, you would invoke the second reaction. There is more hydroxide (you are adding sufficient KOH to raise the concentration to 0.01 M) than hydronium (at about pH = 5, the buffer will contain about <span class="math-container">$\pu{e-5}$</span> M). As the hydronium gets neutralized, the weak acid equilibrium (first reaction) will shift forward (not reverse - typo in the question), making more hydronium, which will get neutralized etc.</p>
<p><strong>Hydroxide reacts with acetic acid</strong></p>
<p>In a different explanation that results in the same answer, you could say the hydroxide turns some of the weak acid into weak base (third reaction). As a result, the weak acid / weak base concentration ratio will change, causing a net reverse shift in the weak acid equilibrium (first reaction) and lowering the hydronium concentration.</p>
<p><strong>Which way is better?</strong></p>
<p>It is personal preference. If we knew how fast the two reactions go, we might choose the faster one, but we don't. For figuring out the equilibrium concentrations and the pH, it does not matter.</p>
<blockquote>
<p>Is there a specific rule that an added acid/base will react only with the base/acid in the product side of the reaction?</p>
</blockquote>
<p>No, and you can always combine two equilibrium reactions with overlapping species (like the hydronium ion in the first and second reaction) to arrive at a third that also has to be at equilibrium.</p>
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https://chemistry.stackexchange.com/questions/111957/which-acid-base-does-a-strong-base-acid-react-when-added-to-a-buffer-solution
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Question: <p>Fluorides such as <span class="math-container">$\ce{CaF2}$</span> or <span class="math-container">$\ce{LiF}$</span> are sparingly soluble in water. Does their solubility product constant <span class="math-container">$K_\text{sp}$</span> about the equilibrium of</p>
<p><span class="math-container">$$\ce{CaF2 <<=> Ca^2+ + 2 F^-}$$</span></p>
<p>already include the subsequent basicity of the fluoride anion?</p>
<p><span class="math-container">$$\ce{F- + H2O <=> HF + OH^-}$$</span>.</p>
<p>More generally, consider an insoluble binary ionic salt of type "CA", where the cation <span class="math-container">$\ce{Ca^2+}$</span> is a spectator ion and the anion <span class="math-container">$\ce{A^-}$</span> comes from a generic weak acid (HA). Does the reported <span class="math-container">$K_\text{sp}$</span> value of "CA" takes into account the subsequent hydrolysis equilibrium of <span class="math-container">$\ce{A-}$</span> with water?
I think it is impossible to disantangle experimentally the effect of the solubility and the acid-base reaction with water. Both happen simultaneously. Not possible to separate the effects.</p>
Answer:
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https://chemistry.stackexchange.com/questions/188551/competing-solubility-and-acid-base-equilibria
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Question: <p>The system is an aqueous solution of sulfuric acid and sulfur dioxide. I know how much sulfur there is and I know the pH. The system is in equilibrium. I would like to use Octave to solve the system.</p>
<p>I set up the following equations, ignoring $\ce{SO3^2-}$ because the pH is around 1.5.</p>
<p>By definition:
$$ \mathrm{pH} = -\log [\ce{H+}]$$</p>
<p>Conservation of mass:
$$[\ce{H2SO4}] + [\ce{H2SO3}] + [\ce{HSO4-}] + [\ce{SO4^2-}] + [\ce{HSO3-}] = M $$</p>
<p>Henderson-Hasselbalch equations for the system:
\begin{align}
\mathrm{pH} &= \mathrm{p}K_\mathrm{a}(\ce{HSO3-}) + \log \frac{[\ce{HSO3-}]}{[\ce{H2SO3}]}\\
\mathrm{pH} &= \mathrm{p}K_\mathrm{a}(\ce{SO4^2-}) + \log \frac{[\ce{SO4^2-}]}{[\ce{HSO4-}]}\\
\mathrm{pH} &= \mathrm{p}K_\mathrm{a}(\ce{HSO4-}) + \log \frac{[\ce{HSO4-}]}{[\ce{H2SO4}]}\\
\end{align}</p>
<p>I then defined the solution vector to have the following order:
$$ x = [ [\ce{H+}], [\ce{HSO4-}], [\ce{SO4^2-}], [\ce{HSO3-}], [\ce{H2SO4}], [\ce{H2SO3}] ]$$</p>
<p>and my code for Octave is attached. For some reason I always get negative values for some of the results. </p>
<pre><code>function y = phCalc(x)
pH = 1.87;
pKa1 = -5; % this is a guess to drive H2SO4 to ionization in this system
pKa2 = 1.987;% from a textbook
pKa3 = 1.857;% from a text book
M = 0.0012 / 32 * 1000 ; % total sulfur moles per volume
y(1) = 10^(-pH) - x(1);
y(2) = x(5) * 10^(pH-pKa1) - x(2);
y(3) = x(2) * 10^(pH-pKa2) - x(3);
y(4) = x(1) - x(2) - x(3) - x(4);
y(5) = x(2) + x(3) + x(4) + x(5) + x(6) - M;
y(6) = x(4) / 10^(pH-pKa3) - x(6);
endfunction
</code></pre>
Answer: <p>Your code seems to be calculating the <em>differences</em> between a guessed concentration vector <code>x</code> and a calculated concentration vector. These differences in concentrations between guessed and calculated are <code>y</code>. For example, the guessed pH is apparently <code>x(1)</code>, and the calculated pH is <code>10^(-pH)</code>. The difference between them is <code>y(1)</code>.</p>
<p>Thus, its no surprise that the <code>y</code> values are sometimes negative. To solve the problem you need to feed your function to a numerical solver such as <a href="https://www.gnu.org/software/octave/doc/interpreter/Solvers.html" rel="nofollow"><code>fsolve</code></a>. That function will take your guessed solution <code>x</code>, other parameters such as <code>pH</code>, the various <code>pKa</code>s, and <code>M</code>, and your function <code>phCalc</code> as input, and return the value for <code>x</code> which makes all the <code>y</code> variables zero. That solution is the one you want.</p>
<p>On the chemistry side, you mention sulfur dioxide, but I don't see any equations for the solubility of sulfur dioxide gas in water. Additionally, sulfur dioxide is reduced relative to sulfuric acid, so there would need to be an additional reaction, and parameter (extent of reaction or degree of reduction or something) that describes the relative amounts of reduced sulfur (as sulfur dioxide, bisulfite, or sulfite) vs. oxidized sulfur (as sulfate, bisulfate, sulfuric acid, or sulfur trioxide).</p>
<p>There is also the issue of charge balance: what concentration of inert cations (e.g. sodium or potassium or whatever) are you assuming? Charge balance must hold in bulk aqueous solution. If you don't have any sodium / potassium etc., then you should include the self ionization of water as an equation, hydroxide as a species, and <code>y(7) = x(1) - x(2) - 2*x(3) - x(4) - x(OH)</code> as an equation. If you do have sodium/potassium/whatever at significant concentration, you can neglect hydroxide and just go with the equation <code>y(7) = x(1) - x(2) - 2*x(3) - x(4) + x(Na)</code> where <code>x(Na)</code> is the cation concentration, which I have assumed is sodium.</p>
<h1>#</h1>
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https://chemistry.stackexchange.com/questions/34924/how-to-find-the-concentrations-of-the-acid-base-equilibrium-between-sulfuric-aci
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Question: <blockquote>
<p>Acetic acid has a <span class="math-container">$K_\mathrm{a}$</span> of <span class="math-container">$\pu{1.8e-5}$</span>. What is the equilibrium constant for the neutralization of this acid with <span class="math-container">$\ce{NaOH}$</span>?</p>
</blockquote>
<p>Given acetic acid</p>
<p><span class="math-container">$$\ce{HC2H3O2 + H2O <=> C2H3O2- + H3O+} \qquad K_\mathrm{a} = \pu{1.8e-5}$$</span></p>
<p><span class="math-container">$$\ce{HC2H3O2 + OH- <=> C2H3O2- + H2O}$$</span></p>
<p>So, if we do <span class="math-container">$K_\mathrm{w} = K_\mathrm{a}K_\mathrm{b}$</span>, then we get <span class="math-container">$K_\mathrm{b} = \pu{5.55e-10}$</span>. How do I use this to find <span class="math-container">$K_\mathrm{eq}$</span>?</p>
<p>I know how to find <span class="math-container">$K_\mathrm{eq}$</span> using concentration, but I am unsure how to approach this further. The hint says consider the ion product of water, but what does this mean?</p>
Answer: <p>Sodium hydroxide is a strong base and is supposed to be fully dissociated.
You then should've started by writing down the neutralization reaction itself for which you have to determine the equilibrium constant <span class="math-container">$K$</span> and unravel a tangle from there:</p>
<p><span class="math-container">$$\ce{HOAc + OH- <=> OAc- + H2O}$$</span></p>
<p><span class="math-container">$$K' = \frac{[\ce{OAc-}][\ce{H2O}]}{[\ce{HOAc}][\ce{OH-}]}$$</span></p>
<p>Since <span class="math-container">$[\ce{H2O}] = \text{const}$</span> (reaction medium), <span class="math-container">$K'[\ce{H2O}] = K = \text{const}$</span>:</p>
<p><span class="math-container">$$K = \frac{[\ce{OAc-}]}{[\ce{HOAc}][\ce{OH-}]}$$</span></p>
<p>By multiplying both numerator and denominator by <span class="math-container">$[\ce{H+}]$</span>, you can find out that the constant for the neutralization of a weak acid solely depends on the relation between its dissociation constant <span class="math-container">$K_\mathrm{a}$</span> and ionic product of water <span class="math-container">$K_\mathrm{w}$</span>:</p>
<p><span class="math-container">$$K = \frac{\color{red}{[\ce{OAc-}][\ce{H+}]}}{\color{red}{[\ce{HOAc}]}[\ce{OH-}][\ce{H+}]} = \frac{\color{red}{K_\mathrm{a}}}{K_\mathrm{w}}$$</span></p>
<p>For acetic acid:</p>
<p><span class="math-container">$$K = \frac{\pu{1.8e-5}}{\pu{1e-14}} = \pu{1.8e9}$$</span></p>
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https://chemistry.stackexchange.com/questions/109948/equilibrium-constant-for-the-neutralization-of-weak-acid-by-strong-base
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Question: <p>Today I stumbled across a website which defined the titration of weak acid and strong base as the following:</p>
<p>"The titration of a weak acid with a strong base involves the direct transfer of protons from the weak acid to the hydoxide ion. The reaction of the weak acid, acetic acid, with a strong base, NaOH, can be seen below. In the reaction the acid and base react in a one to one ratio." (<a href="https://chem.libretexts.org/Ancillary_Materials/Demos_Techniques_and_Experiments/General_Lab_Techniques/Titration/Titration_of_a_Weak_Acid_with_a_Strong_Base" rel="nofollow noreferrer">https://chem.libretexts.org/Ancillary_Materials/Demos_Techniques_and_Experiments/General_Lab_Techniques/Titration/Titration_of_a_Weak_Acid_with_a_Strong_Base</a>)</p>
<p>CH₃COOH + OH⁻→ CH₃COO⁻ + H₂O</p>
<p>But, I clearly remember my professor for analytical chemistry saying that the neutralization of a weak acid and a strong base occurs due to the shift of equilibrium:</p>
<p>CH₃COOH + H₂O → CH₃COO⁻ + H₃O⁺</p>
<p>Where the strong base (NaOH) would continue to remove hydronium ions from the equilibrium, so that more and more of conjugated base, acetate, would form.</p>
<p>So the first part of my question is; what definition is correct?</p>
<p>Furthermore, indeed we usually formulate following equation to approximately quantify the change in concentrations of acid and base:</p>
<p>CH₃COOH + NaOH → CH₃COO⁻ + H₂O + Na+</p>
<p>This way, we find the moles of acid/base and divide by the total volume to get the pH using Hendersson-Hasselbalch equation. But is it correct this way? Are there more accurate methods? The weak acid should dissociate first before being consumed, is that correct?</p>
<p>Anyway, this is not a question about how to calculate the titration curve of this example, rather it is about understanding the nature of the conversion. Thanks in advance!</p>
Answer: <p>Your second equation is an equilibrium. Both acetic acid and acetate ions are always present in a solution, although their proportion may vary. When <span class="math-container">$\ce{CH3COOH}$</span> is dissolved into water, a fraction of the molecules are transformed by reaction with water, producing some <span class="math-container">$\ce{H3O+}$</span> But if these ions are destroyed by adding <span class="math-container">$\ce{OH-}$</span> ions, more molecules <span class="math-container">$\ce{CH3COOH}$</span> have to react with water. Acetic acid is transformed into acetate ions to compensate for the consumption of the first <span class="math-container">$\ce{H3O+}$</span> ions. So more and more molecules of acetic acid are transformed into ions. The final result is similar to the first equation : more and more <span class="math-container">$\ce{CH3COOH}$</span> is consumed by adding <span class="math-container">$\ce{NaOH}$</span> and more and more acetate ion and water is produced from this process.</p>
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https://chemistry.stackexchange.com/questions/158751/understanding-weak-acid-strong-base-titration
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Question: <p>Since $\ce{HCl}$ is more acidic than $\ce{HF}$, the left hand side of a reaction has a stronger acid than the right hand side.</p>
<p>Since $\ce{NaF}$ is more basic than $\ce{NaCl}$, the left hand side has a stronger base relative to the right hand side.</p>
<p>$$\begin{align}\text{Stronger Acid/Base} & \ce{->} \text{Weaker Acid/Base}\\ \text{(Less Stable)} & \ce{->} \text{(More Stable)}\end{align}$$</p>
<p>So shouldn't the equilibrium constant be larger than $1$? </p>
<p>My textbook says the equilibrium constant is less than $1$.</p>
Answer: <p>We can entirely ignore the sodium cation; it is only a spectator ion. This means, the reaction we are observing is:</p>
<p>$$\ce{HCl + F- <=> Cl- + HF}\tag{1}$$</p>
<p>The equilibrium constant for this reaction is:</p>
<p>$$K = \frac{[\ce{Cl-}][\ce{HF}]}{[\ce{HCl}][\ce{F-}]}\tag{2}$$</p>
<p>We can expand this equation:</p>
<p>$$K = \frac{[\ce{Cl-}][\ce{H+}][\ce{HF}]}{[\ce{HCl}][\ce{H+}][\ce{F-}]}\tag{2'}$$</p>
<p>And then we realise that that is nothing else than:</p>
<p>$$K = \frac{[\ce{Cl-}][\ce{H+}][\ce{HF}]}{[\ce{HCl}][\ce{H+}][\ce{F-}]} = \frac{[\ce{Cl-}][\ce{H+}]}{[\ce{HCl}]} \cdot \frac{[\ce{HF}]}{[\ce{H+}][\ce{F-}]} = \frac{K_\mathrm{a}(\ce{HCl})}{K_\mathrm{a}(\ce{HF})} \tag{2''}$$</p>
<p>So the equilibrium constant is the fraction of the acidity constants of $\ce{HCl}$ and $\ce{HF}$. We know that $\ce{HCl}$ is a much stronger acid than $\ce{HF}$. This is reflected by the inequation:</p>
<p>$$K_\mathrm{a}(\ce{HCl}) > K_\mathrm{a}(\ce{HF}) \tag{3}$$</p>
<p>Since the numerator is larger than the denominator, the value of the fraction must be larger than $1$. Therefore, the product side is preferred.</p>
<p>If your book arrives at any other conclusion, it is disregarding experimental results and should be replaced.</p>
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https://chemistry.stackexchange.com/questions/39406/how-to-estimate-the-equilibrium-constant-of-the-reaction-between-naf-and-hcl
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Question: <p>To be honest, I just don't get it.
I had this question while reading a Chemguide article, on this topic:
<img src="https://i.sstatic.net/AMrse.jpg" alt="enter image description here"></p>
<p>I think my confusion is caused by the way I look at pKa of acid-base indicator. Should you think of it more as a reaction quotient. Because technically, at equilibrium, the two concentration aren't the same so cannot be cancelled out.</p>
Answer: <p>Look at the following equation:</p>
<blockquote>
<p>$$\mathrm{pH} = \mathrm pK_{\mathrm a} + \log\frac{[\ce{A^-}]}{[\ce{HA}]}$$</p>
</blockquote>
<p>At the <em>half equivalence point</em>, say we have 10 moles of $\ce{WA},$ and so there will be 5 moles of $\ce{SB}$ as the name suggests ("half equivalence"). So we know that the 10 moles will be neutralized to 5 moles of $\ce{WA}$ by the $\ce{SB}.$ If we do this then the $\ce{CB}$ is also now equal to the $\ce{WA}$ since it will increase by 5 moles. </p>
<p>$$\ce{[WA]~ =~ [CB]}.$$</p>
<p>Thus, </p>
<p>\begin{align}\mathrm{pH} &= \mathrm pK_{\mathrm a} + \log 1\\ &=\mathrm pK_{\mathrm a}\;.\end{align}</p>
<p><strong>So this is why the the $\mathrm{pH} = \mathrm pK_{\mathrm a}$ at the Half-Equivalence Point.</strong></p>
<hr>
<blockquote>
<p><strong>Vocabulary</strong>: </p>
<ul>
<li>SB : Strong Base. (eg. $\ce{NaOH}$)</li>
<li>CB : Conjugate Base. (eg. $\ce{Ac^{-}}$)</li>
<li>WA : Weak Acid. (eg. $\ce{HAc}$)</li>
</ul>
</blockquote>
<hr>
<p><strong>NOTE</strong>: We can never be 100% accurate here, this is just used to derive that $\mathrm{pH} = \mathrm pK_{\mathrm a}$</p>
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https://chemistry.stackexchange.com/questions/29697/why-does-pka-of-a-acid-base-indicator-equal-to-the-ph-when-the-equivalence-point
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Question: <p>a) Differences in solvent</p>
<p>b) Acid base equilibrium (differences of pH)</p>
<p>c) Parasitic or stray radiation</p>
<p>d) Monochromatic radiation</p>
<p>I have some doubts about pH in spectrophotometry. I'm not sure if it is correct but here is what I thought:</p>
<p>a)Differences in solvent. It can be a source of error because disociation, reaction or asociation of the analyte with solvent can generate absorption products different of the analyte.</p>
<p>b)Acid base equilibrium. I don't know why pH is a variable to take into account in spectrophotometry. I thought maybe the case of the determination of phenolphthalein that depends of pH medium. Spectrophotometry doesn´t work when phenolphtalein is in acidic medium. So pH is a multiplicative interference in this case.</p>
<p>c)Parasitic radiation. It is a source of error because it is radiation that doesn't come from the analyte and it generates signal.</p>
<p>d)Monochromatic radiation. I think that is the wrong answer because it is what we need to avoid big changes in the absorbance as a function of the wavelength.</p>
Answer: <p>It is a very open ended question, but first three choices will cause errors. Regarding your pH issue, you are right, it is multiplicative error. Take the example of phenolphthalein, as you say that its color (=absorption spectrum) is different in acidic or basic medium. So pH change can shift the lambda max of analytes, hence the cause of error.
Phenolphthalein still absorbs light in the acidic medium. I cannot find phenolphthalein spectra, but let us take phenol red indicator, you can see drastic changes in absorbances at the lambda max as a function of pH.</p>
<p>If by some random chance, you were measuring at the isosbestic point, then pH will not affect absorbance but in all other cases pH will change the A values drastically.</p>
<p><a href="https://i.sstatic.net/AqrDc.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/AqrDc.png" alt="enter image description here" /></a></p>
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https://chemistry.stackexchange.com/questions/145857/which-of-the-following-can-be-a-source-of-error-in-spectrophotometry-mark-the-w
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Question: <p>So we know that $K_\mathrm w$ is $10^{-14}$. But what if I am working with non-aqueous systems? Suppose I were to add LDA to liquid ammonia, or dissolve HCl in concentrated sulfuric acid. How would I calculate the percentage of dissociation? I have some inkling of how to do this but I don't really see discussions online and want to make sure my procedure is correct. Furthermore, I don't know of what good keywords to find the self-dissociation constant of ammonia or sulfuric acid are and this is really frustrating. </p>
<p>To emphasize: I am looking for solutions for <em>non-aqueous</em> systems.</p>
Answer: <p>For all solvents that can dissociate into ions, a dissociation constant can indeed be measured. Expect them to be unpredictable in each direction.</p>
<p>Furthermore, the $\mathrm pK_\mathrm a$ values you know are typically those measured in water, although some organic chemists rely more heavily on the DMSO tables e.g. the one by <a href="http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf" rel="nofollow noreferrer">Evans</a>. Just a quick look at the first column shows you that there is no way you can predict the $\mathrm pK_\mathrm a$ value in one solvent from the one in a different solvent:</p>
<p>$$\begin{array}{lcc}\hline
\text{compound} & \mathrm pK_\mathrm a(\ce{H2O}) & \mathrm pK_\mathrm a(\ce{DMSO}) \\ \hline
\ce{HF} & 3.17 & 15\phantom{.0}\\
\ce{HCN} & 9.4\phantom{0} & 12.9\\ \hline\end{array}$$</p>
<p>While $\ce{HF}$ is the stronger acid in water, $\ce{HCN}$ is the stronger acid in $\ce{DMSO}$.</p>
<p>Couple this with a different dissociation constant means that you can calculate <em>nothing</em> ab initio.</p>
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https://chemistry.stackexchange.com/questions/85644/how-to-calculate-he-equilibrium-constant-change-for-acid-base-reactions-in-a-dif
|
Question: <p>Take the following simple equilibrium:
$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$</p>
<p>Ethanoic acid will act as a typical Brønsted-Lowry acid and donate a proton to water, forming a conjugate acid-base pair. In this case, H<sub>3</sub>O<sup>+</sup> is the conjugate acid, and CH<sub>3</sub>COO<sup>-</sup> is the conjugate base.</p>
<p>Logic tells us that if we have a solution of ethanoic acid and water, increasing the amount of acid will decrease the solution's pH. This makes sense, since there'd literally be a higher acid concentration.</p>
<p>However, the interpretation I get due to the equilibrium is the opposite. I see it as increasing the concentration of ethanoic acid will mean there's a larger concentration of conjugate base, hence the solution will become more basic.</p>
<p>Is this flawed logic or simply a misunderstanding of Le Chatelier's principle?</p>
Answer: <p>You are neglecting the hydronium ions produced during dissociation of acetic acid.</p>
<p>We can consider the following sequence of events when we add more acetic acid to solution. In reality they occur simultaneously, but it's easier to think about it sequentially.</p>
<ol>
<li>We have more than the equilibrium concentration of acetic acid, so some of it dissociates to produce $\ce{H+}$ and acetate ion.</li>
<li>Now we have more than the equilibrium concentration of acetate ion, so some of it reacts with $\ce{H+}$ to form acetic acid again.</li>
</ol>
<p>Your argument is that step 2 makes the solution more basic, but you are neglecting the effect of step 1. Every extra acetate ion produced in step 1 <em>also involves the production of a $\ce{H+}$.</em> Furthermore, step 2 has to occur to a lesser degree than step 1, because step 2 consumes the acteate ions produced in step 1. The acidity of the solution must therefore go up as a whole.*</p>
<hr>
<p>*Of course it's also possible that steps 1 and 2 both occur at nearly the same rates, and then the pH of the solution doesn't really change much as a result. This is what happens in a buffer.</p>
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https://chemistry.stackexchange.com/questions/99890/a-simple-equilibrium-problem
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Question: <p>I'm going back through my notes from a chemistry class, which I did well in despite not having a very good conceptual understanding, in order to hopefully develop that understanding, but there's something I can't find in the book or online. Why is it that, when solving for the pH of a solution that results from the mixture of a strong acid with a weak base or a weak acid with a strong base, we can treat the acid-base reaction as going to completion before solving any resulting equilibrium problem, or completely ignoring any equilibrium in the pre-mixed solutions.</p>
<p>For a concrete example, when mixing 40.0 mL of 0.50 M ammonia with 25.0 mL of 0.30 M hydrochloric acid, the approach the class would have me use is to first solve the limiting reaction problem as if the reaction goes to completion. Doing so, we see that there are 0.0075 mol of chloride ions, 0.0075 mol of ammonium, and 0.013 mol of ammonia. Then, because we don't have excess hydrogen ions, nor excess hydroxide, nor only neutral ions, we solve the equilibrium problem of ammonium with ammonia.</p>
<p>What I don't understand is how we can ignore the fact that the ammonia solution would have already been in equilibrium when mixed with the hydrochloric acid, and as I understand it should be constantly arriving towards equilibrium as the reaction is taking place.</p>
<p>Alternatively, if there had been excess hydrogen ions, I was told to use the concentration of hydrogen ions from the completed reaction, but that would also be ignoring the fact that the ammonium contributes to the pH (because it again ignores it's equilibrium).</p>
Answer: <p><strong>The strategy of major species</strong></p>
<p>You correctly describe the strategy. First, let reactions go to completion (in the direction that makes sense, i.e. weak base and strong acid forms weak acid and spectator ion - never the other way around). Then, check what the major species are and estimate the pH. Finally (and this step is often omitted), see if equilibria involving minor species (such as hydroxide for acidic solutions) need adjustment.</p>
<p><strong>Why does the strategy of major species work?</strong></p>
<p>For the example, the OP already calculated the estimated amount of species: 0.0075 mol of ammonium, and 0.013 mol of ammonia. With a total volume of 65 mL, that comes out as:</p>
<p><span class="math-container">$$c_\ce{NH4+} = \pu{0.115M}$$</span>
<span class="math-container">$$c_\ce{NH3} = \pu{0.20M}$$</span></p>
<p>From this, we can estimate the pH to be a bit more basic than the <span class="math-container">$\mathrm{p}K_\mathrm{a}$</span>, i.e. 9.49. Now we can think about the minor species, hydroxide and hydronium. At pH = 9.49, there should be an excess of hydroxide over hydronium, by about 0.00003 M. Dissociation of water does not give an excess, so it has to come from ammonia turning into ammonium:</p>
<p><span class="math-container">$$\ce{NH3 + H2O <=> NH4+ + OH-}$$</span></p>
<p>However, when this reaction makes hydroxide, it also changes the ratio of ammonium to ammonia, changing the pH. The only reason we don't keep running in circles, adjusting one equilibrium and then the other, is that this adjustment is tiny compared to the concentration of the major species. It amounts to decreasing the ammonia concentration by 0.00003 M and increasing the ammonium concentration by the same amount. The shift in pH is so small that it makes no difference when the pH is written with appropriate number of significant figures.</p>
<p><strong>Alternate strategy</strong></p>
<p>For the calculators we use, this is a good strategy to deal with multiple equilibria (in your case, that of the weak acid/base pair and autodissociation of water).</p>
<p>If you had an analog computer that adjusts the concentrations based on a "pH slider" and reports out whether all the constraints are met, there would be no reason to do this in multiple steps. Instead, you would just push your pH slider from one extreme to the other and stop when all equations are satisfied.</p>
<p>In your case, you could calculate the ratio of ammonia to ammonium from the pH and the <span class="math-container">$\mathrm{p}K_\mathrm{a}$</span>, and the hydroxide concentration from the autodissociation constant of water. Then, you vary the pH until the charge balance (chloride plus hydroxide has to match hydronium plus ammonium) is achieved. It is a fun exercise to program this into a spread sheet. </p>
<p><strong>What actually happens?</strong></p>
<p>All reactions (any of the acids - water, hydronium or ammonium - reacting with any of the bases - water, ammonia or hydroxide) go on at the same time. The further from equilibrium a reaction is, the faster is the net change. Once all the reactions are at equilibrium, there is no more net change. The details depend on the kinetics but are not relevant to the equilibrium state that is finally reached.</p>
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https://chemistry.stackexchange.com/questions/134540/what-actually-happens-in-strong-acid-weak-base-reactions
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Question: <p>For example, in the reaction $\ce{NaOH +HCl->NaCl +H_2O}$, the products are favoured by a factor of approximately $10^{24}$.
Is this a general rule? Is it because the products are more stable than the reactants?</p>
Answer: <p>Here is a simple intuitive explanation. In your above example, there is actually 2 reactions occurring, the forward and the reverse reaction. Let us consider the forward reaction. Here, both $\ce{NaOH}$ and $\ce{HCl}$ are strong bases and acids. What this means is that HCl has a strong ability to donate a proton and NaOH has a strong ability to accept protons. This means that when NaOH and HCl react together, the reaction will basically go to completion. So you will end up with up with very little NaOH and HCl and the equilibrium will favour the products.</p>
<p>Now lets look at the reverse reaction which is: $$\ce{Na+ + Cl- + H2O -> NaOH + HCl}$$ Here $\ce{Cl-}$ is the conjugate base of HCl which is a strong acid. This means that $\ce{Cl-}$ is a really weak base, meaning that it is pretty bad at pulling protons from molecules. Also water is the conjugate acid of NaOH, which is a strong base. This means that water is a really weak acid, meaning that it is also pretty bad at kicking out one of its protons. This means that in this reaction, barely any of the products will react and you will mainly end up with NaCl and water.</p>
<p>As you can see in both reaction the equilibrium will favour the weak acid and base as they aren't able to react that well together. This effect can also be explained by using equilibrium constants but I won't go into that.</p>
|
https://chemistry.stackexchange.com/questions/38831/why-does-equilibrium-favour-weak-acid-or-weak-base
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Question: <p>How would you know when an acid or base is paired with <span class="math-container">$\ce{H2O}$</span> that it will form an <span class="math-container">$\ce{OH-}$</span> ion or a <span class="math-container">$\ce{H3O+}$</span> ion? I just started the acid and base equilibrium unit, and I'm just confused on what conditions <span class="math-container">$\ce{OH-}$</span> or <span class="math-container">$\ce{H3O+}$</span> ions form.</p>
Answer: <p>Acids are hydrogen ion donors. When the react with water, they can give a hydrogen ion to form <span class="math-container">$ \ce{H3O+}$</span>.<br/>
For example:<br/></p>
<p><span class="math-container">$ \ce{HCl(aq) + H2O(l) \rightarrow Cl-(aq) + H3O+(aq)}$</span></p>
<p>Simple acids, such as <span class="math-container">$ \ce{HCl}$</span> or <span class="math-container">$\ce{H2SO4}$</span>, can be recognized as acids by the H at the start of the formula. Other more complex acids may be written with <span class="math-container">$ \ce{COOH}$</span> at the end of the formula, which denotes a carboxylic acid. An example of this is acetic acid: <span class="math-container">$\ce{CH3COOH}$</span>. Otherwise, the name of the compound should give it away.</p>
<p>Bases are hydrogen ion acceptors, and generate <span class="math-container">$\ce{OH-}$</span> when they react with water. Strong bases are generally soluble metal hydroxides, such as <span class="math-container">$\ce{NaOH}$</span>. You should be able to recognize them from their formulas.</p>
<p>Simple weak bases often contain nitrogen, like ammonia: <span class="math-container">$\ce{NH3}$</span>.</p>
<p><span class="math-container">$\ce{NH3(aq) + H2O(l) \rightleftharpoons NH4+(aq) + OH-(aq) }$</span></p>
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https://chemistry.stackexchange.com/questions/147865/determine-which-ion-h3o-or-oh-is-in-excess-at-the-end-of-the-reaction
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Question: <p>When a buffer (for example, a mixture of 100 mM acetic acid and 100 mM acetate), is diluted 10-fold, all the concentrations change drastically. Surprisingly, after re-establishment of equilibrium, the pH is not much different from the starting pH.</p>
<p>How can you visualize how diluting a buffer disturbs the equilibrium of a buffer and how the system returns to equilibrium? To make things simple in a thought experiment, we will posit that dilution is faster than acid/base chemistry, i.e. the first step is dilution without any reactions, and the second step is acid base chemistry without further dilution.</p>
<p>The explanation/visualization could be at the molecular or bulk level, and could use kinetic or K vs. Q or Gibbs free energy arguments. </p>
Answer: <p>Consider an acid buffer with <span class="math-container">$pK_a=5$</span>. Ignore the hydroxyl concentration, assuming it is negligible.</p>
<p>Dilution leads to an identical (proportional) reduction in the concentration of all solutes: undissociated acid, complementary base (anion), and protium. Because the concentration of anion and protium is reduced, encounters between the two will occur less frequently. On the other hand, dissociation of the acid into component ions continues unabated, since this process is (we are ignoring activities) a first order reaction. The result is an increase in the concentrations of anion and protium, thus reestablishing a value of the protium concentration close to that prior dilution.</p>
<p>Other important factors should be remembered in the context of ideal buffers and the Henderson-Hasselbach equation: (1) the concentrations of <span class="math-container">$\ce{A^-}$</span> and <span class="math-container">$\ce{HA}$</span> are set equivalent in an ideal buffer (2) the <span class="math-container">$\ce{H^+}$</span> concentration is set equal to the <span class="math-container">$K_a$</span> and (3) the concentrations of <span class="math-container">$\ce{A^-}$</span> and <span class="math-container">$\ce{HA}$</span> are much higher than those of <span class="math-container">$\ce{H^+}$</span> . That means that in both absolute and relative amounts, the concentrations of <span class="math-container">$\ce{A^-}$</span> and <span class="math-container">$\ce{HA}$</span> need to change by only a small amount in order to bring the pH back near the desired set point. This is very clear from the following plots (concentrations are molar and <span class="math-container">$pK_a = 5$</span>). Every 0.1 time units the concentration is diluted by 17%. The pH spikes up but then settles back again as the acid dissociates. You practically don't see the changes in the undissociated acid and anion on the plot during the recovery because they are tiny. </p>
<p><a href="https://i.sstatic.net/XLgkW.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/XLgkW.jpg" alt="enter image description here"></a></p>
<p>You can compare this to the case when your buffer is not as good (same <span class="math-container">$pK_a$</span>, but lower capacity, 100 <span class="math-container">$\mu M$</span> concentration). I altered the kinetics too so the pH would stabilize after each dilution:
<a href="https://i.sstatic.net/Tf3Tz.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Tf3Tz.jpg" alt="enter image description here"></a></p>
<p>I should add: ignoring <span class="math-container">$[OH^-]$</span> is admittedly bad near neutral pH/low buffer capacity, but then I was attempting to provide a visualization. Adding terms to cope with the hydroxyl equilibrium etc would complicate things without adding much value or accuracy (since the kinetics are not meant to be accurate anyway, just illustrative).</p>
<hr>
<p>Addendum: Here's the code:</p>
<pre><code>K = 1e-5; % M
kd=10; % arbitrary, set to alter convergence speed, numerical stability
dt = 1e-7; % arbitrary, alter to change convergence speed, numerical stability
v0 = [0.2; 0.2 ; K]; % M [HA], [A], [H],
%% for dilute conditions
%kd=80;
%dt = 1e-5;
%v0 = [0.0001; 0.0001 ; K]; % [HA], [A], [H]
Nstep = 10000;
t = [0:dt:(Nstep-1)*dt];
tdilute = t(round(Nstep*[0.1:0.1:0.9]));
dv_dt = @(kd,K,v) [ -kd*v(1)+kd/K*v(2)*v(3); kd*v(1)-kd/K*v(2)*v(3)] ;
dv_ = @(v,dt) dv_dt(kd,K,v)*dt;
dv = @(v) dv_(v,dt);
v = v0;
conc = v0;
for ii=2:Nstep
vstep = dv(v);
v = v + [vstep(1); vstep(2); vstep(2)];
conc(:,ii) = v;
if any(t(ii) == tdilute)
v = v/1.2; % <-- change factor to dilute by here
end
end
figure
subplot(2,1,1)
plot(t,conc(1,:),'r')
hold on
plot(t,conc(2,:),'g--')
plot(t,conc(3,:),'b--')
ylabel('[c]')
legend('[HA]','[A^-]','[H^+]')
subplot(2,1,2)
plot(t,-log10(conc(3,:)),'b--')
ylabel('pH')
xlabel('time (au)')
</code></pre>
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https://chemistry.stackexchange.com/questions/109968/what-are-the-concentration-changes-when-diluting-an-equimolar-acetic-acid-aceta
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Question: <p>I have a question concerning acid/base reactions. Is it true that an alcohol is a better base than water?
My Professor listed the basicity (because we talked about E1/E2 and Sn1/Sn2 reactions) of the following compounds the following way: RO- > OH- > ROH > H2O (with stronger base being the RO- and the weakest here H2O). I would like to understand that with respect to pka/pkb values...</p>
<p>Now I came across the following problem:
Bronstedt Bases/Acids are determined by their equilibrium constant in WATER (pka/pkb). Alcohols (like ethanol have a pka about 16-19), water has a pks of 14, which makes alcohol a base by definition.</p>
<ol>
<li>if that happens <strong>-> ROH + H2O -> ROH2(+) + OH(-)</strong>, wouldn't you calculate the <strong>pkb</strong> for <strong>ROH</strong> and the pks for ROH2+? The alcohol is the base here? Unfortunately I haven't found any pkb value for ROH on the internet... (I don't mean the conjugated base of ROH!) - alcohols are even considered amphoteric? how is that possible? why do I only find pka (and no pkb) values then, like for ammonia?</li>
<li>Ammonia is considered amphoteric and has a pka and a pkb value. But if these equilibrium constants are determined in water, how can it have a pka value, if it only acts as a base there? Shouldn't be water the only molecule that can be considered amphoteric in water?</li>
<li>If we would say molecules like H2O and NH3 can "<strong>generally</strong>" donate a proton or take up a proton and because of that are considered amphoteric, aren't there many more molecules than can take up and donate a proton when a strong acid/base is present? wouldn't there be many more amphoteric molecules?</li>
</ol>
<p>So I'm really trying to understand why if alcohols can act as base AND as acid, why there are not two values for that (like for water (pka and pkb=14) and ammonia (pka=23 and pkb= 4,75), even though that would require two different reactions in alcohols too:
ROH -> ROH2(+)
ROH -> RO(-)</p>
<p>I hope I haven't confused you too much and explained my problem at least somewhat clearly. :)</p>
Answer:
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https://chemistry.stackexchange.com/questions/145185/base-acid-reaction-alcohol-amphoteric-why-no-pkb-value
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Question: <p>Suppose I had an acid HA, and it reaches equilibrium in water. Then I remove the hydronium ions. I don't know how I would do this, maybe adding some hydroxide ions. My question is can I do this and will I be able to eventually create a 50/50 mixture of acid and conjugate base? Is this a very good buffer?</p>
Answer: <p>You can't just remove hydronium ions because the solution needs to remain electrically neutral.</p>
<p>You can't just add hydroxide ions, but you could add sodium hydroxide until the HA/A- ratio was 50/50.</p>
<blockquote>
<p>Is this a very good buffer?</p>
</blockquote>
<p>It is the optimum ratio of HA and A- for buffering. </p>
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https://chemistry.stackexchange.com/questions/27146/what-happens-if-i-have-an-acid-base-equillibrium-and-remove-hydronium-ions-until
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Question: <blockquote>
<p>An example of a solution not in equilibrium<br>
(A) Chemical ph
indicator<br>
(B) Acid/base buffer<br>
(C) Anhydrous solution<br>
(D)
hypotonic solution<br>
(E) Supersaturated solution</p>
</blockquote>
<p>I eliminated A and B. Then, I chose E, which is a right answer. Nonetheless, I read then that theoretically any solution or system can be in equilibrium. So, why is the answer E but not C or D?</p>
Answer:
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https://chemistry.stackexchange.com/questions/81550/what-solution-is-not-in-equilibrium
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Question: <p>For example, is 3M HCl a stronger acid than 1M HCl? </p>
<p>I would reason that the concentration of an acid/base does not influence its strength. Strength is determined by the pKa, and, as per Le Chatelier's Principle, the initial concentration does not influence the equilibrium constant. </p>
<p>It may shift the equilibrium concentrations (meaning that the pH is higher for the 3M HCl), but it will not change <span class="math-container">$K_a=\frac{[H_3O^+][Cl^-]}{[HCl]}$</span> at equilibrium.</p>
<p>Is this right?</p>
Answer: <p>The term ‘strong acid’ is sometimes used in a rather fuzzy way and you ran into problems doing so. I prefer to use the term ‘strong acid’ <em>only</em> with respect to an acid’s <span class="math-container">$\mathrm pK_\mathrm{a}$</span> value and disregard all other influences. This gives a clearly defined measure of acid strength and we can easily sort various acids by their strength into stronger or weaker acids.</p>
<p>However, this is looking at the acid as a molecule. In real-life applications you are typically more interested in the property of a <em>solution</em>. To give a real-world example, imagine a <span class="math-container">$\pu{1M}\ \ce{HBr}$</span> solution and a <span class="math-container">$\pu{12M}\ \ce{HCl}$</span>. solution. Obviously, <span class="math-container">$\ce{HBr}$</span> is the stronger acid, but the concentration of <span class="math-container">$\ce{HCl}$</span>—also a strong acid and thus fully deprotonated—is higher. Therefore, the <span class="math-container">$\ce{HCl}$</span> solution is more concentrated or, as some would say, more acidic. It can do greater harm and it is able to protonate more Brønsted base molecules than its <span class="math-container">$\ce{HBr}$</span> counterpart.</p>
<p>If instead of examining a <span class="math-container">$\pu{1M}$</span> and a <span class="math-container">$\pu{12M}$</span> solution I had examined a <span class="math-container">$\pu{e-3M}$</span> and a <span class="math-container">$\pu{1M}$</span> solution, we could even base the discussion around the solution’s resulting pH: the stronger acid is much more diluted and will result in a solution of pH 3 while the weaker acid results in a solution of pH 0.</p>
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https://chemistry.stackexchange.com/questions/105242/does-concentration-or-pka-define-acid-strength
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Question: <p>If I dissolve the weak acid HCN into water, I get a solution of oxonium and CN ions and since there is an equilibrium there is more HCN left.</p>
<p>So I thought CN ions can consequently react with water to form hydroxide-ions, which is a weak base-reaction, by which we also get the original HCN molecules.
I am confused because if its true what I am saying, there are two equilibriums. </p>
<p>Could anyone explain this, and tell me in general whether the bases that are formed when you dissolve an acid can react reversibly to an acid?</p>
Answer: <p>There are 3 equilibriums, described by 3 equilibrium constants:</p>
<p>The first one is the equilibrium of water dissociation:</p>
<p><span class="math-container">$$\ce{H2O <<=>H+ + OH-}$$</span></p>
<p><span class="math-container">$$K_\mathrm{w}=[\ce{H+}][\ce{OH-}]$$</span></p>
<hr />
<p>The second one is for the ( very weak ) hydrogencyanide acid dissociation:</p>
<p><span class="math-container">$$\ce{HCN <<=>H+ + CN-}$$</span></p>
<p><span class="math-container">$$K_\mathrm{a}=\frac{[\ce{H+}][\ce{CN-}]}{[\ce{HCN}]}$$</span></p>
<hr />
<p>The third one is for the cyanide anion hydrolysis:</p>
<p><span class="math-container">$$\ce{CN- + H2O <=> HCN + OH-}$$</span></p>
<p><span class="math-container">$$K_\mathrm{b}=\frac{[\ce{HCN}][\ce{OH-}]}{[\ce{CN-}]}=\frac{[\ce{HCN}] \cdot K_\mathrm{w}}{[\ce{CN-}][\ce{H+}]}=\frac{ K_\mathrm{w}}{K_\mathrm{a}}$$</span></p>
<hr />
<p>All 3 equilibriums are therefore related by equation</p>
<p><span class="math-container">$$K_\mathrm{w}=K_\mathrm{a}\cdot K_\mathrm{b}$$</span></p>
<p>Remember all 3 equilibriums are dynamic, with chemical reactions ongoing simultaneously in both directions at equal rate.</p>
<p>In the case the rates were not equal, the dominant reaction shifts the system toward the equilibrium for the rates to be equal again.</p>
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https://chemistry.stackexchange.com/questions/123428/how-do-you-know-is-a-solution-is-acid-or-basic-when-you-dissolve-a-weak-acid-bas
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Question: <p>This is from Laitinen and Harris, "Chemical Analysis, An Advanced Text & Reference" which describes a simple acid-base equilibrium. They derive a relation in the end directly. I haven't seen this expression in many books which teach solution equilibria. The authors cite an article from the Journal of Chemical Education for the expressions (4 & 5). I didn't find article very useful. How should we derive the equations 4 and 5 in a better way than the author's explanation in italics?</p>
<p>The authors start from very simple equations:</p>
<blockquote>
<p><span class="math-container">$\ce{HA + H2O <=> H3O^+ + A-}\tag{1}$</span></p>
<p>Conventionally,</p>
<p><span class="math-container">$\ce{HA <=> H+ + A^-}\tag{2}$</span></p>
<p>and the ionization of water,
<span class="math-container">$\ce{H2O <=> H+ + OH-}\tag{3}$</span></p>
<p>If <span class="math-container">$C_{HA}$</span> and <span class="math-container">$C_{A^-}$</span> are the analytical concentrations of <span class="math-container">$\ce{HA}$</span> and <span class="math-container">$\ce{A^-}$</span>, and if <span class="math-container">$\ce{[HA]}$</span> and <span class="math-container">$\ce{[A^-]}$</span> are the equilibrium concentrations, the relations...</p>
<p><span class="math-container">$\ce{[HA]} = C_{HA} - \ce{([H^+] - [OH^-])}\tag{4}$</span> </p>
<p><span class="math-container">$\ce{[A^-]} = C_{A^-} + \ce{([H^+] - [OH^-])}\tag{5}$</span></p>
</blockquote>
<p>The authors explain ...</p>
<p>"<em>result because the analytical concentration of HA is diminished by the amount of hydrogen ion produced in Reaction (2), which in turn is the total hydrogen ion concentration minus the hydroxyl ion concentration.</em>"</p>
Answer: <p>There is only one source of hydroxyl ions, ionization of water (Eq. 3 in the OP). Ionized water contributes equal amounts of hydroxyl and hydronium ion. Similarly, dissociation of acid <span class="math-container">$\ce{HA}$</span> results in equal amounts of hydronium and of conjugate base <span class="math-container">$\ce{A-}$</span>. Unless a salt of the conjugate base is explicitly added then the acid is the only source of <span class="math-container">$\ce{A-}$</span>. In absence of salt the total hydronium concentration in solution is therefore </p>
<p><span class="math-container">$$\ce{[H+]}=\ce{[OH-]}+\ce{[A-]}$$</span> </p>
<p>Note that this equation reflects the charge balance condition.</p>
<p>Solving for <span class="math-container">$\ce{[A-]}$</span></p>
<p><span class="math-container">$$\ce{[A-]} = \ce{[H+]}-\ce{[OH-]}\tag{1}$$</span> </p>
<p>We also know that the original amount of acid equals the sum of undissociated acid and conjugate base: </p>
<p><span class="math-container">$$C_\ce{HA}=\ce{[HA]}+\ce{[A-]}$$</span> </p>
<p>Note that this equation reflects the mass balance condition.</p>
<p>Solving for <span class="math-container">$\ce{[HA]}$</span></p>
<p><span class="math-container">$$\ce{[HA]}=C_\ce{HA}-\ce{[A-]} \tag{2}$$</span> </p>
<p>Substituting Eq. (1) into Eq. (2) we obtain </p>
<p><span class="math-container">$$\ce{[HA]}=C_\ce{HA}-(\ce{[H+]}-\ce{[OH-]}) \tag{3}$$</span></p>
<p>The derivation that lead to this expression is not general. It was assumed that no conjugate salt was added to the solution (<span class="math-container">$C_\ce{A-}=0$</span>). If a completely dissociating salt <span class="math-container">$\ce{MA}$</span> is added at concentration <span class="math-container">$C_\ce{A-}$</span> then it is necessary to modify the mass balance expression:</p>
<p><span class="math-container">$$C_\ce{HA}+C_\ce{A-}=\ce{[HA]}+\ce{[A-]}$$</span> </p>
<p>and the charge balance expression: </p>
<p><span class="math-container">$$\ce{[H+]}+\ce{[M+]}=\ce{[OH-]}+\ce{[A-]}$$</span> </p>
<p>or since <span class="math-container">$\ce{[M+]}=C_\ce{A-}$</span></p>
<p><span class="math-container">$$\ce{[H+]}+C_\ce{A-} =\ce{[OH-]}+\ce{[A-]}$$</span> </p>
<p>Solving for <span class="math-container">$\ce{[A-]}$</span></p>
<p><span class="math-container">$$\ce{[A-]} = C_\ce{A-} + \ce{[H+]}-\ce{[OH-]}$$</span> </p>
<p>Combining the two balance expressions leads once again to Eq. (3):</p>
<p><span class="math-container">$$\ce{[HA]}=C_\ce{HA}-(\ce{[H+]}-\ce{[OH-]})$$</span></p>
<p>By the way, the quote you cite should read:</p>
<p>"<em>result because the analytical concentration of HA is diminished by the amount of hydrogen ion produced in Reaction <strong>(2)</strong>, which in turn is the total hydrogen ion concentration minus the hydroxyl ion concentration.</em>"</p>
<p>It might be that "typo" ("3" written instead of "2") that led to the confusion.</p>
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https://chemistry.stackexchange.com/questions/131643/algebraic-treatment-of-equilibrium
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Question: <p>Some substances disproportionate. This means a species with an intermediate oxidation state yields two species with higher and lower oxidation state. For example:</p>
<p><span class="math-container">$$\ce{Hg2Cl2 -> Hg + HgCl2}$$</span></p>
<p>My question is if there are substance that have the analogous acid/base behavior, where an intermediate protonation state yields the conjugate acid and the conjugate base. You could write it like this:</p>
<p><span class="math-container">$$\ce{2AH -> A- + AH2+}\tag{1}$$</span></p>
<p>An example would be water:
<span class="math-container">$$\ce{2H2O <- OH- + H3O+}$$</span></p>
<p>What I am looking for, however, is a case where the mixture of conjugate acid and the conjugate base (of the same amphoteric starting material) are the major species, i.e. a reaction of the type (1) with an equilibrium constant larger than 1. This would mean that the pKa values are in the opposite order than usual, i.e. it is "easier" to lose the second proton than it is to lose the first. For disproportionation, there is a similar feature of the reduction potentials (it is easier to accept the second electron than it is to accept the first).</p>
<p>Are there any such examples?</p>
<p><strong>Update</strong>: To clarify my question, I am looking for substances that skip a protonation state, i.e. pKa2 < pKa1. If you could isolate the intermediate protonation state, it would yield the higher and lower protonation state. This is similar to species that skip an oxidation number (or have the corresponding species at very low concentration) because the reduction potentials favor disproportion of the (hypothetical or barely detectable) species with intermediate oxidation state.</p>
Answer: <p>As I mentioned in the comments, I know of one case where a polyprotic acid has a second proton which is more acidic than the first: the aqueous pervanadyl complex, <span class="math-container">$\ce{[VO_2(H_2O)_4]^+}$</span>. <a href="https://en.wikipedia.org/wiki/Acid_dissociation_constant#Polyprotic_acids" rel="noreferrer">According to Wikipedia</a>, </p>
<p><span class="math-container">$$
\begin{array}{rcl}
\ce{[VO_2(H_2O)_4]^+ &<=>& H3VO4 + H+ + 2 H2O} & \mathrm{pK_{a_1}} = 4.2\\
\ce{H3VO4\ \ &<=>& H2VO4^- + H+} & \mathrm{pK_{a_2}} = 2.60\\
\ce{H2VO4- &<=>& HVO4^2- + H+} & \mathrm{pK_{a_3}} = 7.92\\
\ce{HVO4^2- &<=>& VO4^3- + H+} & \mathrm{pK_{a_4}} = 13.27
\end{array}
$$</span></p>
<p>Though it certainly feels quite odd that <span class="math-container">$\mathrm{pK_{a_2}<pK_{a_1}}$</span>, I suspect this feature is not absurdly uncommon in larger/more complex systems, where structural reorganisation upon deprotonation can be severe. Perhaps what makes the pervanadyl complex unusual is its comparative simplicity.</p>
<p>However, Wikipedia provides no source for those acidity constants. I looked around for an article discussing the speciation of vanadium ions in aqueous solution, and discovered this nice reference:</p>
<blockquote>
<p>Huang, J.-H.; Huang, F.; Evans, L.; Glasauer, S. Vanadium: Global
(Bio)Geochemistry. <em>Chemical Geology</em> <strong>2015</strong>, 417, 68–89.
DOI:<a href="https://doi.org/10.1016/j.chemgeo.2015.09.019" rel="noreferrer">10.1016/j.chemgeo.2015.09.019</a></p>
</blockquote>
<p>The relevant portion in this article is section 4.1. (Aqueous speciation chemistry of vanadium). The authors use thermodynamic data obtained elsewhere to establish the following equilibria:</p>
<p><span class="math-container">$$
\begin{array}{rcl}
\ce{[VO_2(H_2O)_4]^+ &<=>& H3VO4 + H+ + 2 H2O} & \mathrm{pK_{a_1}} = 3.67\\
\ce{H3VO4\ \ &<=>& H2VO4^- + H+} & \mathrm{pK_{a_2}} = 3.40\\
\ce{H2VO4- &<=>& HVO4^2- + H+} & \mathrm{pK_{a_3}} = 8.06\\
\ce{HVO4^2- &<=>& VO4^3- + H+} & \mathrm{pK_{a_4}} = 13.28
\end{array}
$$</span></p>
<p>As you can see, the first two constants are rather different from the values quoted on Wikipedia, and now they've become difficult to distinguish. However, clearly this is still an unusual case. They take these data and then calculate speciation curves to obtain the following graph:</p>
<p><a href="https://i.sstatic.net/CI8gE.png" rel="noreferrer"><img src="https://i.sstatic.net/CI8gE.png" alt=""></a></p>
<p>Evidently, the speciation curves don't really look anything out of the ordinary. It's likely this is partly due to the proximity of the values of <span class="math-container">$\mathrm{pK_{a_1}}$</span> and <span class="math-container">$\mathrm{pK_{a_2}}$</span> used, and also because the solution is highly dilute. It would be nice to know what happens in more concentrated solutions, but unfortunately we get something like this:</p>
<p><a href="https://i.sstatic.net/3eR2O.jpg" rel="noreferrer"><img src="https://i.sstatic.net/3eR2O.jpg" alt=""></a></p>
<p>Like some other transition metals, vanadium has a tendency to oligomerise in solution, forming complex mixtures of <a href="https://en.wikipedia.org/wiki/Polyoxometalate" rel="noreferrer">polyoxometalates</a>. Therefore, it's basically impossible to get a good visualisation of the effects of <span class="math-container">$\mathrm{pK_{a_2}<pK_{a_1}}$</span>.</p>
<p>Fortunately, we can forget the chemistry and just study how the mathematics behaves. For simplicity, let's consider a diprotic acid <span class="math-container">$\ce{H2A}$</span>, where we choose the values of <span class="math-container">$\mathrm{pK_{a_1}}$</span> and <span class="math-container">$\mathrm{pK_{a_2}}$</span>. Also for simplicity, let us safely ignore the autodissociation of water by working with relatively concentrated solutions. Therefore, the mole fractions of the species <span class="math-container">$\ce{H2A}$</span>, <span class="math-container">$\ce{HA-}$</span> and <span class="math-container">$\ce{A^2-}$</span> will be:</p>
<p><span class="math-container">$$X_{H_2A} = \frac{[H^+]^2}{[H^+]^2 + K_{a_1}[H^+] + K_{a_1}K_{a_2}}$$</span></p>
<p><span class="math-container">$$X_{HA^-} = \frac{K_{a_1}[H^+]}{[H^+]^2 + K_{a_1}[H^+] + K_{a_1}K_{a_2}}$$</span></p>
<p><span class="math-container">$$X_{A^{2-}} = \frac{K_{a_1}K_{a_2}}{[H^+]^2 + K_{a_1}[H^+] + K_{a_1}K_{a_2}}$$</span></p>
<p>Let's keep <span class="math-container">$\mathrm{pK_{a_1}}=5$</span> and decrease <span class="math-container">$\mathrm{pK_{a_2}}$</span>. Here are the individual speciation curves for <span class="math-container">$\mathrm{pK_{a_2}=10}$</span> to <span class="math-container">$\mathrm{pK_{a_2}=-1}$</span>, in decrements of 1 unit of <span class="math-container">$\mathrm{pK}$</span> (click on graphs for a larger version).</p>
<p><a href="https://i.sstatic.net/P0nDl.png" rel="noreferrer"><img src="https://i.sstatic.net/P0nDl.png" width="200" height="139"></a><a href="https://i.sstatic.net/GscDb.png" rel="noreferrer"><img src="https://i.sstatic.net/GscDb.png" width="200" height="139"></a><a href="https://i.sstatic.net/Ti8ly.png" rel="noreferrer"><img src="https://i.sstatic.net/Ti8ly.png" width="200" height="139"></a></p>
<p><a href="https://i.sstatic.net/fm5kv.png" rel="noreferrer"><img src="https://i.sstatic.net/fm5kv.png" width="200" height="139"></a><a href="https://i.sstatic.net/sKkf1.png" rel="noreferrer"><img src="https://i.sstatic.net/sKkf1.png" width="200" height="139"></a><a href="https://i.sstatic.net/GLWwq.png" rel="noreferrer"><img src="https://i.sstatic.net/GLWwq.png" width="200" height="139"></a></p>
<p><a href="https://i.sstatic.net/tniIb.png" rel="noreferrer"><img src="https://i.sstatic.net/tniIb.png" width="200" height="139"></a><a href="https://i.sstatic.net/64yHx.png" rel="noreferrer"><img src="https://i.sstatic.net/64yHx.png" width="200" height="139"></a><a href="https://i.sstatic.net/10308.png" rel="noreferrer"><img src="https://i.sstatic.net/10308.png" width="200" height="139"></a></p>
<p><a href="https://i.sstatic.net/4p4yf.png" rel="noreferrer"><img src="https://i.sstatic.net/4p4yf.png" width="200" height="139"></a><a href="https://i.sstatic.net/Au4M7.png" rel="noreferrer"><img src="https://i.sstatic.net/Au4M7.png" width="200" height="139"></a><a href="https://i.sstatic.net/eigcc.png" rel="noreferrer"><img src="https://i.sstatic.net/eigcc.png" width="200" height="139"></a></p>
<p>As you can see, there is no sudden catastrophe as the second proton becomes more acidic than the first. As the second ionisation becomes more favourable, it smoothly and continuously suppresses the formation of <span class="math-container">$\ce{HA-}$</span> . Something nice does happen at the crossover point, where <span class="math-container">$\mathrm{pK_{a_1} = pK_{a_2}}$</span>, namely all species are present in the same concentration when <span class="math-container">$\mathrm{pH = pK}$</span>. However, this not surprising if you look at the equations above.</p>
<p>For good measure, here are all the plots once again, now with a logarithmic y-axis. It's nice to see that even with the suppression of <span class="math-container">$\ce{HA-}$</span>, it still exists as a trace in the background, and still behaves smoothly.</p>
<p><a href="https://i.sstatic.net/5rjHM.png" rel="noreferrer"><img src="https://i.sstatic.net/5rjHM.png" width="200" height="139"></a><a href="https://i.sstatic.net/c1Fko.png" rel="noreferrer"><img src="https://i.sstatic.net/c1Fko.png" width="200" height="139"></a><a href="https://i.sstatic.net/jr17z.png" rel="noreferrer"><img src="https://i.sstatic.net/jr17z.png" width="200" height="139"></a></p>
<p><a href="https://i.sstatic.net/vsw2B.png" rel="noreferrer"><img src="https://i.sstatic.net/vsw2B.png" width="200" height="139"></a><a href="https://i.sstatic.net/EslYF.png" rel="noreferrer"><img src="https://i.sstatic.net/EslYF.png" width="200" height="139"></a><a href="https://i.sstatic.net/m7QYC.png" rel="noreferrer"><img src="https://i.sstatic.net/m7QYC.png" width="200" height="139"></a></p>
<p><a href="https://i.sstatic.net/4nYmw.png" rel="noreferrer"><img src="https://i.sstatic.net/4nYmw.png" width="200" height="139"></a><a href="https://i.sstatic.net/dDlCF.png" rel="noreferrer"><img src="https://i.sstatic.net/dDlCF.png" width="200" height="139"></a><a href="https://i.sstatic.net/BlLwQ.png" rel="noreferrer"><img src="https://i.sstatic.net/BlLwQ.png" width="200" height="139"></a></p>
<p><a href="https://i.sstatic.net/qUpYV.png" rel="noreferrer"><img src="https://i.sstatic.net/qUpYV.png" width="200" height="139"></a><a href="https://i.sstatic.net/VNcj2.png" rel="noreferrer"><img src="https://i.sstatic.net/VNcj2.png" width="200" height="139"></a><a href="https://i.sstatic.net/mBXYP.png" rel="noreferrer"><img src="https://i.sstatic.net/mBXYP.png" width="200" height="139"></a></p>
<p>And to finish things off, now here are graphs showing the mole fraction of each unique species as <span class="math-container">$\mathrm{pK_{a_2}}$</span> varies from 10 to -6, in decrements of 1 unit of <span class="math-container">$\mathrm{pK}$</span>. I rather like the subtle way the curves for the fully protonated and fully ionised species change, as the intermediate ion stops being a significant contributor in the equilibria.</p>
<p><a href="https://i.sstatic.net/ZcdOJ.png" rel="noreferrer"><img src="https://i.sstatic.net/ZcdOJ.png" width="200" height="139"></a><a href="https://i.sstatic.net/jjYtM.png" rel="noreferrer"><img src="https://i.sstatic.net/jjYtM.png" width="200" height="139"></a><a href="https://i.sstatic.net/sDPdj.png" rel="noreferrer"><img src="https://i.sstatic.net/sDPdj.png" width="200" height="139"></a></p>
<p><a href="https://i.sstatic.net/6ddvK.png" rel="noreferrer"><img src="https://i.sstatic.net/6ddvK.png" width="200" height="139"></a><a href="https://i.sstatic.net/ZAlzy.png" rel="noreferrer"><img src="https://i.sstatic.net/ZAlzy.png" width="200" height="139"></a><a href="https://i.sstatic.net/ZQH7z.png" rel="noreferrer"><img src="https://i.sstatic.net/ZQH7z.png" width="200" height="139"></a></p>
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https://chemistry.stackexchange.com/questions/133202/in-acid-base-chemistry-are-there-amphoteric-substances-that-undergo-something-t
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Question: <p>You perform a titration of an acid (let's say <span class="math-container">$\pu{20 ml}$</span> <span class="math-container">$\ce{CH_3COOH}$</span>) with a base (let's say <span class="math-container">$\pu{0.5 M}$</span> <span class="math-container">$\ce{NaOH}$</span>). The burette is marked at <span class="math-container">$\pu{0.10 ml}$</span> intervals. From what I read in a <a href="https://chemistry.stackexchange.com/questions/131402/indicator-error-in-acid-base-titration">previous thread</a>, the largest source of error in acid-base titrations comes from endpoint determination (i.e. adding one drop too much or little). However, I want to investigate if this is the case and compare this error to the indicator error. Below are my reasoning that I need help to correct and fill in.</p>
<p>One drop has approximately a volume of <span class="math-container">$\pu{0.05 ml}$</span>. In the example titration given above, adding one extra drop would lead to overestimating the amount of substance of <span class="math-container">$\ce{CH_3COOH}$</span> with <span class="math-container">$\pu{0.05ml} \cdot \pu{0.5 M} = \pu{0.05E-3} \cdot \pu{0.5 mol} = \pu{2.5E-5 mol}$</span>. This would increase the concentration with <span class="math-container">$\frac{\pu{2.5E-5 mol}}{\pu{0.020 l}} = \pu{0.00125 M}$</span>. This error appears very small.</p>
<p>For the phenolphthalein indicator, we only know the effective pH range (<span class="math-container">$8.3$</span> - <span class="math-container">$10.0$</span> according to <a href="https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/14%3A_Ionic_Equilibria_in_Aqueous_Solutions/14.09%3A_Indicators" rel="nofollow noreferrer">this source</a>) with two significant figures. Furthermore, the pH value of the solution of <span class="math-container">$\ce{CH_3COOH}$</span> and <span class="math-container">$\ce{NaOH}$</span> (which is obtained after the titration) depends on the concentration of <span class="math-container">$\ce{CH_3COOH}$</span>. Let's say the <span class="math-container">$c_{\ce{CH_3COOH}} = \pu{0.1 M}$</span>. When this reaction</p>
<p><span class="math-container">$$\ce{CH3COOH + OH^- ⇌ CH3COO^- + H2O}\tag1\label{eq:one}$$</span></p>
<p>is in equilibrium, the concentration of <span class="math-container">$\ce{CH_3COOH}$</span> and <span class="math-container">$\ce{OH^-}$</span> will have decreased with <span class="math-container">$x$</span>:</p>
<p><span class="math-container">$$x/(0.1-x)^2 = 1.7\times10^9$$</span></p>
<p>The pH of this solution will thus be <span class="math-container">$8.9$</span> (two significant figures). If instead <span class="math-container">$c_{\ce{CH_3COOH}} = \pu{0.2 M}$</span> then the pH of the solution at the equivalence point would have been <span class="math-container">$9.0$</span> (two significant figures). This already presents a difficulty since we don't know <span class="math-container">$c_{\ce{CH_3COOH}}$</span> before the titration. Combined with the broad effective range of the phenolphthalein indicator (which starts to turn pink already at pH <span class="math-container">$8.3$</span>), the indicator error appears larger to me.</p>
<p>Let's investigate this claim. Assume that <span class="math-container">$c_{\ce{CH_3COOH}} = \pu{0.1 M}$</span>. At the equivalence point, the pH will be <span class="math-container">$8.9$</span>. Assume that you saw the faintest colour of pink when the pH was <span class="math-container">$8.3$</span> and you stopped titrating. In this solution, we have <span class="math-container">$c_{\ce{OH^-}} = 10^{-(14-8.3)}$</span>. The equilibrium expression for reaction \eqref{eq:one} is:</p>
<p><span class="math-container">$$\frac{x}{((0.1-x) \cdot c_{\ce{OH^-}})} = 1.7 \times 10^9,$$</span></p>
<p>where <span class="math-container">$x$</span> is the decrease in the concentration of <span class="math-container">$\ce{CH_3COOH}$</span> and <span class="math-container">$\ce{OH^-}$</span> compared to the initial concentrations. The initial concentration of <span class="math-container">$\ce{OH^-}$</span> is <span class="math-container">$10^{-(14-8.3)} + 0.09997\ldots = \pu{0.09997\dots M}$</span>. Thus one would determine <span class="math-container">$c_{\ce{CH_3COOH}} = \pu{0.09997\dots M}$</span> instead of <span class="math-container">$c_{\ce{CH_3COOH}} = \pu{0.1 M}$</span>, which is an error of <span class="math-container">$\pu{0.000027 M}$</span>. Apparently, based on this example, the indicator error is less than the error of adding one drop too much. If this conclusion generalizable?</p>
<p>So if the endpoint determination error is the biggest source of error, the error in the above titration should be <span class="math-container">$\pu{0.00125 M}$</span>. It seems like the error in titration should be larger.</p>
Answer:
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https://chemistry.stackexchange.com/questions/171766/calculating-error-in-acid-base-titration
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Question: <p>Say you have Ammonium Formate in solution, the salt formed by the reaction of Formic Acid and Ammonia. Would both the Formate ion and Ammonium ion protonate and deprotonate, respectively, to form their conjugates until equilibrium is reached? I would also assume that one of the ions would form more of their conjugate than the other depending on which ion in the pair is a stronger acid/base. My question arose when researching the Leuckart reaction, which wikipedia states that if ammonium formate is used, it dissociates to formic acid and ammonia before participating in the reaction. I also found it peculiar that ammonium formate is listed as soluble in diethyl ether, which typically doesnt dissolve salts. This behavior would make more sense if ammonium formate dissociated into formic acid and ammonia.</p>
Answer:
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https://chemistry.stackexchange.com/questions/82959/how-do-salts-of-a-weak-acid-weak-base-behave-in-solution
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Question: <p>When a neutralisation reaction happens, for example, <span class="math-container">$ \pu{100 mol l^-1}$</span> of <span class="math-container">$\ce{HCl}$</span> with <span class="math-container">$\pu{100 mol l^-1}$</span> <span class="math-container">$\ce{NH_3}$</span>, why does <em>all</em> of the base and acid get converted to salt? Why isn't there an equilibrium established between salt and acid-base? </p>
Answer: <p>Acids and bases don't necessarily completely react with each other. Consider your hypothetical reaction between hydrochloric acid and ammonia. </p>
<p><span class="math-container">\begin{align}
\ce{HCl + H3N &-> H4N+ + Cl-}&
\text{Extent:}& \approx100\%
\end{align}</span></p>
<p>The products are <strong>ammonium ion</strong> and <strong>chloride ion</strong> - the <strong>conjugate acid</strong> and <strong>conjugate base</strong>, respectively. So if the forward (above) reaction is complete, how can the reverse reaction between the conjugate acid and conjugate base be anything but insignificant? I.e. </p>
<p><span class="math-container">\begin{align}
\ce{H4N+ + Cl- &<<=> HCl +H3N}&
\text{Extent:}& \approx0\%
\end{align}</span></p>
<p>An equilibrium is established whenever the extent of reaction is not complete. Another good example would be an acetic acid solution. Acetic acid and water exist in equilibrium with acetate anion and hydronium ion. So no not all acid-base reactions go to an extent of 100%. Things to consider: stoichiometry and acid and base strengths. </p>
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https://chemistry.stackexchange.com/questions/22208/why-do-acids-usually-completely-react-with-bases
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Question: <blockquote>
<p>Which of the following statements best characterizes the difference between the titration of a strong acid with a strong base and that of the titration of a weak acid with a strong base?</p>
<p>A) Both the strong acid and the strong base are completely dissociated in solution whereas the weak acid dissociation equilibrium constant is small enough that there is a measurable amount of the un-dissociated acid present in aqueous solution.</p>
<p>B) The equivalence point of a titration of a strong acid with a strong base may be observed with an indicator whereas the equivalence point of a titration of a weak acid with a strong base can only be observed with a pH meter.</p>
<p>C) The titration of a strong acid with a strong base is a fast reaction whereas the titration of a weak acid with a strong base is generally a very slow reaction.</p>
</blockquote>
<p><a href="http://www.gcsescience.com/aa7.htm" rel="nofollow noreferrer">This website</a> seems to indicate that choice C is correct.</p>
<p>For B, all I know is that the pH at the equivalence point does not equal 7.00 (pH > 7.00) for the weak acid titration.</p>
<p>For A, I know the first part of that sentence is true. Strong acids and bases are characterized by completely dissociating whereas weak acids do not dissociate to the same extent. But is it true that there's a "measurable amount" of the un-dissociated acid present in aqueous solutions?</p>
Answer: <p>The answer is A.</p>
<p>Let's look at a weak acid: acetic acid. It has a $K_a$ of $1.8\cdot 10^{-5}$. From this you can calculate how much it will dissociate in water. Let's say we have a concentration of $1~\mathrm{M}$:</p>
<p>\begin{align}
K_a &= \frac{[\ce{CH3COO^{-}}][\ce{H+}]}{[\ce{CH3COOH}]}\\
1.8 \cdot 10^{-5} &= \frac{x \cdot x}{1~\mathrm{M} - x}\\
x &= 4.233 \cdot 10^{-3}\\
\end{align}</p>
<p>This means, at equilibrium, $[\ce{CH3COOH}] = 1~\mathrm{M}- 4.233 \cdot 10^{-3} = 0.995~\mathrm{M}$. As you can see, by far the largest part is not dissociated!</p>
<p>B is not true, you can still observe the equivalence point. Have a look at the following picture, that shows the titration curve for weak / strong acid titrated with strong base. When choosing an indicator for colorimetric titration select one so that the pH jump at the equivalence point contains the interval $\mathrm{p}K_a\pm1$. Phenolphthalein has a $\mathrm{p}K_a \approx 9$, so to decide if it is a suitable indicator you would check if the pH jumps from 8 to 10 at the equivalence point. It does:</p>
<p><img src="https://www.chemicool.com/img1/graphics/titration-strong-weak.gif" alt="Titration curve"></p>
<p>C is not true either. If it was a very slow reaction, you could not titrate accurately, as you would have to wait a long time between every drop of $\ce{NaOH}$.</p>
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https://chemistry.stackexchange.com/questions/4081/what-is-the-difference-between-the-titration-of-a-strong-acid-with-a-strong-base
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Question: <p>I'm wondering if reactions that involve a weak acid and a weak base go to completion. For example, say we have equal amounts and equal volumes of acetic acid and ammonia being mixed together. Would this neutralisation reaction go to completion or would it reach a state of equilibrium?</p>
<p>$\ce{NH_3 + CH_3COOH <=> NH_4^+ + CH_3COO^-}$ </p>
<p>Instinctively, I'm thinking that the reaction would be an equilibrium, since proton transfer should be able to occur between the products ($\ce{NH_4^+ + CH_3COO^-}$) to reform the reactants. Can anyone clarify if this is true?</p>
Answer: <p>Weak acids and bases do indeed remain in an equilibrium state with a certain concentration of the free acid and the free base alongside a certain concentration of the deprotonated acid anion and the protonated base cation.</p>
<p>In general, you can consider any acid-base reaction to be essentially the following:</p>
<p><span class="math-container">$$\ce{HA <=> H+ + A-}\tag{1}$$</span></p>
<p>Which in turn means that the <span class="math-container">$\mathrm pK_\mathrm a$</span> value is defined as in <span class="math-container">$(2)$</span>:</p>
<p><span class="math-container">$$K_\mathrm a = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}\tag{2}$$</span></p>
<p>Reactions <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span> are usually written out in water with hydronium in place of a naked proton but the same principle applies. Of course, a base reaction can be considered essentially the reverse, meaning that the acid constant typically given is actually <span class="math-container">$\mathrm pK_\mathrm a (\ce{HB+})$</span> rather than anything directly corresponding to the free base. If we now take the reaction of a (weak) acid and a
(weak) base, we get equation <span class="math-container">$(3)$</span> and the equilibrium constant as given in <span class="math-container">$(4)$</span>:</p>
<p><span class="math-container">$$\begin{align}\ce{HA + B &<=> A- + HB+}\tag{3}\\[0.7em]
K &= \frac{[\ce{A-}][\ce{HB+}]}{[\ce{HA}][\ce{B}]}\tag{4}\end{align}$$</span></p>
<p>We can now perform simple mathematics with <span class="math-container">$(4)$</span> to arrive at the modified equation <span class="math-container">$(5)$</span> as below:</p>
<p><span class="math-container">$$\begin{align}K &= \frac{[\ce{A-}][\ce{HB+}]}{[\ce{HA}][\ce{B}]}\tag{4}\\[0.7em]
&= \frac{[\ce{A-}][\ce{HB+}][\ce{H+}]}{[\ce{HA}][\ce{B}][\ce{H+}]}\\[0.7em]
&= \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} \times \frac{[\ce{HB+}]}{[\ce{B}][\ce{H+}]}\\[0.7em]
K &= \frac{K_\mathrm a(\ce{HA})}{K_\mathrm a (\ce{HB+})}\tag{5}\end{align}$$</span></p>
<p>We have thus arrived at a way to determine the equilibrium constant of the proton transfer reaction just from the acidity constants of both participants. If both the acid and the base are weak and have similar <span class="math-container">$\mathrm pK_\mathrm a$</span>/<span class="math-container">$\mathrm pK_\mathrm b$</span> values — as ammonium and acetic acid do — the equilibrium constant will be close to <span class="math-container">$1$</span> and thus both sides of the equation will have similar concentrations.</p>
<p>Using <span class="math-container">$(5)$</span>, you can also predict a ‘degree of completion’ for any acid-base reaction.</p>
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https://chemistry.stackexchange.com/questions/83429/do-weak-acid-weak-base-neutralisation-reactions-go-to-completion
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Question: <p>Compounds, commonly used as <a href="https://en.wikipedia.org/wiki/PH_indicator" rel="nofollow noreferrer">pH-indicators</a>, are soluble in aqueous media and polar solvents (DMSO, DMF, NMP etc.) and are functional pretty much across all these solvents. However, practically none these indicators are soluble in non-polar solvents (hexane, chloroform, dioxane etc.).</p>
<p>I briefly searched for the suitable available systems and so far there is only an article about oxoporphyrinogen derivatives used as pH-indicators in $\ce{CH2Cl2}$ [1]. Are there any other <em>common</em> substances that are typically used for the visualization of acid-base equilibrium in solvents with low polarity? </p>
<h3>Reference</h3>
<ol>
<li>Shundo, A.; Ishihara, S.; Labuta, J.; Onuma, Y.; Sakai, H.; Abe, M.; Ariga, K.; Hill, J. P. <em>Chem. Commun.</em> <strong>2013,</strong> <em>49</em> (61), 6870–6872. <a href="https://doi.org/10.1039/C3CC42859A" rel="nofollow noreferrer">DOI: 10.1039/C3CC42859A</a>.</li>
</ol>
Answer:
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https://chemistry.stackexchange.com/questions/87521/what-ph-indicators-are-commonly-used-for-non-polar-medium
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Question: <blockquote>
<p><strong>Question</strong><br>
Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the equations
$$\ce{NO2- + H2O -> ?}$$</p>
</blockquote>
<p>I thought it would be $\ce{NO2- + H2O -> HNO2 + OH-}$, but my textbook says that the correct products are $\ce{HNO3}$ and $\ce{OH-}$. What am I doing wrong here? The answer that they give doesn't even seem to balance.</p>
Answer: <p>It should form nitrous acid... If it was $NO_3^{-}$, nitrate ion, then it would presumably form nitric acid $HNO_3$.</p>
<p>Nitrous acid reacts with water to form nitrite and hydronium:</p>
<p>$$\ce{HNO_2 + H_2O -> NO_2^- + H_3O^+} $$</p>
<p>Water is an amphoteric species. Assuming we just have water the nitrite ion, the nitrite ion accepts the hydrogen forming nitrous acid:</p>
<p>$$\ce{NO_2^- + H_2O -> HNO_2 + OH^-} $$</p>
<p>This reaction is interesting as well since nitrous acid will also dissociate again into nitrite ions and water as well:</p>
<p>$$\ce{HNO_2 + H_2O -> NO_2^- + H_3O^+} $$</p>
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https://chemistry.stackexchange.com/questions/8862/why-isnt-nitrous-acid-a-product-of-nitrite-and-water
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Question: <p>Doesn’t the weak acid dissociate to produce the conjugate base on its own? </p>
<p>I understand that when a strong base is added to the buffer solution, the hydroxide ions will react with the hydrogen ions formed via the dissociation of the weak acid. The equilibrium position will shift rightwards to replenish these hydrogen ions, ensuring that few hydroxide ions remain in the solution. </p>
<p>Conversely, when a strong acid is added to the buffer solution, the hydrogen ions will react with the conjugate base to form the weak acid molecule, thereby ensuring that most of the additional hydrogen ions are removed from the solution, and the pH of the solution remains stable. In the literature I’ve been reading so far, it is mentioned that a salt solution containing the conjugate base ensures that there is a continuous supply of the ions, but can’t the ions simply be replenished by the dissociation of more weak acid molecules, as in the previous case?</p>
Answer: <p>You are correct that a weak acid will only partially dissociate in water. Taking for example the weakly acidic salt <span class="math-container">$\ce{NH4Cl}$</span> of which the ammonium cation is acidic, in water we get reaction (1).</p>
<p><span class="math-container">$$\ce{NH4+ + H2O <=> NH3 + H3O+}\tag{1}$$</span></p>
<p>To analyse this reaction we need the equilibrium constant which in the case of acids is defined as an acidity constant <span class="math-container">$K_\mathrm a$</span> as shown in (2). (Wether we use <span class="math-container">$\ce{H+}$</span> or <span class="math-container">$\ce{H3O+}$</span> is irrelevant.)</p>
<p><span class="math-container">$$K_\mathrm a = \frac{[\ce{NH4+}][\ce{H+}]}{[\ce{NH3}]}\tag{2}$$</span></p>
<p>Using the tabulated <span class="math-container">$\mathrm pK_\mathrm{a}$</span> value of 9.25, we can calculate the amount of <span class="math-container">$\ce{NH3}$</span> in a <span class="math-container">$\pu{1M}$</span> solution of <span class="math-container">$\ce{NH4Cl}$</span>:</p>
<p><span class="math-container">$$\begin{align}K_\mathrm a &= \frac{[\ce{NH4+}]^2}{[\ce{NH3}]}\\[0.3em]
K_\mathrm a (c_0 - [\ce{NH4+}]) &= [\ce{NH4+}]^2\\[0.5em]
[\ce{NH4+}]^2 + K_\mathrm a [\ce{NH4+}] - K_\mathrm a c_0 &= 0\\
[\ce{NH4+}] &= \frac{-K_\mathrm a \pm \sqrt{K_\mathrm a^2 + 4 K_\mathrm a c_0}}{2}\\
[\ce{NH4+}] &= \pu{2.37e-5 M}\tag{3}\end{align}$$</span></p>
<p>Whoops. We may have a good concentration of ammmonium chloride but only a very minor amount of it is still ammonium, <span class="math-container">$99.998~\%$</span> are conjugate base. That doesn’t look like a good buffer system. If you were to add an acid, that would be fine since the excess base can capture that. But if you add more base the pH will go up rapidly, because there are no ammonium ions in solution to buffer it.</p>
<p>Only if you add <em>both</em> a significant amount of an acid and its conjugate base will you have significant amounts of both compounds in solution which are then able to buffer by capturing excess protons or hydroxides.</p>
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https://chemistry.stackexchange.com/questions/105646/why-must-a-buffer-solution-contain-both-a-weak-acid-and-a-salt-solution-of-its-c
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Question: <h2>Context</h2>
<p>The Henderson-Hasselbalch equation is as follows. $$pH=-\log\big(K_a\big)+\log\bigg(\frac{[A^-]}{[HA]}\bigg)$$ One may follow <a href="https://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation#Derivation" rel="nofollow noreferrer">its derivation</a> in order to understand how it came to be, yet there are still facets of it that the proof fails to explain. For instance, the equation provides mathematical evidence for why the pH at the midpoint of a monoprotic acid-base titration is equal to the $\ce{pK_a}$. Since $[A^-]=[HA]$ at the midpoint, the fraction reduces to 1. Because $\log(1)=0$, $pH=-\log\big(K_a\big)+0=-\log\big(K_a\big)$, which, of course, gives the pH at the midpoint.</p>
<p>My question is about the same sort of 'insight.'</p>
<h2>My Question</h2>
<p>Given <em>A</em> mL of <em>B</em> M HA with $\ce{K_a}=C$ titrated by <em>D</em> M <em>x</em>OH at 25<sup>o</sup>C, where HA is a monoprotic weak acid and <em>x</em>OH is a strong base akin to NaOH, the following expression equates to the pH at the equilibrium point (try it out on <em>Example #1-e</em> <a href="https://www.chemteam.info/AcidBase/Titration-problems-weak&strong.html" rel="nofollow noreferrer">here</a>!).</p>
<p>$$pH=14+\log\Bigg(\sqrt{\frac{B*D*10^{-14}}{C(D+B)}}\Bigg)$$</p>
<p>While deriving this, it became necessary for me to determine the $[A^-]$ (it is the major pH determining species at the EQPt). I reasoned as follows.</p>
<blockquote>
<p>To find the $[A^-]$, find the mmol of conjugate base and divide it by the total volume.</p>
<p>Since the mmol of conjugate base is equal to the mmol of weak acid used, and the latter quantity is given by the initial volume of acid times its initial molarity ($A*B$), the former quantity is equal to $A*B$.</p>
<p>The total volume, on the other hand, is equal to the initial volume plus what is needed to reach the equivalence point. The initial volume is given by A, so the expression will be $A+V_2$, where $\ce{V2}$ is what’s needed to reach the equivalence point. Take $\ce{M1V1}=\ce{M2V2}$ and plug in the given variables to get $B*A=D*V_2$. Solve for $\ce{V2}$ to get $V_2=\frac{B}{D}*A$. Substitute the value of $\ce{V2}$ into $A+V_2$ to find the total volume, or $A+\frac{B}{D}*A$.</p>
<p>As mentioned before, divide the two expressions to give the final equation, listed below.</p>
<p>$$[A^-]=\frac{A*B}{A+\frac{B}{D}*A}$$</p>
</blockquote>
<p>However, the expression simplifies as follows.</p>
<p>\begin{split}
[A^-]&=\frac{A*B}{A+\frac{B}{D}*A}\\
&=\frac{A(B)}{A(1+\frac{B}{D})}\\
&=\frac{B}{1+\frac{B}{D}}\\
&=\frac{B}{\frac{D}{D}+\frac{B}{D}}\\
&=\frac{B}{\frac{D+B}{D}}\\
&=\frac{B*D}{D+B}\\
&=\frac{B*D}{B+D}\\
\end{split}</p>
<p>I find it interesting that such a complicated concept (at least to me) simplifies down to the product of two molarities over their sum. Similar to the way that I justified my initial expression and a facet of the Henderson-Hasselbalch equation, can anyone straight-up explain why this final expression yields the EQPt concentration of conjugate base in the given scenario? If it is possible to keep the answer to a relatively basic, non-organic point of view, that would be much appreciated!</p>
<p><em>Please note that I have done the best to explain what I am asking -- I think it is a legal but unordinary question for this site. If I can clarify, do not hesitate to comment!</em></p>
Answer:
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https://chemistry.stackexchange.com/questions/99538/eqpt-ph-of-a-monoprotic-acid-base-titration-product-of-molarites-over-their-sum
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Question: <p>Consider a solution which contains a weak acid and the salt of its conjugate base with a strong base.
e.g. $$\ce{CH3COOH + H2O <=>[$K_\rm{a}$] CH3COO- + H3O+}\tag1$$
$$\ce{CH3COO- + H2O <=>[$K_\rm{h}$] CH3COOH + OH-}\tag2$$</p>
<p>Now, $\displaystyle{K_\rm{a} = \frac{[\ce{H3O+}] [\ce{CH3COO-}]}{[\ce{CH3COOH}]}}$ and $\displaystyle{K_\mathrm{h} = \frac{K_\rm{w}}{K_\mathrm{a}}}$.</p>
<p>Now, my notes say that for the equilibrium concentration of undissociated acid, the acid coming due to hydrolysis of the salt is neglected and the equilibrium concentration is approximately equal to the initial concentration.</p>
<p>It is also mentioned in a book that</p>
<blockquote>
<p>Assume that the extent of protonation of acetate ions and
the deprotonation of acetic acid molecules is so small that the concentrations of both
species are nearly the same as their initial values.</p>
</blockquote>
<p>Why is this so? If $K_\rm{a}$ is low, then $K_\rm{h}$ should be high. I understand that the deprotonation of acid will be low as it is a weak acid so that we can neglect the $\ce{CH3COO-}$ coming through the acid dissociation but <strong>how can we neglect the hydrolysis of the $\ce{CH3COO-}$ coming from the salt?</strong></p>
<p>Edit $-$</p>
<p>My calculation for equilibrium concentrations.</p>
<p>$\ce{[CH3COOH]_\mathrm{eq}} =$ initially added acid $-$ deprotonated acid $+$ hydrolysed salt</p>
<p>$\ce{[CH3COO-]_\mathrm{eq}} =$ initially added salt $-$ hydrolysed salt $+$ deprotonated acid</p>
<p>$\ce{[H3O+]}_{\mathrm{eq}} =$ deprotonated acid $-$ hydrolysed salt</p>
<p>Let the concentration of</p>
<p>initially added acid $= c_1$</p>
<p>initially added salt $= c_2$</p>
<p>deprotonated acid at eq. $= x$</p>
<p>hydrolysed salt at eq. $= y$</p>
<p>Thus, $$K_{\mathrm{a}} = \frac{[\ce{H3O+}] [\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \frac{(x - y)(c_2 - y + x)}{(c_1 - x + y)}$$</p>
<p>Now, assuming $x << c_1, \, c_2$,</p>
<p>$$K_{\mathrm{a}} = \frac{(- y)(c_2 - y)}{(c_1 + y)}$$</p>
Answer: <p>$\mathrm{p}K_\mathrm{a} = 4.75$, therefore $K_\mathrm{a} = 10^{-4.75}$. Thus,</p>
<p>$$K_\mathrm{a} = \frac{[\ce{AcO^-}][\ce{H+}]}{[\ce{AcOH}]}$$</p>
<p>So, let's assume $[\ce{AcOH}] = \pu{1.0 M}$. Just to get a feel, how far the acid is dissociated:</p>
<p>$$K_\mathrm{a} = \frac{x \cdot x}{\pu{1.0 M}} \to x = \sqrt{10^{-4.75}}~\pu{M} = 0.00422 \approx 0.4\%$$</p>
<p>where we have assumed that the contribution from the dissociation of water is negligible, and therefore $[\ce{H+}] = [\ce{AcO-}] = x$.</p>
<p>Now, what happens if I add $\pu{1.0 M}$ $\ce{AcO-}$? Then, effectively, we have:</p>
<p>$$K_\mathrm{a} = \frac{[\ce{AcO-}][\ce{H+}]}{[\ce{AcOH}]} = \frac{\pu{1.0 M}
\cdot [\ce{H+}]}{\pu{1.0 M}}$$</p>
<p>Therefore, we have:</p>
<p>$$[\ce{H+}] = K_\mathrm{a} \frac{\pu{1.0 M}}{\pu{1.0 M}} = \pu{10^{-4.75} M}$$</p>
<p>If we compare the two cases, before addition of acetate, we had a $\mathrm{pH}$ of</p>
<p>$$[\ce{H+}] = \sqrt{10^{-4.75}} \to \mathrm{pH} = 2.375$$</p>
<p>and after addition of acetate, we have</p>
<p>$$[\ce{H+}] = 10^{-4.75} \to \mathrm{pH} = 4.75$$</p>
<p>Meaning that the solution got more basic after the addition of acetate, as excpected.</p>
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https://chemistry.stackexchange.com/questions/82963/why-is-the-hydrolysis-of-the-conjugate-base-of-a-weak-acid-neglected-in-buffer-s
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Question: <p>While solving for weak acid and weak base salt hydrolysis, why do we take the degree of hydrolysis for both the anion and cation to be the same?</p>
<p>If we assume they are the same, when solving for $K_\mathrm{a}$ and $K_\mathrm{b}$ for the cation and anion using simultaneous equilibrium, I always get them to be the same, resulting in the solution always being neutral. I feel that both should have different degrees of hydrolysis.</p>
Answer:
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https://chemistry.stackexchange.com/questions/85650/weak-acid-and-weak-base-salt-hydrolysis
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Question: <p>I was solving a question of salt hydrolysis for a mixture of Weak Acid-Base Salts, given the initial amount of salt and Equilibrium constants. I was trying to approach the problem in two ways, Assuming initially that we have all reactants and no products and move the reaction forward OR assuming all the reactants have been converted into product instantaneously and moving the reaction backward for equilibrium.</p>
<p>As concentrations of various species in equilibrium depends on equilibrium constant powers of their coefficients, which does not seem like any linear or straightforward relation. I wonder if both approaches are equivalent.</p>
<p>I.e. What guarantees the uniqueness of an equilibrium point (that is, specific concentrations of various species at eq.), given just the specific values of initial variables (initial concentrations)?</p>
<p>The paper <a href="https://aip.scitation.org/doi/10.1063/1.1669753" rel="nofollow noreferrer">https://aip.scitation.org/doi/10.1063/1.1669753</a> mentions something like that but the math went over my head. I wonder if there is any simple explanation based on the first principles of thermodynamics, like the curve for free energy for the whole system, does have some restrictions (it should be concave or something like that).</p>
Answer: <p>One of the first experimental demonstrations that the same equilibrium point is reached if the reaction starts with all reactants or all products was <span class="math-container">$\ce{2HI<=>I_2 + H_2}$</span> made by Bodenstein in <span class="math-container">$\approx 1894$</span>. I have redrawn the data.</p>
<p><a href="https://i.sstatic.net/BdVsqm.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/BdVsqm.png" alt="2HI=I2+H2" /></a></p>
<p>figure. <span class="math-container">$\ce{ 2HI=I_2 + H_2}$</span>. <span class="math-container">$x$</span> is the fraction of HI at <span class="math-container">$721$</span> K. <span class="math-container">$x=[HI]/([HI]+2[I_2])$</span></p>
<hr />
<p>In the reaction <span class="math-container">$\ce{A<=>B}$</span> the equilibrium constant is determined by the ratio of rate constants forward and back, <span class="math-container">$k_1,\;k_{-1}$</span> respectively. If another reaction is coupled to the first one <span class="math-container">$\ce{B<=>C}$</span> then this has a different equilibrium constant with rate constants <span class="math-container">$k_2,\;k_{-2}$</span> but these are separate reactions and so the rate constants and hence equilibrium constants are unchanged when A, B, C are together in solution. The concentrations are changed, of course, because when C is added it changes the concentration of B and hence A.</p>
<p>From a thermodynamic perspective the free energy determines the equilibrium constant and adding C does not change the enthalpy or entropy leading to reaction between A and B.</p>
<p>The difficulty is to show (as in your reference) that this is true for all combinations of coupled reactions, linear as well as those forming rings e.g. A-B-C-D-A etc. It is the case, however, that only one set of concentrations is found otherwise the system of reactions could flip between one and the other and a perpetual motion machine could be constructed which is impossible.</p>
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https://chemistry.stackexchange.com/questions/152689/is-equilibrium-point-for-a-coupled-equilibria-of-arbitary-number-of-coupled-reac
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Question: <p>Consider a buffer solution containing a weak acid and its conjugate base (<span class="math-container">$c_\mathrm{total} = \pu{0.1 mol L-1}$</span>):
<span class="math-container">$$\ce{HA <=> H^+ + A^-}$$</span></p>
<p>When <span class="math-container">$\ce{H+}$</span> ions are added (from a strong acid), the position of the equilibrium shifts to the right and <span class="math-container">$\ce{A^-}$</span> combines with most of the <span class="math-container">$\ce{H+}$</span>, preventing a significant decrease in <span class="math-container">$\mathrm{pH}$</span>.</p>
<p>Along the same line, my text book says upon the addition of a strong base, the <span class="math-container">$\ce{OH-}$</span> react with <span class="math-container">$\ce{H+}$</span> in the reaction mixture and reduce their concentration, causing the <span class="math-container">$\ce{HA}$</span> to ionize and replace most of them, which prevents a significant decrease in <span class="math-container">$\mathrm{pH}$</span>.</p>
<p>However, If you add <span class="math-container">$\ce{OH-}$</span> ions from a strong base, say <span class="math-container">$c = \pu{0.01 mol L-1}$</span> to a buffer with <span class="math-container">$\mathrm{pH} = 5$</span>, there obviously aren't enough <span class="math-container">$\ce{H+}$</span> ions in the reaction mixture to react with the base. How will the buffer function in this case?</p>
<p>I can think of two ways:</p>
<ol>
<li>The <span class="math-container">$\ce{OH-}$</span> ions will keep depleting <span class="math-container">$\ce{H+}$</span> from the reaction mixture, driving the equilibrium to the right until all <span class="math-container">$\ce{OH-}$</span> ions are neutralized, and then the system will try to regain equilibrium.</li>
<li>The <span class="math-container">$\ce{OH-}$</span> ions react with <span class="math-container">$\ce{HA}$</span> instead, and then the partial increase in pH occurs from the shifting of the equilibrium to the left.</li>
</ol>
<p>Is one of these correct? Or would the buffer just not work at all if <span class="math-container">$\ce{OH-}$</span> added is more than <span class="math-container">$\pu{E-5 mol L-1}$</span>?</p>
Answer: <p>During neutralization of the buffer by a strong base solution, there are ongoing three reversible reactions, maintaining three equilibria:</p>
<p><span class="math-container">$$
\begin{align}
\ce{H2O + A- &<=> OH- + HA} &\quad K_\mathrm{b} &= \frac{\ce{[OH-][HA]}}{\ce{[A-]}} \tag{1}\\
\ce{H2O &<=> OH- + H+} &\quad K_\mathrm{w} &= \ce{[H+][OH-]} \tag{2}\\
\ce{HA &<=> H+ + A-} &\quad K_\mathrm{a} &= \frac{\ce{[H+][A-]}}{\ce{[HA]}} \tag{3}
\end{align}
$$</span></p>
<p><span class="math-container">$$K_\mathrm{w} = K_\mathrm{a} \cdot K_\mathrm{b}\tag{4}$$</span></p>
<p>Therefore, added <span class="math-container">$\ce{OH-}$</span> reacts with both <span class="math-container">$\ce{HA}$</span> and <span class="math-container">$\ce{H+}$</span>, reaching again equilibria according to <span class="math-container">$K_\mathrm{b}$</span> and <span class="math-container">$K_\mathrm{w}$</span>.</p>
<p>Additionally, <span class="math-container">$\ce{HA}$</span> autodissociation equilibrium makes balance according to <span class="math-container">$K_\mathrm{a}.$</span></p>
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https://chemistry.stackexchange.com/questions/153741/how-exactly-does-a-buffer-made-up-of-a-weak-acid-and-its-conjugate-base-work-upo
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Question: <p>Metals in low Oxidation states are usually reducing agents. Is there any example of a metal in zero Oxidation state acting as an oxidizing agent. </p>
Answer: <p>As with pretty much every "is there ______?" in chemistry, the answer is yes, there is.</p>
<p>Of course, using the pure <em>metal</em> itself as an oxidising agent would be really tough. However, if you use certain tricks to stabilise the negative oxidation state, it is possible. For example, π-acceptor ligands such as $\ce{CO}$ withdraw electron density from the metal. So, $\ce{M[(CO)_n]^m-}$ complexes are stable for certain choices of $\ce{M}$, $n$, and $m$. For more details see: <a href="https://chemistry.stackexchange.com/questions/58052/18-electron-rule-for-determining-the-stability-of-transition-metal-complexes">18 Electron Rule For Determining the Stability of Transition Metal Complexes</a></p>
<p>Consider $\ce{[V(CO)6]}$, a 17-electron complex. It would be quite happy to accept another electron to become $[\ce{V(CO)6}]^-$. This reduction can be achieved with sodium metal. So, here vanadium(0) is acting as an oxidising agent; it is itself reduced to vanadium(−1).</p>
<p>These examples are not incredibly common, but it can happen.</p>
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https://chemistry.stackexchange.com/questions/68558/oxidation-states
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Question: <p>which of the statements is true:</p>
<ol>
<li><p>looking at the electronic configuration of an element, can the possible oxidation states/oxidation number of an elemnet be predicted.
for example, the valence electronic configuration of nitrogen is 2s^2 2p^3. From this can we know all the possible oxidation states exhibited by the element nitrogen.</p>
<pre><code> (OR)
</code></pre>
</li>
</ol>
<p>2.We should first know all the compounds nitrogen can form, and then find the oxidation number of nitrogen in each compound, and then give all the possible oxidation states nitrogen can exhibit.</p>
Answer: <p>The oxidation numbers cannot be predicted from the structure of the atom. They are experimentally determined. Not much chance you predict that while nitrogen can get all integer oxidation numbers from <span class="math-container">$−3$</span> to <span class="math-container">$+5$</span>, its neighbor oxygen can only have <span class="math-container">$−2$</span>, <span class="math-container">$−1$</span>, <span class="math-container">$0$</span>, <span class="math-container">$+1$</span>, <span class="math-container">$+2$</span> in known compounds. And then come fractional oxidation states...</p>
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https://chemistry.stackexchange.com/questions/147911/variable-oxidation-states
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Question: <p>I'm given a question:</p>
<blockquote>
<p>Oxidation number of $\ce{O}$ in $\ce{BaO2}$ is $x$ and in $\ce{OF2}$ is $y$; then the value of $x+y$
is what?</p>
</blockquote>
<p>Now my main question is that if $\ce{F}$ has $-1$ valency in $\ce{OF2}$ then $\ce{O}$ must have valency of $+2$. But is that possible since $\ce{O}$ mainly has oxidation states of $-1$ & $-2$ or am I making any mistake?</p>
Answer: <p>The concept of oxidation state simply works on concept of electronegativity.The more electronegative atom acquire a negative charge while less electronegative atom acquire a positive charge.
Depending on this hypothesis oxygen have 5 oxidation states.</p>
<ol>
<li><p>In all the oxides,oxygen has an oxidation state of $-2$. Eg. $\ce{CO2,CO}$</p></li>
<li><p>In all peroxides (oxygen-oxygen linkage), oxygen has an oxidation state of $-1$. For example, consider $\ce{H2O2}$, here $\ce{H}$ is less electronegative so it will acquire a charge of $+1$ and to balance the $2$ positive charge of 2 H-atoms,each oxygen atom will acquire a charge of $-1$.</p></li>
<li><p>In all superoxides ($\ce{KO2,CsO2,RbO2}$), oxygen has an oxidation state of $-\frac{1}{2}$,this is because $\ce{K,Cs,Rb}$, being elements of the first group and less electronegative than oxygen acquire a charge of $+1$, to balance it, each oxygen atom acquires a charge of $-\frac{1}{2}$.</p></li>
<li><p>In one of the exceptions $\ce{OF2}$, the fluorine being more electronegative acquires a charge of $-1$ and to balance the $-2$ charge of 2 fluorine atoms oxygen acquires a charge of $+2$.</p></li>
<li><p>As last, there is $\ce{O2F2}$, similarly here to balance the $-2$ charge on 2 $\ce{F}$-atoms each oxygen atom acquire a charge of $+1$.</p></li>
</ol>
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https://chemistry.stackexchange.com/questions/60961/oxidation-states-of-oxygen
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Question: <p>When I took high school chemistry many years ago, considerable effort was spent on teaching us to <em>compute</em> oxidation states of atoms in various compounds, following a set of rules that looked somewhat arbitrary to me at the time. As far as I remember, we were never told what benefits (other than passing tests) knowledge of oxidation states would give us.</p>
<p>They were used as part of a convoluted procedure for "balancing redox reactions", but I never saw an example of that where it wouldn't give the same results simply to require that the number of nuclei of each element, as well as the total charge, must be the same on both sides of the reaction, and solve the resulting linear Diophantine equations.</p>
<p>Nevertheless, the concept must be useful other than for setting homework exercises -- I see encyclopedias and other sources use much space on classifying compounds based on the oxidation state of this atom or that. Still I don't recall seeing any case where the oxidation states are <em>used for</em> anything (other than computing other oxidation states).</p>
<p><strong>What is this concept actually used for?</strong> Just a basic example or two where knowing the oxidation states helps produce a meaningful prediction.</p>
<p>It doesn't seem that "oxidation state" actually encodes any particular configuration change inside the atom that tends to be preserved across reactions. Or does it somehow require effort (energy?) to change an atom from
one oxidation state to another? In which case, what actually changes?</p>
<p>Sorry if this is too basic a question. It may be something "everybody knows". I've tried looking e.g. at the <a href="https://en.wikipedia.org/wiki/Oxidation_state">Wikipedia article</a>, but that too seems to be entirely focused on how to <em>determine</em> oxidation states and doesn't explain why one would desire to know them. And all the <a href="/questions/tagged/oxidation-state" class="post-tag" title="show questions tagged 'oxidation-state'" rel="tag">oxidation-state</a> questions <em>here</em> seems similarly to be about determining the oxidation states, and it is left unsaid why the asker wants to know ...</p>
<hr>
<p><strong>Edit:</strong> I'm looking for <em>applications</em> of oxidation numbers, by which I mean rules, laws, tendencies, rules-of-thumb or the like of the general form:</p>
<blockquote>
<p>When you have <em>(condition that involves oxidation numbers)</em> then <em>(prediction that can be verified without knowing what oxidation numbers are)</em>.</p>
</blockquote>
Answer: <p>Of course they are useful. Perhaps you have not come across them yet, but being able to determine the oxidation state of an atom allows us to understand the properties of chemicals and how redox reactions work.</p>
<p>Let's just give two very simple examples. There are tons more. (I don't even want to go into organometallic chemistry, where being able to determine the oxidation state is incredibly important.)</p>
<hr>
<p>Potassium permanganate, $\ce{KMnO4}$, has manganese in a +7 oxidation state. This means that it would have an electronic configuration identical to the noble gas argon: $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6}$, and therefore its $\mathrm{3d}$ orbitals are empty.</p>
<p>Now, most transition metal compounds are said to be coloured because of so-called "d-d* transitions". However, this can only occur if the d orbitals are partially filled - if there are no electrons in the d orbitals, like in $\ce{KMnO4}$, then there are no d-d* transitions available.</p>
<p>As such, the intense purple colour of $\ce{KMnO4}$ has to be explained via a different mechanism. In this case, it is <a href="https://chemistry.stackexchange.com/q/39829/16683">explained by ligand-metal charge transfer</a>.</p>
<hr>
<p>Moving to organic chemistry. Let's say you have an acyl chloride, $\ce{RCOCl}$, and you want to convert it to an aldehyde, $\ce{RCHO}$. If you calculate the respective oxidation states of the carbonyl carbons, you get <a href="http://www.masterorganicchemistry.com/2011/07/25/calculating-the-oxidation-state-of-a-carbon/" rel="nofollow noreferrer">+3 and +1 respectively</a>.</p>
<p>So, that tells you what kind of reagent you need to effect this transformation: you need a <em>reducing agent</em>. Hydrogen gas, $\ce{H2}$, is <a href="https://chemistry.stackexchange.com/questions/60624/difference-between-lindlar-and-rosenmund-catalysts">one such example of a reducing agent</a>. Why? Well, that's because of oxidation states again. $\ce{H2}$ has hydrogen in an oxidation state of 0, and when hydrogen forms a bond to carbon, it has an oxidation state of +1.</p>
<p>Oxidation states are key to understanding why redox reactions work and why certain species are reducing agents (e.g. $\ce{LiAlH4}$: H(-1)) and why others are oxidising agents (e.g. DMSO, $\ce{(CH3)2SO}$, in the Swern oxidation).</p>
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https://chemistry.stackexchange.com/questions/61040/what-are-oxidation-states-used-for
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Question: <p>When going through list of oxidation states on <a href="https://en.m.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements" rel="noreferrer">Wikipedia</a> I encounterd there that sodium, potassium, rubidium and caesium exhibit oxidation states of -1, but not lithium, even though its electronegativity is more than that of the others. How's that possible? Can some one provide an example.</p>
Answer: <p>You are talking about alkalides, salts where the anion is an alkali metal. There is a very brief overview on <a href="https://en.wikipedia.org/wiki/Alkalide" rel="noreferrer">Wikipedia</a> which also provides a couple of examples, including <span class="math-container">$\ce{[Na(\text{cryptand[2.2.2]})]+Na-}$</span>. A good university level inorganic text book such as Greenwood and Earnshaw or Housecroft and Sharpe will provide more detail.</p>
<p>If you are interested in unusual oxidation states of alkali metals you might also like to know that in electrochemical experiments there is some evidence for <span class="math-container">$\ce{Cs^3+}$</span>, which is isoelectronic with <span class="math-container">$\ce{Xe^2+}$</span>. Again I think Greenwood and Earnshaw discusses this, but I don't have it to hand at the moment to confirm.</p>
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https://chemistry.stackexchange.com/questions/109846/unusual-oxidation-states-of-alkali-metals
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Question: <p>I downloaded a periodic table app which gives detailed information for each element. The app shows some elements, however, have two different oxidation states. Hydrogen, for instance, can have either a 1+ or 1- state. What do these two states signify? Why do these elements have <em>more</em> than one oxidation state?</p>
<p>Can someone explain this?</p>
Answer: <p>The oxidation state of an element in a compound is simply decided on the basis of electronegativity of other atoms. As the definition of oxidation state says </p>
<blockquote>
<p>Oxidation state of an element in a given compound is the charged acquired by its atom on the basis of electronegativity of other atoms in the compound.</p>
</blockquote>
<p>The basis of calculating oxidation number is that the more electronegative element acquires the negative charge and the less electronegative one acquires the positive charge. So in molecules in which hydrogen is less electronegetive it acquires a $+1$ charge while in molecules in which it is more electronegetive it will acquire a $+1$ charge ($1$ because it has the valency of $1$ and usually oxidation number does not exceed valency).</p>
<p>Let's understand it with the help of examples:<br>
In the $\ce{HCl}$ molecule, $\ce{H}$ is less electronegetive that's why charge of $+1$,for the same reason,$+1$ in $\ce{H2O}$.
In the compound $\ce{ZnH2}$, hydrogen is more electronegetive so it will have a charge of $-1$, for similar reason $-1$ in $\ce{NaH}$.</p>
<p>Usually metals (that make metal hydrides) are less electronegetive than hydrogen, That's why we also say hydrogen has an oxidation state of $-1$ in metal hydrides.</p>
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https://chemistry.stackexchange.com/questions/60679/elements-with-multiple-oxidation-states
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Question: <p>Palladium follows an exception to normal electron filling-up rules and so has the electron configuration of $\ce{[Kr] 4d^{10} 5s^0}$</p>
<p>The oxidation states of palladium are $\ce{+II}$ and $\ce{+IV}$.
How can these oxidation states be explained by looking at the special electron configuration of $\ce{Pd}$? </p>
Answer: <p>Ions of the transition metals do not often occur "naked" the way one might think of a sodium ion. Of course, sodium ions cannot occur by themselves either, but let's compare the two for a moment. </p>
<blockquote>
<p>Sodium nitrate, $\ce{NaNO3}$</p>
</blockquote>
<p>When $\ce{NaNO3}$ dissolves in water, the two ions dissociate. </p>
<p>$$\ce{NaNO3 -> Na+(aq) + NO3- (aq)}$$</p>
<p>Both ions are <strong>solvated</strong> by the solvent. The water molecules (which are polar) surround the ions. The partial charges due to the dipoles in the water molecules stabilize the charges on the ions. However, the interactions between the sodium ions and the water molecules are fleeting. These interactions are not strong enough to be consider "bonds" (there is no covalent electron sharing between water and sodium ion). Thus, we often consider sodium ions in solution (despite their solvation) to be "just" sodium ions.</p>
<p><img src="https://i.sstatic.net/G6d2b.png" alt="enter image description here"></p>
<blockquote>
<p>Palladium nitrate, $\ce{Pd(NO3)2}$</p>
</blockquote>
<p>Transition metal ions are different. They tend to form complex ions by <a href="http://en.wikipedia.org/wiki/Complex_ion" rel="nofollow noreferrer">coordinating</a> solvent molecules. These <a href="http://en.wikipedia.org/wiki/Dative_bond" rel="nofollow noreferrer">dative interactions</a> are far stronger than the ion-dipole interactions in solvated sodium ions. </p>
<p>Palladium nitrate does dissociate, but the palladium ion picks up water molecules that it holds tightly. Thus, we do not worry about the electron configuration of $\ce{Pd^{2+}}$, since it never exists on its own. It is always part of a complex ion or coordination compound, and we care about the <a href="http://en.wikipedia.org/wiki/Ligand_field_theory" rel="nofollow noreferrer">electron configuration of that complex ion</a>.</p>
<p>$$\ce{Pd(NO3)2 + nH2O -> Pd(H2O)6^{2+}(aq) + 2NO3- (aq)}$$</p>
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https://chemistry.stackexchange.com/questions/7705/explaining-the-oxidation-states-of-palladium
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Question: <p><a href="https://i.sstatic.net/5huav.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/5huav.png" alt="enter image description here"></a></p>
<blockquote>
<p>I have noted down the available positive oxidation states of the first
row of transition elements (on the Periodic Table) from the respective Wikipedia articles of
the elements.</p>
<p>$\ce{Sc} - 3, 2,1$</p>
<p>$\ce{Ti} - 4, 3, 2, 1 $</p>
<p>$\ce{V} - 5, 4, 3, 2, 1$</p>
<p>$\ce{Cr} - 6, 5, 4, 3, 2, 1$</p>
<p>$\ce{Mn} - 7, 6, 5, 4, 3, 2, 1$</p>
<p>$\ce{Fe} - 6,5,4,3,2,1 $</p>
<p>$\ce{Co} - 5,4,3,2,1 $</p>
<p>$\ce{Ni}- 4,3,2,1$</p>
<p>$\ce{Cu} - 4,3,2,1$</p>
<p>$\ce{Zn} - 2,1$</p>
</blockquote>
<ul>
<li>Why does $\ce{Fe}$ show not show $+7$ or $+8$ oxidation state?</li>
<li>After $\ce{Fe}$ from $\ce{Co}$ why is there a decrease in number of possible oxidation states?
Why don't $\ce{Fe,Co,Ni,Cu,Zn}$ show higher oxidation states though they have valence electrons left in $3d$ subshell? </li>
</ul>
Answer: <p>Regarding the maximum oxidation state achievable, there are two competing trends. Both have to do with ionisation energies.</p>
<p>In going to a higher oxidation state, the additional IEs are always balanced against the release in energy in the formation of stronger ionic/covalent bonds. However, if the additional IE is too large, then the favourable bonding will not be sufficient to bring out the higher oxidation state. A simple example is sodium, which cannot adopt the +2 oxidation state because its second IE is simply too large, even though the lattice energy of a hypothetical $\ce{NaCl2}$ crystal is larger than that of $\ce{NaCl}$.</p>
<p>Firstly, the maximum oxidation state is limited by the number of valence electrons available. This is exactly analogous to the case of sodium; therefore, manganese does not exhibit the +8 oxidation state, because its eighth IE involves ionisation from the 3p subshell, much lower in energy than either the 3d or 4s. This therefore accounts for the increase in the maximum oxidation state going from Sc to Mn.</p>
<p>Secondly, the ionisation energies generally increase going across the 3d block, because of an increase in effective nuclear charge. This is the factor which leads to the decrease in the maximum oxidation state after Mn. For example, the seventh IE of Fe is larger than that of Mn; therefore, Fe does not exhibit the +7 oxidation state.</p>
<p>You can look up the IEs in any relevant textbook. However, note that since the +2 oxidation state is very common in the later half of the 3d block, it is usually the third and higher IEs that are more important in this discussion. So, looking at the trend of first/second IEs won't get you anywhere.</p>
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https://chemistry.stackexchange.com/questions/59926/anomalous-oxidation-states-of-transition-metals
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Question: <p><a href="https://i.sstatic.net/YjBmj.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/YjBmj.png" alt="enter image description here" /></a></p>
<p>I was given this question and don't understand why the two sulphur atoms have different oxidation states - and how to actually figure it out. Could someone please explain it to me? I've attached the reasoning that was provided but it isn't very clear to me. I understand that oxygen is usually -2 (with exceptions) and that the molecule itself should have a net charge of -2 but I thought that then meant that each sulphur atom would have an oxidation state of +2. Any help would be greatly appreciated!!</p>
<p><a href="https://i.sstatic.net/pz6C2.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/pz6C2.png" alt="enter image description here" /></a></p>
Answer: <p>If the last paragraph in your quotation is meant to be the book's answer, I would stain that text with my intensely colored, anthocyanin-rich cranberry jam*. I would find either <span class="math-container">$-1$</span> and <span class="math-container">$+5$</span> or <span class="math-container">$0$</span> and <span class="math-container">$+4$</span> more defensible.</p>
<p><strong>-1 and +5</strong></p>
<p>In this model we use the raw electronegativities of the atoms to render the sulfur-sulfur bond nonpolar, so the outer sulfur atom would have the same <span class="math-container">$-1$</span> oxidation state as it would in the disulfide ion <span class="math-container">$\ce{S2^{2-}}$</span>. The central sulfur atom, losing two electrons to each oxygen atom to satisfy <span class="math-container">$-2$</span> for the latter, ends up at <span class="math-container">$-1+6=+5$</span>.</p>
<p><strong>0 and +4</strong></p>
<p>Here we take a slightly more advanced approach, arguing that the sulfur-sulfur bond is inductively polarized by the oxygen atoms withdrawing electronic charge from the central sulfur. In response the sulfur-sulfur bond has its electrons polarized towards the central atom as if that atom were more electronegative. Thereby the outer sulfur is rendered with an oxidation state of <span class="math-container">$0$</span> and the inner one with <span class="math-container">$-2+6=+4$</span>. This assignment has the advantage of dovetailing with a well-known chemical property of thiosulfate: a generally redox-inert acid such as hydrochloric acid causes it decompose to elemental sulfur (oxidation state <span class="math-container">$0$</span>) and sulfur dioxide (sulfur oxidation state <span class="math-container">$+4$</span>).</p>
<hr />
<p>*Cranberries are actually white on the inside when fresh. The bulk color turns red only when the anthocyanin dye, which is not colored in its reduced form, is exposed to air as the cranberries are crushed or popped.</p>
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https://chemistry.stackexchange.com/questions/178247/oxidation-states-of-thiosulphate
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Question: <p>Why does plutonium have more oxidation states than samarium?</p>
<p>Electron configuration of Pu: <span class="math-container">$\ce{[Rn] 5f^6 7s^2}$</span></p>
<p>Electron configuration of Sm: <span class="math-container">$\ce{[Xe] 4f^6 6s^2}$</span></p>
<p>I thought that only the valence electrons affected the oxidation states, so why does plutonium have more oxidation states (6,5,4,3) than samarium (2,3) whilst they both have the same valence electrons (including f-orbitals): ..<span class="math-container">$\ce{f^6}$</span>..<span class="math-container">$\ce{s^2}$</span>?</p>
<p>Elements such as scandium (Sc), Yttrium (Y), and Lanthanum (La) all have the same valence electrons (including d-orbitals): ..<span class="math-container">$\ce{d^1}$</span>..<span class="math-container">$\ce{s^2}$</span>, but these elements have the same oxidation state : 3 and that doesn't change, even when going down that group and looking at actinium.</p>
<p>Does this have to do with the ionization energy, the energy level, or...?</p>
Answer: <p>This is yet another interesting effect of <a href="http://web.archive.org/web/20150511024324/http://depa.fquim.unam.mx/amyd/archivero/LecturasobreNodosRadiales_12854.pdf" rel="nofollow noreferrer">the anomalous compactness of orbitals in the first appearance of each type of subshell</a> (<span class="math-container">$1s$</span>, <span class="math-container">$2p$</span>, <span class="math-container">$3d$</span>, <span class="math-container">$4f$</span>, <span class="math-container">$5g$</span>, etc). The solutions to the Schrödinger equation for electron wavefunctions in hydrogen-like atoms are such that these subshells are composed of orbitals with <em>no radial nodes</em>. This means the electrons in these subshells are closer to the nucleus than you might expect, and therefore they are more strongly bound.</p>
<p>The striking chemical similarity of the lanthanides comes from the fact that the <span class="math-container">$4f$</span> subshell is both anomalously compact and subjected to a high effective nuclear charge. The electrons end up being held so closely to the nucleus that they effective behave as core electrons, and participate very little in chemical interactions; lanthanides do not form coordination compounds using their <span class="math-container">$4f$</span> orbitals, and it is very difficult to go past an oxidation state of +3 in most because few conditions can compensate the large energy required to ionize a fourth electron coupled with the relatively weakly bound compounds they would form.</p>
<p>In the actinides, the <span class="math-container">$5f$</span> orbitals do not suffer from the same lack of radial node, and therefore are more chemically available. The first half of the actinides show many compounds with high oxidation numbers and have significant coordination chemistry using <span class="math-container">$f$</span> orbitals. Curiously, as you go from curium to berkelium, there is a sudden increase in reluctance to display high oxidation numbers, and the actinides beyond curium tend to behave similarly (though of course exploration of their chemical properties is significantly hindered by their rarity and instability). This is likely because even though the <span class="math-container">$5f$</span> orbitals have a radial node, in the second half of the actinides they are subjected to such a high effective nuclear charge that become too strongly bound to display significant participation in chemistry.</p>
<p>This same effect explains why the metals in the first row of transition metals do not display as many stable compounds with high oxidation states compared to the second and third rows, as the <span class="math-container">$3d$</span> orbitals are much smaller than <span class="math-container">$4d$</span> or <span class="math-container">$5d$</span> orbitals.</p>
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https://chemistry.stackexchange.com/questions/7704/plutonium-having-more-oxidation-states-than-samarium
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Question: <p>According to <em><a href="https://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements" rel="noreferrer">List of oxidation states of the elements</a></em>, silicon has a possible oxidation state of $-4$.</p>
<p><a href="https://i.sstatic.net/xpktm.jpg" rel="noreferrer"><img src="https://i.sstatic.net/xpktm.jpg" alt="Enter image description here"></a></p>
<p>Now, I've been looking everywhere for a compound that contains Si(-IV), but I cannot find any definitive references. Perhaps one of the synthetic carbonyls of silicon has this OS? I've seen references to $\ce{[Mo(CO)_4]^{-4}}$ and $\ce{[W(CO)_4]^{-4}}$ in my research, but silicon never shows up.</p>
<p>I am also uncertain as to how these carbonyls come together, and whether or not they actually complex in the same way for metalloids as for the transition metals.</p>
Answer: <p>I am not sure if Silane ($\ce{SiH_4}$) can really be considered. </p>
<p>But there are <a href="http://www.wikiwand.com/en/Silicide" rel="noreferrer">silicides</a> which silicon forms with strongly electropositive metals. In these compounds, silicon has a negative oxidation state.</p>
<p>For <a href="http://www.wikiwand.com/en/Magnesium_silicide" rel="noreferrer">magnesium silicide</a> - $\ce{Mg_2Si}$, the oxidation state of silicon would be -4.</p>
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https://chemistry.stackexchange.com/questions/37209/negative-oxidation-states-of-si
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Question: <p>Transition elements are good catalysts because they have multiple oxidation states?Why multiple oxidation states matter?</p>
Answer: <p>The answer has a lot to do with "what is a catalyst in the first place?"
A catalyst is a compound that can affect the rate of a chemical reaction
by providing an alternative and lower energy profile or pathway. That is,
it is a substance that makes it so that A -> B at a lower energy cost. But
it only has an effect on the rate of the reaction. That is, it only changes
the cost of the activation energy. It is not related to the thermodynamics
of the process and hence, the final product distribution. That is, if the
above process gives 90% B via a non-catalyzed pathway in 4 days, then a
catalyst will help you get 90% B in say 4 hours.</p>
<p>Okay, so a catalyst affects the transition state and activation path. How
does it do this? Typically, by complexing one of the reagents. Complexation
by transition metals affords access to a wide variety of oxidation states
for the metal. This has the property of providing electrons or withdrawing
electrons from the transition state of the reaction. That is, if the
transition state is electron rich, then the transition metal might hold
some of that electron density and those prevent too much from building up
on the reagent. This would then facilitate the reaction. Or the transition
metal might undergo formal oxidation/reduction to achieve electron transfer
to a substrate, thereby allowing a reaction to occur. This is "complexation
and electron storage" taken to the extreme but is a common mechanism in
organometallic chemistry. Indeed, a variety of catalytic pathways rely on
a two electron transfer between the metal and the substrate (e.g.
hydroformylation). It is the ability of the transition metal to be in a
variety of oxidation states, to undergo facile transitions between these
oxidation states, to coordinate to a substrate, and to be a good
source/sink for electrons that makes transition metals such good catalysts.</p>
<p>Of course, this is for transition metal complexes. Most industrial used
catalysts are the transition metal in a bed, as a metal or bound structure.
The above considerations are important but also the physical properties of
absorption/adsorption and the electron band structure of the material.</p>
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https://chemistry.stackexchange.com/questions/28828/why-are-multiple-oxidation-states-useful-for-a-catalyst
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Question: <p>So, metal oxides are generally basic, right? But in higher oxidation states their covalent character is predominant. For example in
Mn2O7 the Mn - O bond is covalent. Why is that?</p>
Answer: <p>Metal oxides are not always basic. In many cases where the bonding is strongly ionic, the oxide ions do act as a base. However, if the metal is in a high oxidation state an ionic bond would require a prohibitively high charge on the metal ion; the ionization energy required to put seven positive charges on one manganese atom far exceeds what can be reclaimed from the electron affinity of oxygen and electrostatic attraction alone. The energetically preferred option for <span class="math-container">$\ce{Mn2O7}$</span> is to form polar covalent bonds instead. Such bonds are similar to those in <span class="math-container">$\ce{CO2, SO3, P4O10,}$</span>, etc., and thus <span class="math-container">$\ce{Mn2O7}$</span> will act similarly to those oxides -- as an acid anhydride.</p>
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https://chemistry.stackexchange.com/questions/128634/covalency-of-metal-oxides-in-high-oxidation-states
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Question: <p>The CRC Elecrochemical Series and the solution to our exercise suggest that the oxidation states of borohydride are -V for Boron and +I for Hydrogen.</p>
<p>Since H (2.2) is more electronegativ than B (2.0), I would have expected the oxidation states to be +III for B and -I for H.</p>
<p>What am I missing here?</p>
<p>The solution to our exercise lists this (WARNING: it turned out that the solution was wrong):</p>
<p><img src="https://i.sstatic.net/Omazz.png" alt=""></p>
<p>And the CRC Elecrochemical Series lists this (also suggesting -V for B):</p>
<p><img src="https://i.sstatic.net/RZHuZ.png" alt=""></p>
Answer: <blockquote>
<p>I would have expected the oxidation states to be +III for B and -I for
H.</p>
</blockquote>
<p>You're right, boron is in the (+3) oxidation state in these equations.</p>
<p>In $\ce{B(OH)_3}$ each $\ce{OH}$ is (-1) and the boron is (+3), overall the molecule is neutral</p>
<p>In $\ce{BH_{4}^{-}}$ each $\ce{H}$ is (-1) and the boron is (+3), overall the ion is (-1)</p>
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https://chemistry.stackexchange.com/questions/14038/what-are-the-oxidation-states-in-borohydride-bh4
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Question: <p>In p-block elements, higher oxidation states are less stable down the group due to the inert pair effect. This is not the case for transition metals.</p>
<p>Why do heavier transition metals show higher oxidation states than lighter ones? Is the inert pair effect not valid for transition metals also?</p>
Answer: <p>The inert pair effect is based on the fact that main group elements’ oxidation states depend on s and p orbitals (and only them). When going down the periodic table, the energy difference between s and p orbitals changes leading to some elements losing their valence s electrons more easily than others.</p>
<p>Transition metals’ chemistry happens in the d orbitals primarily — unless you count the copper and zinc groups wherein a significant part of the chemistry is in fact only s orbital chemistry. The d orbitals — at first approximation of the free ion — are degenerate, i.e. they all have the same energy. Thus, these electrons are typically much more accessable. Furthermore, going down the periodic table increases the number of electrons counted as core electrons meaning that the outermost valene electrons experience a weaker effective nuclear attraction. It is therefore easier (i.e. requires less energy) to remove valence electrons and higher oxidation states are much more accessible.</p>
<p>If it weren’t for the inert pair effect, this would also be visible in main group chemistry — except it is, if you compare the relative stabilities of iodate and chlorate.</p>
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https://chemistry.stackexchange.com/questions/67510/why-do-heavier-transition-metals-show-higher-oxidation-states
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Question: <p>USNCO 2004 Question 45 states:</p>
<blockquote>
<p>Which element exhibits the greatest number of oxidation
states in its compounds?</p>
<p>(A) Ca (B) V (C) Cu (D) As</p>
</blockquote>
<p>I ruled out Ca, as I know it only exists as +2. I then drew electron configurations for V, Cu and As. I ruled out Cu as it had half filled 4s and full 3d orbitals, also because I've only ever seen copper in a +2 oxidation state.</p>
<p>However, I wasn't able to differentiate between V and As. They both have 3 unpaired electrons, albeit As has an extra full 3d orbital.</p>
<p>I was tempted to go with As, as being less metallic than V, it is more likely to have negative oxidation states, but the answer states Vanadium has the most oxidation states.</p>
<p>Is there any way I can work this question out by looking at electron configurations, or will I have to memorize the oxidation states?</p>
Answer:
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https://chemistry.stackexchange.com/questions/107718/comparing-the-number-of-oxidation-states-of-vanadium-and-arsenic
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Question: <p>For example, how could I calculate the oxidation state of lead in <span class="math-container">$\ce{PbS}$</span>? Or for that matter how would I calculate it for sulfur? Or I have two unknowns, for example sulfur in <span class="math-container">$\ce{PbSO4}$</span>? Do I assume that lead has a 2+ charge and therefore a +2 oxidation state and since it has ? Or is there something that I am missing? Additionally, from all the reading that I've done I've only found a rule for bromine, chlorine, and iodine in binary compounds, what happens if they are in a compound with more than two elements? For example <span class="math-container">$\ce{COCl2}$</span>. How would someone calculate the oxidation state of carbon? This stuff is really confusing for me and I apologize if this seems really basic or I have missed something.</p>
<p>I have read the introduction to oxidation states post, and it was a very well written answer, however it still did not help things click. I still feel like I am missing something</p>
Answer: <p>For compounds where you have a non-metal with a metal, there is likely an ionic bond. The oxidation states are just the charges on the ion. PbS has Pb2+ and S2-, so those charges are simply the oxidation states.</p>
<p>For PbSO4, you know sulfate is a 2- ion, so it must be bonded to a Pb2+. Therefore, again, Pb has a +2 oxidation number. When you look at the sulfate ion, you know that the sum of the oxidation states of the elements must equal the ion's overall charge. Oxygens are -2, so set up the equation: (X = oxidation number of sulfur)</p>
<p>X + 4(–2) = -2
X = +6</p>
<p>Calculating the oxidation state of carbon can be more complex. If you draw a dot and cross diagram showing the covalent bonds made by carbon, this will help. Look at the electronegativities of the elements in the bonds. Carbon takes the electron pair if it is the more electronegative, but this is rarely the case. For every electron pair carbon takes, it has a -1 added to the oxidation state, and +1 added if it loses the electron pair. For example, if you draw the methane diagram you find that the oxidation state of carbon is +4. Using this method, you can find the oxidation state of carbon in compounds.</p>
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https://chemistry.stackexchange.com/questions/129501/how-does-one-calculate-oxidation-states
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Question: <p>There are oxidation states from -4 to +8</p>
<p>Why do the negative oxidation states not go all the way down to -8? I mean if an atom is hypervalent meaning that it can have more than an octet than that means it can have potentially 8 bonds(counting a double bond as 2 bonds and a triple bond as 3 bonds respectively) to other atoms more electronegative than itself and thus Xenon can have a +8 oxidation state and some metals can as well.</p>
<p>Doesn't that also mean that a hypervalent atom can have 8 bonds to other atoms less electronegative than itself and thus have a -8 oxidation state?</p>
Answer: <blockquote>
<p>Why do the negative oxidation states not go all the way down to -8?</p>
</blockquote>
<p>Let's think about what we would need for this to occur - i.e. an oxidation state of -8.</p>
<p>We would need a central atom that has considerable electronegativity - one that is at least more electronegative than its attached atoms. This already excludes numerous molecules; most molecules' central atoms are, on a relative basis, more <em>electropositive</em> rather than electronegative. </p>
<p>In addition, this atom would need to have one of the following configurations; "allocated" refers to the assignment of oxidation state - since oxidation states are assigned on the premise that all bonding is ionic - the more electronegative element in a bonding configuration is "allocated" all the bonding electrons in the calculation of oxidation states. I.e. in H2O, the oxygen would be allocated both pairs of bonding electrons in the H-O bonds, and oxygen would then "have" 8 electrons, and its oxidation state would be -2. </p>
<p>1 valence electron and be "allocated" 9 electrons </p>
<p>2 valence electrons and be "allocated" 10 electrons</p>
<p>3 valence electrons and be "allocated" 11 electrons,</p>
<p>and on the other extreme:</p>
<p>6 valence electrons and be "allocated" 14 electrons,</p>
<p>7 valence electrons and be "allocated" 15 electrons,</p>
<p>8 valence electrons and be "allocated" 16 electrons. </p>
<p>As we can see, the first set of criteria are basically incompatible. Elements only having 1 or 2 valence electrons are going to be small (on a relative basis). These elements would also <em>much rather</em> be oxidized than reduced. Oxidation allows these elements to easily become isoelectronic with a noble gas. </p>
<p>The other set of criteria for a negative 8 oxidation state is also hard to fulfill. None of the group 6 elements except the possibility of sulfur and elements below it are hypervalent - i.e. they can accommodate more than 8 valence electrons. Even if we consider sulfur as able to accommodate <em>12</em> valence electrons (as some depictions of sulfuric acid suggest) we must remember that sulfur is not very electronegative and unlikely to be allocated any electrons. </p>
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https://chemistry.stackexchange.com/questions/15337/why-do-negative-oxidation-states-not-extend-to-8
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Question: <p>Why are fluorides of transition metals unstable in low oxidation states? </p>
<p>I would think that since fluorine and oxygen are highly electronegative, it's obvious that they be stable at high oxidation states. However, $\ce{Cl}$ is also highly electronegative, but it does exist in low oxidation states with transition metals.</p>
<p>For example $\ce{CuCl}$ exists but not $\ce{CuF}$ and $\ce{TiCl2}$ exists but not $\ce{TiF_{2}}$. Why is it so? </p>
<blockquote>
<p><em>Fluorides of transition metals unstable in low oxidation states</em> </p>
</blockquote>
<p>This is the statement given in my high school chemistry text book but no explanation provided. I am interested to know the reason for it. </p>
Answer: <p>You can take the example of $\ce{CuF}$ which is not known.</p>
<p>Reason: $\ce F$ is such an electronegative element that it will always oxidise $ \ce{Cu}$ to $\ce{Cu^2+}$ and not $\ce{Cu^1+}$. Hence, whenever $\ce{Cu}$ reacts with flourine, copper(II) flouride is formed. The following reaction illustrates this:</p>
<p>$$\ce{Cu + F2 -> CuF2}$$</p>
<p>$\ce{CuF2}$ loses fluorine at temperatures above $\pu{950 °C}$. $\ce{CuF}$ will be formed initially and as your question says, it is highly unstable so it will be transformed to some other stuff and the reactions are:</p>
<p>$$\ce{2CuF2 -> 2CuF (unstable) + F2}$$</p>
<p>$$\ce{2CuF -> CuF2 + Cu}$$</p>
<p>Similarly, you can illustrate this for $\ce{VF2}$.</p>
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https://chemistry.stackexchange.com/questions/55778/why-are-fluorides-of-transition-metals-unstable-in-low-oxidation-states
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Question: <p>I have never seen anything other than a set of rules like these when textbooks present how to assign oxidation numbers. Such as these: </p>
<p><img src="https://i.sstatic.net/YlxeX.png" alt="enter image description here"></p>
<p>However, if we keep in mind that oxidation numbers are simply imaginary numbers which suppose all bonding to be ionic - i.e. not electron sharing - and if we keep in mind simply <em>relative</em> electronegativity, we can easily work out the oxidation state of <em>any</em> element in <em>any</em> compound. </p>
<p>For example, take water. Bonding order: $\ce{H-O-H}$. Oxygen has two lone pairs. Oxygen is more electronegative than hydrogen. So we suppose that oxygen takes both electrons in both bonding pairs. Oxygen has 8 electrons. Its valence is 6. 6-8 is -2; oxygen has two more electrons assigned to it than what it has in its valence, so naturally its oxidation state is <em>negative</em> 2. No rules needed. </p>
<p>Now hydrogen peroxide. $\ce{H-O-O-H}$. Here we have an oxygen-oxygen bond so in this case, neither element wins the electronegativity battle. Electrons are split between the oxygen and the oxygen. However, since the oxygens are much more electronegative than the hydrogens, we assign both electrons in both the $\ce{O-H}$ bonding pairs to oxygen. Oxygen has 7 electrons assigned to it; oxygen has a valence of 6; oxidation state: -1. No need to memorize exceptions as stated in the table above. </p>
<p>We could go on. However, I am curious:</p>
<p>1) Has anyone been taught to assign oxidation states this way?
2) If not, what do you think of this method? </p>
Answer: <p>In the IUPAC Recommendations 2016 the definition of oxidation state underwent a significant and comprehensive change. It does now resemble the version wich was proposed be Hans-Peter Loock and is quoted in the earlier version of this answer, parts of which are included below.</p>
<p>The electronegativity battle scheme is most helpful for all kinds of compounds since it is one of the most generic ways to derive oxidation states. The table represents just a cheat sheet that might be very helpful in the beginning. If you are spending most of your time with chemistry, this table will be present as some sort of muscle memory - usually referred to as chemical intuition.</p>
<p>The IUPAC <a href="http://goldbook.iupac.org/O04365.html" rel="nofollow noreferrer">gold book</a> now defines oxidation state as follows:</p>
<blockquote>
<p><strong>oxidation state</strong><br />
gives the degree of oxidation of an atom in terms of counting electrons. The higher the oxidation state (OS) of a given atom, the greater is its degree of oxidation. Definition:<br />
<em>OS of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds.</em></p>
</blockquote>
<blockquote>
<p>The underlying principle is that the ionic sign in an <span class="math-container">$\ce{AB}$</span> molecule is deduced from the electron allegiance in a LCAO-MO model: The bond’s electrons are assigned to its main atomic contributor. Homonuclear <span class="math-container">$\ce{AA}$</span> bonds are divided equally. In practical use, the ionic-approximation sign follows Allen electronegativities (see Source). There are two general algorithms to calculate OS:</p>
</blockquote>
<blockquote>
<ol>
<li>Algorithm of assigning bonds, which works on a Lewis formula showing all valence electrons in a molecule: OS equals the charge of an atom after its heteronuclear bonds have been assigned to the more electronegative partner (except when that partner is a reversibly bonded Lewis-acid ligand) and homonuclear bonds have been divided equally:<br />
<a href="https://i.sstatic.net/bzfzQ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/bzfzQ.png" alt="O04365-1" /></a></li>
</ol>
</blockquote>
<blockquote>
<ol start="2">
<li>Algorithm of summing bond orders: Heteronuclear-bond orders are summed at the atom as positive if that atom is the electropositive partner in a particular bond and as negative if not, and the atom’s formal charge (if any) is added to that sum, yielding the OS. This algorithm works on Lewis formulas and on bond graphs of atom connectivities for an extended solid:<br />
<a href="https://i.sstatic.net/bHvBW.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/bHvBW.png" alt="O04365-2" /></a></li>
</ol>
</blockquote>
<blockquote>
<p><em>NOTE:</em></p>
</blockquote>
<blockquote>
<ol>
<li><em>Specific uses may require modified OS values: Electrochemical OS is nominally adjusted to represent a redox-active molecule or ion in Latimer or Frost diagrams. Nominal OS values may also be chosen from close alternatives for systematic-chemistry descriptions.</em></li>
<li><em>Some OS may be ambiguous, typically when two or more redox-prone atoms enter bonding compromises and nearest integer values have to be chosen for the OS.</em></li>
<li><em>The caveat of reversibly bonded Lewis-acid ligands originates from the simplifying use of electronegativity instead of the MO-based electron allegiance to decide the ionic sign. Typical examples are the transition-metal complexes with so called Z ligands in the CBC ligand-classification scheme (see Source).</em></li>
<li><em>When used in chemical nomenclature as a symbol, the OS value is in roman numerals.</em></li>
<li><em>At the introductory teaching level, prior to the bonding-based definition and algorithms: OS for an element in a chemical formula is calculated from the overall charge and postulated OS values for all the other atoms. For example, postulating <span class="math-container">$\text{OS} = +1$</span> for <span class="math-container">$\ce{H}$</span> and <span class="math-container">$−2$</span> for <span class="math-container">$\ce{O}$</span> yields correct OS in oxides, hydroxides, and acids like <span class="math-container">$\ce{H2SO4}$</span>; with coverage extended to <span class="math-container">$\ce{H2O2}$</span> via decreasing priority along the sequence of the two postulates. Additional postulates may expand the range of compounds to fit a textbook’s scope.</em></li>
</ol>
</blockquote>
<blockquote>
<p><em><strong>Source:</strong></em></p>
</blockquote>
<blockquote>
<ol>
<li>Karen, P.; Mcardle, P.; Takats, J. Toward a comprehensive definition of oxidation state (IUPAC Technical Report). <em>Pure Appl. Chem.</em> <strong>2014,</strong> <em>86</em> (6), 1017–1081 <a href="https://doi.org/10.1515/pac-2013-0505" rel="nofollow noreferrer">DOI: 10.1515/pac-2013-0505</a>.</li>
<li>Karen, P.; Mcardle, P.; Takats, J. Comprehensive definition of oxidation state (IUPAC Recommendations 2016). <em>Pure Appl. Chem.</em> <strong>2016,</strong> <em>88</em> (8), 831–839 <a href="https://doi.org/10.1515/pac-2015-1204" rel="nofollow noreferrer">DOI: 10.1515/pac-2015-1204</a>.</li>
<li>Nomenclature of Inorganic Chemistry. IUPAC Recommendations 2005. p. 34. Available as pdf from <a href="https://old.iupac.org/publications/books/rbook/Red_Book_2005.pdf" rel="nofollow noreferrer">old.iupac.org</a></li>
</ol>
</blockquote>
<p>I recommend having a look at the sources cited above, they contain a lot more information. (Note that the link from the gold book to the red book on the IUPAC website is broken.)</p>
<p>Prior to 2016 the IUPAC defined oxidation states in their <a href="https://web.archive.org/web/20141102154648/http://goldbook.iupac.org/O04365.html" rel="nofollow noreferrer">gold book (via the Internet Archive)</a> differently. I am including this definition, since it will be still present in prominent text books and it is what lead to the postulates given in the table of the original question.</p>
<blockquote>
<p><strong>oxidation state</strong> (deprecated, pre-2016 definition)
A measure of the degree of oxidation of an atom in a substance. It is defined as the charge an atom might be imagined to have when electrons are counted according to an agreed-upon set of rules:</p>
<ol>
<li>the oxidation state of a free element (uncombined element) is zero;</li>
<li>for a simple (monatomic) ion, the oxidation state is equal to the net charge on the ion;</li>
<li>hydrogen has an oxidation state of <span class="math-container">$1$</span> and oxygen has an oxidation state of <span class="math-container">$-2$</span> when they are present in most compounds. (Exceptions to this are that hydrogen has an oxidation state of <span class="math-container">$-1$</span> in hydrides of active metals, e.g. <span class="math-container">$\ce{LiH}$</span>, and oxygen has an oxidation state of <span class="math-container">$-1$</span> in peroxides, e.g. <span class="math-container">$\ce{H2O2}$</span>;</li>
<li>the algebraic sum of oxidation states of all atoms in a neutral molecule must be zero, while in ions the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion. For example, the oxidation states of sulfur in <span class="math-container">$\ce{H2S}$</span>, <span class="math-container">$\ce{S8}$</span> (elementary sulfur), <span class="math-container">$\ce{SO2}$</span>, <span class="math-container">$\ce{SO3}$</span>, and <span class="math-container">$\ce{H2SO4}$</span> are, respectively: <span class="math-container">$-2$</span>, <span class="math-container">$0$</span>, <span class="math-container">$+4$</span>, <span class="math-container">$+6$</span> and <span class="math-container">$+6$</span>. The higher the oxidation state of a given atom, the greater is its degree of oxidation; the lower the oxidation state, the greater is its degree of reduction.</li>
</ol>
</blockquote>
<p>What we see is, that the table in the original question actually reflects some of these rules. However, this set is not generic at all and it lacks a definition for compounds that do not contain oxygen or hydrogen, are elemental or monoatomic ions. With only these rules it is impossible do determine the oxidation states for <span class="math-container">$\ce{BF3}$</span> and many, if not most, compounds.</p>
<p>The lack of the definition is very well known, but it took the IUPAC until 2016 to actually change the official set. Hans-Peter Loock proposed a much simpler concept in <a href="http://dx.doi.org/10.1021/ed1005213" rel="nofollow noreferrer">Expanded Definition of the Oxidation State</a>, which resembles the currently used definition quite well.</p>
<blockquote>
<p>The oxidation state of an atom in a compound is given by the hypothetical charge of the corresponding atomic ion that is obtained by heterolytically cleaving its bonds such that the atom with the higher electronegativity in a bond is allocated all electrons in this bond. Bonds between like atoms (having the same formal charge) are cleaved homolytically.</p>
</blockquote>
<p>So he basically came to the same conclusion as Dissenter in the question. Another statement from this article is</p>
<blockquote>
<p>This is not a new definition, but it predates the IUPAC rules by several decades. For example, Linus Pauling provided a similar definition of the oxidation state in his 1947 edition of General Chemistry (3).</p>
<p>(3):Pauling, L. General Chemistry; Freeman: San Francisco, CA, 1947. <a href="http://store.doverpublications.com/0486656225.html" rel="nofollow noreferrer">Republished by Courier Dover Publications, 2012</a>.</p>
</blockquote>
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https://chemistry.stackexchange.com/questions/10944/electronegativity-considerations-in-assigning-oxidation-states
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Question: <p>How do you balance this using the ion-electron method;</p>
<p><span class="math-container">$\ce{CrO5 + H2SO4 -> Cr2(SO4)3 + H2O + O2}$</span></p>
<p>The oxygen has multiple oxidation states in <span class="math-container">$\ce{CrO5}$</span>, and none of the sites I looked this up on dealt with that.</p>
Answer: <p>This is a rather unusual case of what is discussed in answers like <a href="https://chemistry.stackexchange.com/a/91664/17175">this one</a>, where we circumvent problems with multiple atoms being oxidized or reduced by considering whole compounds as oxidizing or reducing agents.</p>
<p>Here, the whole-compound redox-active material is <span class="math-container">$\ce{CrO5}$</span>, and as in peroxide disproportionations generally this is both an oxidizing agent and a reducing agent. We thereby render</p>
<p><span class="math-container">$\ce{CrO5 -> Cr^{3+} + (5/2) O2 + 3 e^-}$</span></p>
<p><span class="math-container">$\ce{CrO5 + 10H^+ + 7 e^- -> Cr^{3+} + 5 H2O}$</span></p>
<p>We then apply the usual method of multiplying the first reaction by <span class="math-container">$7$</span> and the second one by <span class="math-container">$3$</span> to balance the electrons leading to</p>
<p><span class="math-container">$\ce{10 CrO5 + 30 H^+ -> 10 Cr^{3+} + (35/2) O2 + 15 H2O}$</span></p>
<p>and reducing to "lowest (whole number) terms"</p>
<p><span class="math-container">$\ce{4 CrO5 + 12 H^+ -> 4 Cr^{3+} + 7 O2 + 6 H2O}$</span></p>
<p>All that remains now is to add the spectator ions if desired, and we're done.</p>
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https://chemistry.stackexchange.com/questions/129467/balancing-redox-equations-with-oxygens-in-multiple-oxidation-states
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Question: <p>As I understand, Roman numerals are used for oxidation states, whereas Arabic numerals are used for the charge of full ions, with some ambiguity allowed when a metal centre is all but ionised, such as in hexaaqua compounds.</p>
<p>But I have never seen a Roman numeral with a negative sign in front of it. Would I write the oxidation state of sulphur in hydrogen sulphide (H<span class="math-container">$_2$</span>S) as (-II)?</p>
Answer: <p>According to IUPAC guidelines, roman numerals are used to denote oxidation states when used in the name of a compound. This is used when the cationic group can have varying oxidation states. For example, <span class="math-container">$\ce{FeSO4}$</span> is written as iron (II) sulphate, and <span class="math-container">$\ce{Fe2(SO4)3}$</span> is written as iron (III) sulphate. Although anionic groups can also have variable oxidation states, we omit mentioning the oxidation state in the name of the compound. Instead, we use standard prefixes to identify the oxidation states. For example, potassium oxide (<span class="math-container">$\ce{K2O}$</span>), potassium peroxide (<span class="math-container">$\ce{K2O2}$</span>), and potassium superoxide (<span class="math-container">$\ce{KO2}$</span>) have different oxidation states for oxygen: -2, -1, and -1/2 respectively. But instead of mentioning this explicitly, we denote these with different standard names.</p>
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https://chemistry.stackexchange.com/questions/171876/can-you-write-oxidation-states-with-negative-roman-numerals
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Question: <p>I'm hoping someone can help me tighten up my understanding of the origins of the differences between Ni and Pd catalysis. </p>
<p>I understand that Ni has higher charge density and this is the reason for it's larger capacity for oxidative addition for say an aryl halide. It also has access to the oxidtaion states (0), (I), (II), (III) and sometimes (IV). Palladium conversely only shuttles between (0) and (II) in a standard cross-coupling reaction. I believe the lower first ionisation energy can account for the increased tendency of Ni to do SET but I don't know how best to explain why all the additional oxidation states are available to Ni and not for Pd. </p>
<p>thanks in advance, Cat</p>
Answer:
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https://chemistry.stackexchange.com/questions/111829/reactivity-of-ni-vs-pd-available-oxidation-states
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Question: <p>What are the oxidation states of the sulfur in the tetrathionate ion <span class="math-container">$\ce{S4O6^{2-}}$</span>?</p>
<p><a href="https://i.sstatic.net/l5kDW.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/l5kDW.png" alt="Structure"></a></p>
Answer: <p>Consider the structure of <a href="http://en.wikipedia.org/wiki/Tetrathionate">tetrathionate</a>. The two central sulfurs each have two lone pairs and are assigned half of the electrons from the two bonds they make, since the electrons of bonds between atoms of the same element must be distributed evenly (due to there being, by definition, an electronegativity difference of zero between two atoms of the same element). A neutral sulfur atom has six valence electrons, so the oxidation state of the central sulfurs can be calculated as follows:</p>
<p>$$6 - 4 - \frac{1}{2}(4) = 0$$</p>
<p>That is, six electrons in neutral sulfur, minus four from the lone pairs, minus half of the four sulfur-sulfur bonding electrons, gives zero.</p>
<p>The terminal sulfurs, on the other hand, have no lone pairs, and all the electrons from the sulfur-oxygen bonds are assigned to oxygen, since it's the more electronegative element. The only electrons assigned to those sulfurs are half of those from the sulfur-sulfur single bonds. Hence, their oxidation state is:</p>
<p>$$6 - \frac{1}{2}(2) = +5$$</p>
<p>That is, six electrons in neutral sulfur, minus half of the two sulfur-sulfur bonding electrons, giving an oxidation state of +5.</p>
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https://chemistry.stackexchange.com/questions/5643/what-are-the-oxidation-states-of-sulfur-in-the-tetrathionate-ion
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Question: <p><a href="https://i.sstatic.net/Szzjs.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Szzjs.jpg" alt="enter image description here"></a></p>
<p>I) According to my logic, sulfur-sulfur linkage (linkage between same atoms) result in no charge separation.So, the oxidation number of the sulfur atom linked to only sulfur should be 0(zero) and other sulfur being +4 as it is connected to 3(more electronegative) oxygen atoms.</p>
<p>Thus, the oxidation number of sulfur atoms should be +4,0.</p>
<p>II)However, <a href="http://www.sciencedirect.com/science/article/pii/S138614259800153X" rel="nofollow noreferrer">Using X-ray absorption to probe sulfur oxidation states in complex molecules</a> claims that oxidation states are +6 and -2.</p>
<p>III)Again,<a href="http://przyrbwn.icm.edu.pl/APP/PDF/121/a121z2p71.pdf" rel="nofollow noreferrer">Determination of Changes in Sulfur Oxidation States
in Prostate Cancer Cells</a> proved that that oxidation states are +5,-1.</p>
<p>These three sets of results left thoroughly confused.<em>Which one of them is correct and why?</em></p>
<p>Following the same sequence of logic:
My counter-question for the claims (I) and (II) is
<strong><em>"Why the oxidation number of sulfur atoms of Disulfur monoxide is [ +2 and 0 ] instead of [+4 and -2] or [+3 and -1] ?"</em></strong></p>
<p><a href="https://i.sstatic.net/Z5IBp.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Z5IBp.png" alt="Disulfur monoxide"></a></p>
Answer:
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https://chemistry.stackexchange.com/questions/71924/why-oxidation-states-of-suphur-atoms-of-disuphur-monoxide-is-2-and-0
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Question: <p>In transition elements the oxidation states differ by one (+2 to +7 in Mn). However, in the p-block elements, the oxidation states differ by two (-1,+1,+3,+5 in the halogen group). Why is this so?</p>
Answer: <p>In d block, variable valency occurs due to small differences in successive ionization enthalpy. Hence, it shows valency with difference of 1</p>
<p>While in p block, variable valency occurs due to inert pair effect. Due to this 2 electrons remain paired and become inert for bonding. Hence, in p block the difference of 2 occurs.</p>
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https://chemistry.stackexchange.com/questions/44220/difference-in-the-change-in-oxidation-states-of-transition-elements-and-p-block
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Question: <p>Why do the elements in the middle of the transition series show more number of common oxidation states than others?</p>
<p><img src="https://2.bp.blogspot.com/-gl6AodOkf5c/VLPvPFYxBPI/AAAAAAAAA1M/YFFFFmXKHU0/s1600/variable-oxidation-states-of-transition-elements.JPG" alt="" /></p>
Answer:
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https://chemistry.stackexchange.com/questions/168510/why-middle-elements-of-transition-series-show-more-number-of-oxidation-states
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Question: <p>Both osmium and iridium have 12 oxidation states. Indeed, iridium has the highest oxidation state of all elements at +9. see <a href="https://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements" rel="nofollow">https://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements</a></p>
<p>I'm just curious as to why these elements have the most oxidation states and whether this confers on them special properties or different reactivities or any special uses. Thank you!</p>
Answer:
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https://chemistry.stackexchange.com/questions/61353/why-do-osmium-and-iridium-have-the-most-oxidation-states-of-all-the-elements
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Question: <blockquote>
<p>Why is it that there are fixed oxidation states for most elements out there, with no to very rare exceptions? </p>
</blockquote>
<p>I was taught that oxidation state is a totally artificial concept we have invented for book-keeping purposes. How is it that its applicable so widely? I understand the case of formal charges, because that is what keep into consideration when making bonds in the first place.</p>
<p>But oxidation state is somewhat arbitrary, because we don't think about electronegativity when we are making bonds. So can someone help me understand the intuitive reason behind it, its been bothering me for months now?</p>
Answer:
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https://chemistry.stackexchange.com/questions/70463/why-are-oxidation-states-such-a-fundamental-quantity-for-elements
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Question: <p>Paulings electronegativity is a relative scale, based on the difference in electronegativity between X and Y, <span class="math-container">$\Delta EN = 0.102 \sqrt {\Delta}$</span>, where <span class="math-container">$\Delta = (X-Y)_{measured}-(X-Y)_{theoretical}$</span> bond energies.</p>
<p>But what is the oxidation state of X and Y? Is it an average of the common oxidation states for each element, or is it simply the most common oxidation state? Or do I have it all wrong to begin with?</p>
Answer:
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https://chemistry.stackexchange.com/questions/151759/which-oxidation-states-were-used-when-pauling-developed-his-electronegativity-sc
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Question: <p>Take propanoic acid for example. If I assign oxidation states to the 3 carbons in it using the idea that the more electronegative element gets all the electrons in the bonds it makes, the 3 carbons get -3( the one in the CH3), -2( the one in the middle), and +3( the one in the acid group). </p>
<p>Is this conclusion correct? The reason I doubt is that all the examples we've ever been given in class, we've assumed all the atoms of the same element to have the same oxidation state. </p>
Answer:
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https://chemistry.stackexchange.com/questions/111748/can-an-element-have-variable-oxidation-states-in-a-compound
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Question: <p>Here a screenshot from our lecture on bioinorganic molybdenum complexes, dealing with the catalytic cycle of xanthine oxidase:</p>
<p><a href="https://i.sstatic.net/oSChf.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/oSChf.png" alt="enter image description here" /></a></p>
<p>It is said that the only biologically relevant oxidation states of Mo are Mo(IV) and Mo(VI). That said, for the life of me, I cannot figure out why molecule 1 (or the other ones for that matter) in the catalytic cycle should have the oxidation state +VI.. I mean there are two oxo ligands and three thiol residues so, if the molecule is neutral it should be +VII, or am I missing something here?</p>
Answer:
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https://chemistry.stackexchange.com/questions/180636/how-are-oxidation-states-of-bioinorganic-molybdenum-tungsten-complexes-determine
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Question: <p>I have just been looking at <a href="https://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements</a> and found that boron has a -5 oxidation state. I would like to know which boron compounds form this oxidation state.</p>
Answer: <p>As I remember I have read somewhere that <span class="math-container">$\ce{Al3BC}$</span> is konwn to have Boron in -5 oxidation state. </p>
<ul>
<li>you can also reffer to this PDF page no. 139
<a href="https://d-nb.info/995006210/34" rel="nofollow noreferrer">PDF IN GERMAN</a></li>
</ul>
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https://chemistry.stackexchange.com/questions/108521/oxidation-states-of-boron
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Question: <blockquote>
<p>For preservation of meat, sodium nitrite is usually added and as a result <span class="math-container">$\ce{NO}$</span> is then formed. Consequently, <span class="math-container">$\ce{NO}$</span> reacts with the sulfur and iron atoms from decomposition of proteins, forming <span class="math-container">$\ce{[Fe4S3(NO)7]-}.$</span> X-ray crystallography shows that the complex anion has a structure as shown below: </p>
<p><a href="https://i.sstatic.net/x5Qgw.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/x5Qgw.png" alt="anion"></a></p>
<p>a) Blacken all the circles corresponding to iron atoms and add symbols Fe(A),
Fe(B), Fe(C) and Fe(D) beside the circles in the sequence of top → left → right.</p>
<p>b) The configuration of 3d electron shell of the iron atoms has been studied with modern structural analysis. Knowing that the mean oxidation number of the
four iron atoms is <span class="math-container">$–0.5$</span>, give their configurations of 3d shell, respectively.
Assume that each iron atom adopt sp hybridization. </p>
<p><span class="math-container">$\ce{[Fe4S3(NO)7]^{–}}$</span> anion can be reduced and a new complex <span class="math-container">$\ce{[Fe2S2(NO)4]^{2-}}$</span> is formed</p>
<p>c) Give the oxidation state of each iron atom with Arabic numerals.</p>
<p>d) <span class="math-container">$\ce{[Fe2S2(NO)4]2-}$</span> can be converted into <span class="math-container">$\ce{[Fe2(SCH3)2(NO)4]^{n}}$</span>, a carcinogen. Which of the following three species is added to <span class="math-container">$\ce{[Fe2S2(NO)4]^{2-}}$</span> : <span class="math-container">$\ \ce{CH3+}$</span>,<span class="math-container">$\ \ce{•CH3}$</span> or <span class="math-container">$\ \ce{CH3−}$</span>? </p>
</blockquote>
<p><strong>My Attempt</strong></p>
<p>a)</p>
<p><a href="https://i.sstatic.net/lEX37.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/lEX37.png" alt="anion with marked atoms"></a></p>
<p>I don't have any real reason to why I chose those 4 atoms. I chose them since they were the 4 biggest atoms in the ring and made sense they would alternate with the sulfur atoms in the ring. Is there a more solid explanation?</p>
<p>b) Firstly, why do the iron atoms have a negative oxidation state, doesn't it 'give' its electrons to the sulfur and nitrogen atoms since it is more electropositive? I look up negative oxidation states of iron and I found out when bonded to carbonyl compounds, the iron atom usually has a negative oxidation state. So is the <span class="math-container">$\ce{NO}$</span> ligand in this case acting like <span class="math-container">$\ce{CO}$</span>? </p>
<p>The answer is Fe(A) has 3d7 configuration while the others have a 3d9 configuration. Could someone please give an explanation for this.</p>
<p>c)</p>
<p><a href="https://i.sstatic.net/l5W4D.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/l5W4D.png" alt="enter image description here"></a></p>
<p>I have drawn what I think is the structure above, but I have the same problem as part b)</p>
<p>d) I have no idea. I am guessing the methyl cation since the iron atom has a negative oxidation state (I am guessing from part c) hence it will more likely react will a cation rather than an anion or radical.</p>
Answer: <p>I suppose it's asked about formal oxidation numbers of iron as there is not enough information to make assumptions about the ones that reflect electronic structure.
For instance, analysis of interatomic distances could point to a delocalized electron pair, but you are not given any linear or angular geometrical parameters to begin with.</p>
<p>It's only the symmetry that's evident from the crystal structure: the thioferrate cluster <span class="math-container">$\ce{[Fe4S3(NO)7]-}$</span> belongs to <span class="math-container">$C_\mathrm{3v}$</span> point group, hence the only possible arrangement of <em>four</em> iron atoms is a tetrahedral one.
Another conclusion that can be made is that three equatorial iron atoms are equivalent (<span class="math-container">$\ce{Fe(B)}$</span>, <span class="math-container">$\ce{Fe(C)}$</span>, <span class="math-container">$\ce{Fe(D)}$</span>), and the last apical one (<span class="math-container">$\ce{Fe(A)}$</span>) most likely deviates in terms of oxidation state.</p>
<p>Supposing formal oxidation number (<span class="math-container">$\text{O.N.}$</span>) of three equivalent irons and the last one are <span class="math-container">$y$</span> and <span class="math-container">$x$</span>, respectively, and the fact that the average <span class="math-container">$\text{O.N.}$</span> is <span class="math-container">$-0.5$</span>, the following holds true:</p>
<p><span class="math-container">$$x + 3y = -0.5\cdot 4 \implies y = -\frac{x + 2}{3}$$</span></p>
<p>As formal oxidation numbers are integers, there are several pairs giving the solution:</p>
<p><span class="math-container">$$
\begin{array}{l|rrrrr}
\hline
x & \ldots & -5 & -2 & \color{green}{1} & 4 & 7 & \ldots \\
\hline
y & \ldots & 1 & 0 & \color{green}{-1} & -2 & -3 & \ldots\\
\hline
\end{array}
$$</span></p>
<p>among which the least "extreme" (e.g. the one with the absolute lowest negative <span class="math-container">$\text{O.N.}$</span>) for a metal would be the <span class="math-container">$-1;+1$</span> pair; so that for <span class="math-container">$\overset{0}{\ce{Fe}}:~[\ce{Ar}]\,(\mathrm{3d})^6\,(\mathrm{4s})^2$</span> your answer gives two configurations, <span class="math-container">$(\mathrm{3d})^7$</span> for <span class="math-container">$\overset{+1}{\ce{Fe}}\ce{(A)}$</span> and <span class="math-container">$(\mathrm{3d})^9$</span> for <span class="math-container">$\overset{-1}{\ce{Fe}}\ce{(B,C,D)}$</span>.</p>
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https://chemistry.stackexchange.com/questions/53588/oxidation-states-of-iron-in-roussins-salts
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Question: <p>Halogens like $\ce{Cl,Br,I}$ generally show oxidation states between $-1$ and $+7$. However, is it possible that they can show an oxidation state beyond $-1$ (e.g. $-3$)?</p>
<p>I was thinking of some compound in which chlorine is bonded with a less electronegative element and forms more than one bond with such element(s).</p>
<p>Is it theoretically possible?</p>
Answer: <p>Yes, they can show oxidation states lower than -1. For example in <a href="http://onlinelibrary.wiley.com/doi/10.1002/rcm.2712/abstract">hyperlithiated compounds</a>, like $\ce{Li3Cl}$, the chlorine would formally have an oxidation state of -3.</p>
<p>However, a word of caution, oxidation states are a tricky thing. They are determined by adopting the hypothetical view that the compound you are looking at consists only of single-atomic ions, e.g. in the case of $\ce{Li3Cl}$ of 3 $\ce{Li+}$ and $\ce{Cl^{3-}}$. The hypothetical charges of these "ions" are then your oxidation states. Clearly, this view has nothing to do with the real bonding situation in the compound and oxidation numbers correlate very poorly with the real electron distribution. It is simply a useful little device for keeping track of redox reactions. So, the -3 oxidation state of chlorine in $\ce{Li3Cl}$ does not mean that chlorine has a 10-electron valence shell. The real bonding situation will be somewhat more complicated.</p>
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https://chemistry.stackexchange.com/questions/59146/is-it-possible-for-halogens-to-show-oxidation-states-less-than-1
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Question: <p>It is observed that actinides do not exhibit +2 oxidation state, contrary to that of some of the lanthanide elements. Why?</p>
Answer: <p>See also: <a href="https://chemistry.stackexchange.com/q/48971/16683">Why is WF6 stable whereas CrF6 is unknown?</a></p>
<p>Whenever one wants to compare oxidation states, there are a couple main factors to take into account. To reach a <em>higher</em> oxidation state, one obviously has to pay for it in the form of ionisation energy/energies. However, there is a compensatory effect in that elements in higher oxidation states generally get more out of bonding. For example, the lattice energy of $\ce{FeCl3}$ is larger than that of $\ce{FeCl2}$, or the four covalent bonds in $\ce{XeF4}$ are collectively stronger than the two covalent bonds in $\ce{XeF2}$.</p>
<p>The higher ionisation energies of the actinides tend to be smaller than those of the lanthanides, for the same reasons as explained in the linked question (5f orbitals have one radial node; 4f orbitals do not). This generally favours higher oxidation states and disfavours low oxidation states in actinides.</p>
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https://chemistry.stackexchange.com/questions/82722/oxidation-states-of-actinides
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Question: <p>I read that the fluorides of 3d metals in lower oxidation states, e.g. <span class="math-container">$\ce{VF2}$</span>, <span class="math-container">$\ce{TiF2}$</span>, and <span class="math-container">$\ce{CuF}$</span> are thermodynamically unstable. For example, <span class="math-container">$\ce{CuF}$</span> disproportionates to <span class="math-container">$\ce{Cu + CuF2}$</span>, as described on <a href="https://en.wikipedia.org/wiki/Copper(I)_fluoride" rel="nofollow noreferrer">Wikipedia</a>.</p>
<p>On the other hand, their iodides tend to be more stable (so, for example, <span class="math-container">$\ce{CuI}$</span> can be easily obtained from a redox reaction between <span class="math-container">$\ce{Cu^2+}$</span> and <span class="math-container">$\ce{I-}$</span>). Why is this the case?</p>
Answer: <p>In general, iodides stabilise lower oxidation states and fluorides stabilise higher oxidation states, e.g. <span class="math-container">$\ce{CuF2}$</span> versus <span class="math-container">$\ce{CuI}$</span>. This can be explained with some thermodynamics. Consider</p>
<p><span class="math-container">$$\ce{M(s) + $\frac{n}{2}\,$X2 (g/l) -> MX_n (s)}$$</span></p>
<p>In order to form a halide salt with a higher oxidation state (i.e. larger <span class="math-container">$n$</span>), you need to pay some energetic costs to get the metal to the higher oxidation state, namely: </p>
<ul>
<li>more electrons are removed from the metal, so you need more <em>ionisation energies</em></li>
<li>you need to generate more <span class="math-container">$\ce{X-}$</span> anions, so you need to break more <span class="math-container">$\ce{X-X}$</span> bonds (<em>bond dissociation energies</em>) and add more electrons to them (<em>electron affinities</em>).</li>
</ul>
<p>This seems bad, so why would any higher oxidation state ever be formed? The answer is that when you pay these costs, you recoup some of the energy by forming more bonds between <span class="math-container">$\ce{M^n+}$</span> and <span class="math-container">$\ce{X-}$</span>. If this bonding is strong enough, this can outweigh the costs described previously, and that makes the higher oxidation state more stable than the lower one.</p>
<p>For fluorine, the <span class="math-container">$\ce{M-F}$</span> bond is strong (regardless of whether you consider it to be ionic or covalent) and hence this energetic <em>gain</em> is large. On the other hand, the formation of fluoride ion is not particularly costly: the <span class="math-container">$\ce{F-F}$</span> bond is easy to break and fluorine loves picking up electrons, so the electron affinity works in your favour.</p>
<p>Overall, this means that there is a thermodynamic force for formation of fluorides in high oxidation states. Conversely, if you consider <em>iodides</em> instead, the formation of iodide ions is not as easy as the formation of fluoride. And the <span class="math-container">$\ce{M-I}$</span> bonding is weaker, too, so there isn't even much incentive to go to higher oxidation states.</p>
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https://chemistry.stackexchange.com/questions/102326/stability-of-3d-metal-fluorides-and-iodides-in-different-oxidation-states
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Question: <p>A long time ago I was researching the effect of the self regulatory response in Fe and Co. I found that my results made sense based on the idea of the electronegativity of the ions considered. I found a webpage that listed the relationship between the different electronegativities for different oxidation states of Fe and Co. </p>
<p>My problem is I didn't save the webpage and I don't know where I found it. I am not a chemist so I don't know in which kind of books I could find a list of values of electronegativities for different oxidized ions. </p>
<p>It basically said that the relationship between the electronegativities was something like</p>
<p><span class="math-container">$$\ce{Co^2+} < \ce{Fe^2+} < \ce{Fe^3+} < \ce{Co^3+}$$</span></p>
<p>Note: I don't remember it well, so what I wrote could be lies. The important thing was that there was a flip of order in electronegativities for Fe and Co when changing oxidation state.</p>
<p>Does anyone know where I could find that kind of information? I tried googling, but I am not finding the webpage and don't remember the terms I used to find it in the first place.</p>
<p>I never understood the why of the order either, and I have seen chemist webpages that tell you which ion is more electronegative by just looking at it. If someone could explain that too I would appreciate it.</p>
<hr>
<p>Edit:</p>
<p>I forgot to add that it was in octahedral complexes. I was able to find <a href="http://alpha.chem.umb.edu/chemistry/ch611/documents/Lec2-TheTransitionMetalspostlecture" rel="noreferrer">this pdf online</a> (page 32) but it doesn't state a reference. <a href="https://i.sstatic.net/BG1fI.png" rel="noreferrer"><img src="https://i.sstatic.net/BG1fI.png" alt="enter image description here"></a> </p>
Answer: <p>Pearson conveniently lists cumulative experimental data in the 1988 paper [<a href="https://doi.org/10.1021/ic00277a030" rel="noreferrer">1</a>], referrring to the earlier work of Moore [2].
Selected values of <span class="math-container">$I$</span> (ionization potential), <span class="math-container">$A$</span> (electron affinity), <span class="math-container">$χ$</span> (absolute electronegativity – probably, that's what you are looking for) and <span class="math-container">$η$</span> (absolute hardness) for iron and cobalt cations are:</p>
<blockquote>
<p><strong>Table I.</strong> Experimental Parameters for Monatomic Cations (eV)</p>
<p><span class="math-container">$$
\begin{array}{lcccc}
\hline
\text{ion} & I & A & χ & η \\
\hline
\ce{Fe^2+} & 30.65 & 16.18 & 23.42 & 7.24 \\
\ce{Fe^3+} & 54.8 & 30.65 & 42.73 & 12.08 \\
\ce{Co^2+} & 33.50 & 17.06 & 25.28 & 8.22 \\
\ce{Co^3+} & 51.3 & 33.50 & 42.4 & 8.9 \\
\hline
\end{array}
$$</span></p>
</blockquote>
<p>So it looks like the relation is a bit different:</p>
<p><span class="math-container">$$\ce{Fe^2+} < \ce{Co^2+} < \ce{Co^3+} < \ce{Fe^3+}$$</span></p>
<p>Complete table as a screenshot:</p>
<blockquote>
<p><a href="https://i.sstatic.net/jnLPR.png" rel="noreferrer"><img src="https://i.sstatic.net/jnLPR.png" alt="Table I - complete" /></a></p>
</blockquote>
<h3>References</h3>
<ol>
<li>Pearson, R. G. Absolute Electronegativity and Hardness: Application to Inorganic Chemistry. <em>Inorganic Chemistry</em> <strong>1988</strong>, 27 (4), 734–740. <a href="https://doi.org/10.1021/ic00277a030" rel="noreferrer">https://doi.org/10.1021/ic00277a030</a>.</li>
<li>Moore, C. E. "Ionization Potentials and Ionization Limits"; <em>Natl. Stand. Ref. Data Ser.</em> (<em>U.S. Natl. Bur. Stand.</em>); <strong>1970</strong>, NSRDS-NBS 34. (<a href="https://nvlpubs.nist.gov/nistpubs/Legacy/NSRDS/nbsnsrds34.pdf" rel="noreferrer">NIST - PDF</a>)</li>
</ol>
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https://chemistry.stackexchange.com/questions/115469/reference-for-electronegativities-of-different-metal-oxidation-states
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Question: <p>To get the oxidation state of carbon (denoted $x$) in octane, $\ce{C8H18}$, I used the fact that hydrogen has an oxidation state of $+1$:</p>
<p>$$8x + 18\cdot 1 = 0 \qquad \Longrightarrow \qquad x = -\frac{18}{8} = -2.25$$</p>
<p>But in that way, I get an oxidation number that is fractional, or not an integer. Is that normal?</p>
Answer: <p>The "oxidation number" is a theoretical value used to do electron bookkeeping and is one way of comparing the number of electrons "owned" by an atom in a molecule or ion versus how many valence electrons present in the atom as depicted on the periodic table.</p>
<p>Your calculation is assuming that all of the carbon atoms in octane have the same oxidation state, which is incorrect. The two terminal carbons have an oxidation state of -3 while the inner 6 have oxidation states of -2. Taking the average of these values gives you an average oxidation state, which in your case is a fractional number.</p>
<p>It is doubtful that the oxidation state calculated in this manner is of any use, since it is based on the assumption that all the carbons in the molecule are the same.</p>
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https://chemistry.stackexchange.com/questions/5357/are-fractional-oxidation-states-possible
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Question: <p>How to determine the oxidation states of the metal atoms when both cation and anion are coordination complexes?</p>
<p>For example, how to determine the oxidation states of <span class="math-container">$\ce{Pt}$</span> in the following compound?:</p>
<p><span class="math-container">$[\ce{Pt(NH3)4}][\ce{PtCl6}]$</span></p>
Answer: <p>There is something called an average oxidation state, which is typically not very meaningful chemically, but can (usually) easily be calculated by just knowing the structural formula. In your case, since chlorine is the most electronegative element (save nitrogen), you can assume it to be in $\mathrm{-I}$ state while nitrogen of course would have $\mathrm{-III}$ and hydrogen $\mathrm{+I}$. Thus to get platinum’s average oxidation state just sum up to the compound’s charge:</p>
<p>$$\begin{align}2 x + 4 \cdot (-3) + 3\cdot 4 \cdot (+1) + 6 \cdot (-1) = 0\\
2x - 12 + 12 - 6 = 0\\
2x -6 = 0\\
2x = 6\\
x = 3\end{align}$$</p>
<p>Which gives you an average oxidation state of $\mathrm{+III}$.</p>
<p>But as I said before, average oxidation states do not carry any physical or chemical meaning and can only serve as exercises to students and pupils. Take for example tetrathionate’s ($\ce{S4O6^2-}$) average sulphur oxidation state of $+\frac{5}{2}$.</p>
<p>Instead, one should always attempt to allocate per-atom oxidation states you cannot do that without knowing the charge of each of those complexes, and you need to do it per complex; ideally by drawing it, heterolytically cleaving all bonds (or homolytically for homoatom bonds) and then counting electrons. In the case of your compound — tetraamminplatinum(II) hexachloridoplatinate(IV) — you should arrive at the numbers in brackets in the IUPAC name I gave you, knowing that the cation has $2+$ charge and the anion $2-$.</p>
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https://chemistry.stackexchange.com/questions/59089/how-to-determine-the-oxidation-states-of-the-metal-atoms-when-both-cation-and-an
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Question: <p><a href="https://i.sstatic.net/je5ua.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/je5ua.png" alt="enter image description here"></a></p>
<p>I was doing some research on Wikipedia and I came across the article on <a href="https://en.wikipedia.org/wiki/Non-innocent_ligand" rel="nofollow noreferrer">non-innocent ligands</a>. It talked about the chemical $$[\ce{Ni(S_2C_2Ph_2)_2}]^z$$ as an example. It mentioned that the transition metal complex was diamagnetic when it was neutral(z=0). I decided to figure out how many unpaired electrons the central nickel atom would have at various formal oxidation states. I found the following: </p>
<p>$$
\begin{array}{ccc}
\ce{Ni(II)}&(z=-2)&=2~\text{unpaired}\\
\ce{Ni(III)}&(z=-1)&=3~\text{unpaired}\\
\ce{Ni(IV)}&(z=0)&=4~\text{unpaired}\\
\end{array}$$
I don't understand why the complex is diamagnetic when its charge is 0. The radical ligands would be coupled with each other anti-ferromagneticly but there would still be 4 unpaired electrons in the central nickel atom. Wouldn't they make the complex paramagnetic? Could someone please explain to me why it is diamagneic and tell me whether the other 2 oxidation states are para or dia-magnetic.</p>
Answer: <p>$\ce{Ni^{2+}}$ is a $d^8$ metal. In a square planar orbital splitting, all $8$ electrons are paired, hence it is diamagnetic. Even when the complex is neutral, the electron loss is at the ligands (by definition, non-innocent), and they are antiferromagnetically coupled, which means that their spins "cancel" each other. Since there are two, the net moment is zero.</p>
<p>All complexes are $\ce{Ni(II)}$, so assuming all complexes are square planar, all $d$ electrons are paired up and the magnetic moment from these is 0. Both the $z=-2$ (no radicals) and $z=0$ (antiferromagnetic coupling) area diamagnetic. The $z=-1$ complex is paramagnetic. There is a single unpaired electron in one of the ligands that can't couple to the other one, where all of them are paired up.</p>
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https://chemistry.stackexchange.com/questions/91903/what-are-the-magnetic-properties-of-nis2c2ph22zs-3-oxidation-states
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Question: <p>I looked at the flames of copper (I) chloride and of copper (II) chloride through a spectroscope and they looked the same. The flame colour is the same too. But since they have different oxidation states, and therefore different electron configurations, should their emission spectra be slightly different?</p>
<p>Would this logic also apply, for example, to elemental sodium in a discharge tube, vs sodium chloride burning in a flame? Would their emission spectra differ?</p>
Answer: <p>Interesting question. Keep in mind that the elemental emission spectrum in a flame or plasma and even a discharge does not remember its history in solution or a solid phase.</p>
<p>The punchline is that the emission spectrum is dependent on the elements <em>gas phase chemistry</em> in the flame/plasma/discharge. Prof. Ed has explained you the example of sodium. Let us say we have the following</p>
<p>(a) a block of element sodium
(b) a block of sodium chloride crystal
(c) solution of sodium chloride in water</p>
<p>If you introduce (a), (b) and (c) in the flame, the flame will be colored yellow in each case, which means that the emission is coming from a common emitter. That emitter is a elemental sodium atom excited by high temperature in the gas phase. Thus atomic emission spectrum is a fingerprint of the element.</p>
<p>You may ask that you introduced Na(+) in the flame in the case of b and c. Flames can easily reduce an ion to the elemental state. </p>
<p><strong>How to see the sodium ion spectrum</strong>: As we just said, the emission spectrum is independent of this original state. You can only cause ionization by increasing the temperature. This is the way to see the spectrum of an ion. If we were using a high temperature flame/electrical discharge, <em>we would start seeing Na(+) spectrum along with elemental sodium spectrum.</em></p>
<p>Coming to your particular example: You introduced copper (I) and copper (II) into a flame and they all colored it beautiful blue-green. The reason is that if the flame temperature is low, compounds cannot fully dissociate into atoms (not enough energy to break the bonds). In such cases, very simple diatomic or triatomic molecules are formed in the flame which emit their characteristic colors. In the case of copper, CuCl is formed in the flame. CuOH may be formed as well. Whether you introduced Cu(I) or Cu(II), as a chloride, it does not remember its solution phase or solid phase history.</p>
<p>If you were indeed using a high temperature flame, you will never ever such a blue green coloration, because this time, the emission is from Cu atoms (in UV). Hope that clarifies your confusion.</p>
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https://chemistry.stackexchange.com/questions/119671/do-different-oxidation-states-of-the-same-element-have-the-same-emission-spectru
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Question: <p>I am a student in theoretical chemistry and I am confused about the paper: Trends in adsorption of electrocatalytic water splitting intermediates on cubic $\ce{ABO3}$ oxides (Montoya, J. H.; Doyle, A. D.; Nørskov, J. K.; Vojvodic, A.; <em>Phys. Chem. Chem. Phys.</em> <strong>2018,</strong> <em>20</em> (5), 3813–3818. <a href="https://doi.org/10.1039/C7CP06539F" rel="nofollow noreferrer">DOI: 10.1039/C7CP06539F</a>), where the authors report DFT calculations on the oxides: </p>
<ul>
<li>$\ce{MgBaO3}$ ($\ce{Mg^3+}$ and $\ce{Ba^3+}$)</li>
<li>$\ce{NaLaO3}$ ($\ce{La^5+}$)</li>
<li>$\ce{CaBO3}$ and $\ce{SrBO3}$ ($\ce{B^4+}$)</li>
<li>$\ce{ZrNaO3}$ ($\ce{Zr^5+}$)</li>
<li>$\ce{ScSrO3}$ ($\ce{Sc^4+}$)</li>
<li>$\ce{CaCuO3}$ ($\ce{Cu^4+}$)</li>
</ul>
<p>... and many more. Are these oxidation states even possible?</p>
Answer: <p>Oxidation states are just numbers, a bookkeeping tool for chemistry. These hardly ever correspond to anything observable. Point in case: hypofluorous acid $\ce{HOF}$, see for example <a href="https://chemistry.stackexchange.com/q/15579/4945">the oxidation state of oxygen</a> and <a href="https://chemistry.stackexchange.com/q/58188/4945">of fluorine</a> in the confines of the definition. Maybe also of interst in that regard is the <a href="https://chemistry.stackexchange.com/q/69798/4945">introduction to "oxidation state"/"oxidation number"</a> in general. </p>
<p>Therefore instead of asking whether the oxidation states are stable, one should ask if the bulk/ molecular structure is stable.</p>
<p>When it comes to computationally aided catalyst design, you can pretty much use the entire periodic table to play with. It is a matter of interpreting the results that count, and obviously if experiment can reproduce such predictions. Where you start becomes a matter of taste and starting with perovskites is as good a guess as any.</p>
<p>The authors of the publication make quite an effort, calculating hundreds of potential catalysts. However, they also indicate problems within their methodology. Quoting from the supporting information:</p>
<blockquote>
<p><strong>3 Oxygen evolution data</strong><br>
[...]
\begin{array}{lcl}
\text{formula} & \text{values of }\Delta G, \eta, \dots & \text{warnings}\\
\hline
\ce{MgBaO3} & [\cdots] & \text{a,b}\\
\ce{NaLaO3} & [\cdots] & \text{a,b}\\
\ce{CaBO3} & [\cdots] & \text{b}\\
&\vdots&\\
\hline
\end{array}
<strong>a</strong>: One or more runs did not converge or failed in Quantum Espresso, missing adsorbates are reconstructed from successful runs using scaling<br>
<strong>b</strong>: Changes in atomic positions of greater than 5.5 angstrom found in the slab or adsorbate during optimization, typically indicates structural instability<br>
[...]</p>
</blockquote>
<p>And there are a lot more unstable species.</p>
<p><strong>TL;DR:</strong> Basically, in these extreme cases, where you would observe abnormal oxidation states, the calculation also predicted that the bulk/ molecular structure is not stable.</p>
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https://chemistry.stackexchange.com/questions/99065/are-such-high-oxidation-states-as-reported-by-vojovodic-et-al-possible
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Question: <p>I believe similar questions have been asked but this is different and the answers I have looked at don't answer this question. I have read that oxidation states of heavier transition elements (Ru, Os etc) are more <strong>stable</strong> than first row transition metals.</p>
<p>I understand how they can get to the oxidation state (which is the usual answer to this question) but im asking why they are more stable then the oxidation state is reached. And although an element like Os is bigger meaning the ionisation energy should be small for the first few electrons doesn't the lanthanide contraction mean the valence electrons have a stronger effective nuclear attraction.</p>
<p>By more stable I mean why is RuO4 explosive but OsO4 on the other hand is stable and FeO4 doesn't exist.</p>
Answer:
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https://chemistry.stackexchange.com/questions/139680/why-heavier-transition-metals-can-stabilise-higher-oxidations-states
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Question: <p>I noticed that the metals I learned were "type 1" back in early chemistry class as they have one oxidation state including silver, aluminum, zinc, and cadmium. However, those metals weren't reported as "type 1" when I looked online. I noticed that although silver has three oxidative states, it is still listed as "type 1" metal. Does anyone know why we still categorize it as "type 1" or have an insight into what is still considered a "type 1" metal?</p>
Answer: <p>I haven't heard of this classification of metals before. Anyways, the below image is from <a href="https://www.amherst.edu/media/view/226859/original/Chem11%2B1011F%2BD1%2BNomenclature%2Bhandout.pdf" rel="nofollow noreferrer">a nomenclature handout</a> where silver has been considered "type 2" metal:</p>
<p><a href="https://i.sstatic.net/fvBqI.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/fvBqI.jpg" alt="enter image description here" /></a></p>
<p>It's true that silver has oxidation states other than +1 but those are rarely encountered. So, it can get confusing and most of time, we may tend to consider that silver has only one oxidation state leading to it falling into "type 1" metal.</p>
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https://chemistry.stackexchange.com/questions/167988/why-is-silver-considered-a-type-1-metal-although-it-has-multiple-oxidation-sta
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Question: <p>Referring to the series of oxoacids of chlorine: <span class="math-container">$\ce{HClO, HClO2, HClO3},$</span> and <span class="math-container">$\ce{HClO4}$</span>, tabulated in <a href="https://en.wikipedia.org/wiki/Electronegativity#Variation_of_electronegativity_with_oxidation_number" rel="nofollow noreferrer">the Wikipedia article on electronegativity</a>, the article states, in the section near the end titled "Variation of electronegativity with oxidation number":</p>
<p>“<em>As the oxidation state of the central chlorine atom increases, more electron density is drawn <strong>from the oxygen atoms</strong> onto <strong>the chlorine, reducing</strong> the partial negative charge on the oxygen atoms …</em>” and illustrates it with a table. (NB There is an ambiguity in the use of the word "reducing" which is the opposite of "oxidising", but the intended meaning in terms of increasing or decreasing electron density is clear in the Wiki statement.)</p>
<p>But as I understand it (and as also given by the <a href="https://www.thoughtco.com/rules-for-assigning-oxidation-numbers-607567" rel="nofollow noreferrer">rules for assigning oxidation numbers</a>), insofar as oxidation state represents the number of electrons that an atom can gain, lose, or share when chemically bonding with an atom of another element, the table is correct, but shouldn’t the textual statement be <strong>the converse</strong>:</p>
<p>“As the oxidation state of the central chlorine atom increases, more electron density is <strong>drawn from the central chlorine atom</strong> onto <strong>the oxygen atoms, increasing</strong> the partial negative charge on the oxygen atoms ...”</p>
<p><strong>I reason it thus</strong> And just to be clear, let's go back one step to start with the even simpler case of <span class="math-container">$\ce{HCl}$</span>, where clearly the oxidation states must be (<span class="math-container">$\ce{H} = +1, \ce{Cl} = -1$</span>, respectively; total 0):</p>
<p>When we move on to <span class="math-container">$\ce{HClO}$</span>, then it becomes (+1, +1, -2 respectively) as the single oxygen with stronger electronegativity than chlorine is assigned the oxidation state -2, so that balancing the total means the electronegativity of the central chlorine must change from -1 to +1 (in agreement with the table on the Wiki page, but <strong>not with the article’s statement</strong> about the direction of change of electron density.</p>
<p>And as we move along the series to <span class="math-container">$\ce{HClO4}$</span>, the central chlorine is having more and more electron density sucked off it by the four more electronegative oxygen atoms, resulting in the final oxidation states (+1, +7, -8), in agreement with the table in the above-cited Wiki article, but not with the article's textual explanation.</p>
<p>So I conclude that the text statement in the Wiki article is currently wrong (and if so I will amend the Wikipedia article) - or else I'm wrong, and would appreciate someone pointing out my own error…?</p>
Answer: <p>Please do not edit the Wikipedia article, as your wording is incorrect, and the existing wording, while awkward, is correct.</p>
<p>What the article is trying to say is that the electron density on <em>each oxygen</em> is less as the oxidation state of the Cl increases. We can understand this by imagining progressive oxidation starting with <span class="math-container">$\ce{HClO}$</span>. Here, the Cl has oxidation state +1 and the O is -2. But of course the oxygen doesn't have a full negative 2 charge. Some of that charge is shared with the chlorine and the hydrogen, neither of which has a full charge of +1.</p>
<p>To get to <span class="math-container">$\ce{HClO2}$</span>, we can imagine reacting <span class="math-container">$\ce{HClO}$</span> with a neutral oxygen atom (oxidation state 0). After the reaction, the oxygen will have oxidation state -2, and the Cl will now be +3. Again, these are not the formal charges, but give us a hint about where charge will be located. The oxygen will have a partial negative charge, which means the chlorine will have to lose some of its electron density, increasing its actual charge. The original oxygen from <span class="math-container">$\ce{HClO}$</span> is still there and feels a stronger "pull" on its electron density, so it gives more to the chlorine, and the oxygen's partial negative charge becomes less negative.</p>
<p>Completely making numbers up, we could imagine that in <span class="math-container">$\ce{HClO}$</span>, the charge on oxygen is -1, on chlorine is +0.25 and on H is +0.75. After adding the second oxygen to make <span class="math-container">$\ce{HClO2}$</span>, the charge on the new oxygen might be -0.5, on the already attached O might be -0.75, +0.75 on H and +0.5 on Cl. Do you see how the negative charge on <em>each</em> O gets smaller in magnitude, but the <em>total</em> negative charge on oxygen(s) gets larger in order to offset the increased positive charge of the more oxidized chlorine? The Wikipedia statement refers to the individual O's not the sum.</p>
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https://chemistry.stackexchange.com/questions/136697/is-there-an-error-in-a-wikipedia-article-explaining-the-influence-of-oxidation-s
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Question: <p>I have read that <span class="math-container">$\ce{NOCl}$</span> dissociates to <span class="math-container">$\ce{NO+}$</span> and <span class="math-container">$\ce{Cl-},$</span> and the various reasons stated as such seem fine.</p>
<p>But what I don't get is this: if we follow the actual rules of assigning oxidation numbers, we find that nitrogen has got an unit positive charge with an unit positive charge on chlorine too (if we follow the structure of <span class="math-container">$\ce{NOCl}$</span> closely).</p>
<p>So, am I wrong in judging the oxidation number of nitrogen in <span class="math-container">$\ce{NOCl}$</span> to be <span class="math-container">$+1$</span> instead of <span class="math-container">$+3,$</span> as the dissociation of <span class="math-container">$\ce{NOCl}$</span> seems to go against my interpretation? What should be the oxidation number of nitrogen in <span class="math-container">$\ce{NOCl}$</span> in that case?</p>
Answer:
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https://chemistry.stackexchange.com/questions/127893/oxidation-states-the-elements-in-nitrosyl-chloride
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Question: <p>A well known experiment is where a basic sodium hydroxide and sucrose solution is added to a KMnO4 solution. The colors change from purple permanganate to green manganate to yellow/orange manganese dioxide (sometimes blue hypomanganate is formed as an intermediate). This common demo is called the 'Chemical Chameleon Reaction'. After doing this, I am left with the yellow MnO2 final solution. How can I convert the yellow MnO2 back to the original purple KMnO4. I am wondering how I can do a 'Reverse Chameleon'.</p>
Answer: <p>Probably your best approach is to use the methods reported by <a href="https://en.wikipedia.org/wiki/Potassium_permanganate" rel="nofollow noreferrer">Wikipedia</a>. The most commonly used method in industry involves these steps:</p>
<ul>
<li><p>Dissolve solid <span class="math-container">$\ce{MnO2}$</span> in molten potassium hydroxide plus an oxidizing agent. Air, potassium chlorate and potassium nitrate are reported as possibilities for the oxidizer. This forms <span class="math-container">$\ce{K2MnO4}$</span>, the green manganese(VI) compound.</p>
</li>
<li><p>Electrolyze the <span class="math-container">$\ce{K2MnO4}$</span> as an alkaline aqueous solution, forming <span class="math-container">$\ce{KMnO4}$</span> anodically. This has an advantage over a disproportionation approach of recovering all the manganese (theoretically) in the +7 oxidaton state.</p>
</li>
</ul>
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https://chemistry.stackexchange.com/questions/181885/manganese-oxidation-states-reverse-chameleon
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Question: <p>I have read answer to this question on various sites, but was not satisfied by the answers, because I am not able to understand, that if we remove initially 1 electron from scandium, then 4s would be ome unstable, but if we remove 1 more electron, there remains only one electron in 3d subshell. But we have already observed in many compounds, that heavier metallic elements and highly electronegative atoms force stability of unstable configurations. Also, I read somewhere that in [Sc (H20)6 ]2+ Scandium possesses +2 oxidation state, but when I tried to research more on it, some sites displayed the same compound but a +3 complex instead of +2. I am confused on all of this. Please throw some light on this, and please explain why Scandium cannot exhibit +2 oxidation state, is it just because of instability of 3d subshell? And what are the reasons for instability of 3d subshell, is it only the presence of a single electron? Please detail about this.</p>
<p>P.S. : It was also mentioned on some sites that after removal of one electron from 4s subshell of Sc, very high amount of energy is required for removal of second electron (i.e. I.E. 2 >> I.E. 1). How is this possible?</p>
Answer: <p>The claim that the second ionization energy of scandium far exceeds the first is simply not true, or as least not specifically true of scandium. A quick look at the <a href="https://en.wikipedia.org/wiki/Scandium" rel="nofollow noreferrer">Wikipedia page</a> reveals that the first ionization energy is 633 kJ/mol and the second is 1235, a ratio of about two to one. This is quite normal for metals with two outer <span class="math-container">$s$</span> electrons and the figures are actually lower than those for magnesium. Moreover, the third ionization energy is low enough for that to occur readily (unlike magnesium and all other alkaline earth metals), explaining why +3 is the most common (but not exclusive, as evident from the Wikipedia article) oxidation state for scandium.</p>
<p>Scandium in oxidation states lower than +3 is known primarily in organometallic compounds, where most other transition metals also tend to exhibit lower than "normal" oxidation states. One inorganic compound reported by Wikipedia is <span class="math-container">$\ce{KScCl3}$</span>, in which the scandium has oxidation state +2 but does not consist of separate <span class="math-container">$\ce{Sc^{2+}}$</span> ions. Instead this compound retains scandium-scandium metallic bonding. The <span class="math-container">$\ce{Sc(H2O)6^{2+}}$</span> ion suggested in the question is probably a misprint of <span class="math-container">$\ce{Sc(H2O)6^{3+}}$</span>.</p>
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https://chemistry.stackexchange.com/questions/167718/why-scandium-sc-doesnt-show-variable-oxidation-states-especially-2-while
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Question: <p>From Sc, to Mn, the number of oxidation states increases from one (Sc) to seven (Mn). The explanation for this is because the unpaired 3d electrons can be lost along with the 4s electrons during bonding.</p>
<p>Should this be the case? Wouldn't it be easier to lose an electron when it is paired because of interelectronic repulsion in the orbital? Could someone explain this? Thanks!</p>
Answer: <p>The oxidation state of a transition metal is an accounting device. It's not the number of "lost electrons". In potassium permanganate, which has Mn (VII), the manganese atom shares its seven 3d electrons with the ligand (oxygen) in what are essentially covalent bonds, and the "positive charge" associated with the metal is not very different than what it is for ionic calcium compounds.</p>
<p>Regarding "paired electrons are less likely to be involved in bonding than unpaired electrons" -- both paired and unpaired electrons in a transition metal can participate in bonding, provided they are in the outermost shell. The paired electrons in inner shells are too deep in energy to take part in bonding.</p>
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https://chemistry.stackexchange.com/questions/14539/explanation-for-increase-in-the-number-of-oxidation-states-for-transition-elemen
|
Question: <blockquote>
<p>Consider the cell reaction</p>
<p><span class="math-container">$$\ce{2Ag+ + Pb <=> 2Ag + Pb^2+}$$</span>
If <span class="math-container">$\ce{H2S}$</span> gas is passed through the solution, what will be the effect on the EMF of the cell?</p>
</blockquote>
<p><strong>My attempt:</strong></p>
<p>The Nernst equation for the EMF of a cell results in</p>
<p><span class="math-container">$$\mathcal{E}=\mathcal{E}^\circ-\frac{RT}{nF}\ln{\frac{[\ce{Pb^2+}_{(aq)}]}{[\ce{Ag+_{(aq)}}]^2}}$$</span>
Now, I assumed that <span class="math-container">$\ce{Ag2S}$</span> was more covalent and hence less soluble in water than <span class="math-container">$\ce{PbS}$</span>, by Fajan's rule, due to smaller size of cation and pseudo-inert gas configuration and hence it would precipitate out more. hence the ratio
<span class="math-container">$${\frac{[\ce{Pb^2+}_{(aq)}]}{[\ce{Ag+_{(aq)}}]^2}}$$</span></p>
<p>would decrease and hence the EMF of the cell would decrease.</p>
<p>Is my reasoning correct? I do not have the answer with me.</p>
Answer:
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https://chemistry.stackexchange.com/questions/49640/effect-of-passing-hydrogen-sulfide-gas-on-electrode-potential
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Question: <blockquote>
<p>For the galvanic cell: $\ce{Ag|AgCl(s)|KCl(0.2M)||KBr(0.001M)|AgBr(s)|Ag}$, calculate the electromotive force (EMF) generated.<br>
$K_\mathrm{sp}(\ce{AgCl}) = 2.8\times10^{-10}$;
$K_\mathrm{sp}(\ce{AgBr}) = 3.3\times10^{-13}$</p>
</blockquote>
<p>I have tried this forming the cell reaction which is coming out to be
$$\ce{AgCl + Br- <=> AgBr + Cl-}$$
then using the $K_\mathrm{sp}$ of both salts I found the equilibrium constant which should be
$$K_\mathrm{eq} = \sqrt{\frac{K_\mathrm{sp}(\ce{AgCl})}{K_\mathrm{sp}(\ce{AgBr})}}$$
then I put it in the formula
$$E_\mathrm{cell} = -0.059 \log(K_\mathrm{eq})$$ but my answer is coming wrong.</p>
Answer: <p><span class="math-container">$$\begin{align}
E_{\ce{Ag/AgBr}}&=E_{\ce{Ag/Ag+}}^{\circ}+0.059\log\frac{K_{\rm sp, AgBr}}{[\ce{Br-}]} \\
E_{\ce{Ag/AgCl}}&=E_{\ce{Ag/Ag+}}^{\circ}+0.059\log\frac{K_{\rm sp, AgCl}}{[\ce{Cl-}]} \\
EMF&=0.059 \cdot \left| \log \left( {\frac
{ K_{\rm sp, AgCl} \cdot [\ce{Br-}]}
{ K_{\rm sp, AgBr} \cdot [\ce{Cl-}]}
} \right) \right| \\
EMF&=0.059 \cdot \left| \log \left( {\frac
{ 2.8\times10^{-10} \cdot 0.001}
{ 3.3\times10^{-13} \cdot 0.2}
} \right) \right| \\
EMF&=0.059 \cdot \left| \log \left( {\frac { 2.8}{ 0.66} } \right) \right| \\
EMF&=0.037 \rm \ V\\
\end{align}$$</span></p>
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https://chemistry.stackexchange.com/questions/59383/how-to-calculate-the-electromotive-force-of-a-silver-chloride-silver-bromide-ce
|
Question: <p>Reactions for the lead acid battery are: </p>
<p><span class="math-container">$$
\begin{array}{}
\text{Oxidation}&\ce {Pb(s) + HSO4^-(l) &-> PbSO4(s) + H+(l) + 2e-}\\
\text{Reduction}&\ce{PbO2 + HSO4^-(l) + 3H+(l) + 2e- &-> PbSO4(s) + 2H2O}\\
\text{Total reaction}&\ce{Pb(s) + PbO2(s) +2HSO4^- +2H+ &-> PbSO4 + 2H2O}\\
\end{array}
$$</span> </p>
<p>What will be the cell notation for this battery? My attempt:<span class="math-container">$$\ce{Pb(s), Pb^2+(s)| HSO4^-(l)| PbO2(s), Pb^2+(s),Pb(s)}$$</span></p>
<p>The things that I have in mind,</p>
<ol>
<li>In both sides the electrode material is <span class="math-container">$\ce{Pb(s)}$</span></li>
<li>Separate every element that is in the same phase with a comma</li>
<li>Separate every element with different phases with a single bar</li>
<li>Separate the two half reactions with a single bar (as I find no salt bridge here)</li>
</ol>
<p>I am having a tough time learning cell notation. Anyways this was a challenging one for me and I tried my best shot. I am really not sure whether everything is right or not. So please identify my mistakes and show me the right way to do it. </p>
Answer: <p>There are a couple of things wrong here. First off, your final reaction is <strong>unbalanced</strong>. Once you've fixed the balancing, read the other mistakes:</p>
<ol>
<li>The ions do not exist in the <strong>liquid</strong> state! They are solvated/hydrated by the solvent. Since the solvent is water here, we'll say that the ions are in the <strong>aqueous</strong> <strong>(aq)</strong> phase instead. </li>
<li>While there is certainly no salt bridge here, there is still an <strong>electrolyte</strong> - an aqueous solution of sulphuric acid. Hence, you must mention it in your cell notation, between the anode and the cathode.</li>
<li>The <span class="math-container">$\ce{PbSO4}$</span> formed at the anode is in <strong>solid state</strong>. Hence, writing it as <span class="math-container">$\ce{Pb^2+(s)}$</span> is incorrect, as it is <em>not</em> dissociated into the ions <span class="math-container">$\ce{Pb^2+}$</span> and <span class="math-container">$\ce{SO4^2-}$</span>.</li>
</ol>
<p>With all these corrections, your final, correct cell representation should be:</p>
<blockquote class="spoiler">
<p> <span class="math-container">$$\small{\ce{Pb(s), HSO4^-(aq) | PbSO4(s), H+ | H2SO4 ($\pu{x~M}$) | PbO2(s), HSO4-(aq), H+(aq) | PbSO4(s)}}$$</span></p>
</blockquote>
<hr>
<p>Some websites (like <a href="https://youtu.be/PQ48N5jaG2w" rel="nofollow noreferrer">KhanAcademy</a>) and texts (NCERT 12), cite the lead-acid battery reaction as this instead:</p>
<p><span class="math-container">$$
\begin{array}{}
\text{Oxidation}&\ce {Pb(s) + SO4^2-(aq) &-> PbSO4(s) + 2e-}\\
\text{Reduction}&\ce{PbO2 + SO4^2-(aq) + 4H+(aq) + 2e- &-> PbSO4(s) + 2H2O(l)}\\
\text{Total reaction}&\ce{Pb(s) + PbO2(s) +2SO4^2(aq)- + 4H+(aq) &-> 2PbSO4(s) + 2H2O(l)}\\
\end{array}
$$</span> </p>
<p>assuming the bisulphite ion to be further ionized. The cell notation in this case would then be:</p>
<blockquote class="spoiler">
<p> <span class="math-container">$$\small{\ce{Pb(s), SO4^2-(aq) | PbSO4(s) | H2SO4 ($\pu{x~M}$) | PbO2(s), SO4^2-(aq), H+(aq) | PbSO4(s)}}$$</span></p>
</blockquote>
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https://chemistry.stackexchange.com/questions/89562/cell-notation-for-the-lead-acid-battery
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Question: <p>Is the role of potassium hydroxide (<span class="math-container">$\ce{KOH}$</span>) in an alkaline battery to provide hydroxide for the reaction with zinc? Would the battery cease to work if <span class="math-container">$\ce{KOH}$</span> was removed and only <span class="math-container">$\ce{OH-}$</span> from autoionization (<span class="math-container">$\pu{10^{-7} M}$</span>) remained?</p>
Answer: <p>The role of <span class="math-container">$\ce{KOH}$</span> is to provide enough ions for redox reactions. Remember that redox reactions come in pairs, for every reduction there must be oxidation. If <span class="math-container">$\ce{Zn}$</span> reacts with <span class="math-container">$\ce{OH-}$</span> to form a hydroxide, what would be a counter-reaction?</p>
<p>A battery also needs to have certain ionic conductivities to be functional. Could you ever get enough <span class="math-container">$\ce{OH-}$</span> ions through autoionization?</p>
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https://chemistry.stackexchange.com/questions/135351/role-of-potassium-hydroxide-in-alkaline-battery
|
Question: <p>A charged particle like an ion, does its repulsive or attractive effect on other particles change with temperature? Are electrostatic effects temperature dependent?</p>
Answer: <p>The strength of the electric charge does <em>not</em> change with temperature (unless approaching that of the Big Bang). However, <em>effects</em> of the charges do change with temperature. For example, at low temperatures (near 0 K), <a href="https://phys.org/news/2020-06-arrays-strontium-rydberg-atoms-quantum.html" rel="nofollow noreferrer">ions can form arrays in space</a>, though their charges repel. The <a href="https://web.mit.edu/physics/news/physicsatmit/physicsatmit_02_bec.pdf" rel="nofollow noreferrer">history of Bose-Einstein condensates</a> begins with trapped ions, extending to neutral gases.</p>
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https://chemistry.stackexchange.com/questions/135373/are-electrostatic-effects-temperature-dependent
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Question: <p>For the last assignment of the school year, my science teacher gave the class questions about chem, physics, etc. The third question had me confused and I'll be thankful for all the help given. The question is:</p>
<blockquote>
<p>An ionic solution is created and tested to see how much current can go
through it. Two groups are told to create the following solution
consisting of <span class="math-container">$\pu{1.0 +/- 0.1 L}$</span> of water and <span class="math-container">$\pu{2.0 +/- 0.2 g}$</span> [of what?] . The battery used for the test is <span class="math-container">$\ce{8 +/- 1 V}$</span>.</p>
<p>Group A measures a current of <span class="math-container">$\pu{2.1 +/-0.1 mA}$</span>. Group
B measures a current of <span class="math-container">$\pu{3.4 +/- 0.1 mA}$</span>.</p>
<p>Describe 3 possible reasons why Group B measures a current greater
than group A.</p>
</blockquote>
Answer: <p>No offence to you (certainly to the original author of this question), I call such home-work questions such as garbage in-garbage out type questions. No wonder a student will be confused if they see such open ended questions. The reason is that the author is not telling you or us of 2.0 grams of what?</p>
<p>Even if we assume it was 2 grams of salt, several things can affect the measured current</p>
<p>(1) What are the electrode areas?
(2) What are the electrodes made of?
(3) What is the distance between the electrodes?</p>
<p>Last but not the least, the identity of the salt matters a lot. Did both groups use the same salt? Monovalent, divalent, etc</p>
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https://chemistry.stackexchange.com/questions/135389/current-with-an-ionic-solution
|
Question: <p>Activated carbon is less expensive than mercury . They are recyclable . Mercury is toxic. Nowadays activated carbon is being used in supercapacitors,Li-ion battery.</p>
Answer: <p>Reference electrodes must involve a well defined redox system with reproducible potential.</p>
<p>Mercury is an essential part of the redox system of the calomel reference electrode <span class="math-container">$\ce{Hg|Hg2Cl2(s)|Cl-}$</span>:
<span class="math-container">$$\ce{2 Hg <=> Hg2^2+ + 2 e-}$$</span>
<span class="math-container">$$\ce{Hg2^2+ + 2 Cl- <=> Hg2Cl2(s)}$$</span>
and cannot be removed nor replaced, unless you want to use a difference electrode.</p>
<p>A similar, more often used reference electrode is the silver chloride electrode, <span class="math-container">$\ce{Ag|AgCl(s)|Cl-}$</span>, being safer, simpler and more practical:
<span class="math-container">$$\ce{Ag <=> Ag+ + e-}$$</span>
<span class="math-container">$$\ce{Ag+ + Cl- <=> AgCl(s)}$$</span>
This electrode can be very compact and is frequently integrated to measurement electrodes, forming a full electrochemical cell, like the pH "glass" electrode.</p>
<p>If a glassy carbon were used, it would be inert electrode material, similarly as platinum in the primary hydrogen reference electrode. It would need a well defined redox system established within the solution.</p>
<p>Note that Li-Ion cells use graphite, not activated carbon, forming lithium graphite intercalate.</p>
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https://chemistry.stackexchange.com/questions/139774/can-i-use-activated-carbon-electrod-as-a-reference-electrod-in-a-ph-meter-instea
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Question: <p>When we make an electrochemical cell, we dip the electrodes in the salt solution. How exactly does the electrode potential change as we dip the electrodes in the salt solutions? Would different salt solutions change the effect of the electrode potential?</p>
Answer: <p>The short answer is that we want a cathodic reaction so we can use the metals with a positive standard reduction potential (because 0 is for <span class="math-container">$\ce{H2} $</span> reduction).</p>
<p>In reality however to really find a suitable electrolyte cyclic voltammetry is typically used and electrolytes with suitable electrochemical windows for the cell reaction are chosen.</p>
<p><a href="https://i.sstatic.net/1xVSa.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/1xVSa.png" alt="enter image description here"></a>
The electrochemical window is the voltages where the current is zero in the picture above. If the wrong electrolyte is chosen then a full redox reaction happens(with the electrolyte participating) rather than half reactions at each electrode negating the need for the external circuit and creating heat</p>
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https://chemistry.stackexchange.com/questions/123539/how-does-solution-which-electrodes-are-immersed-in-effect-electrode-potential
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Question: <p>As a part of a school investigation I constructed a voltaic cell experiment where for each run I added a different ligands to the copper (II) sulfate half-cell electrolyte in a daniell cell to see if there was a change in the output voltage Ecell. When plotting a graph of this, it seemed that Ecell increased as the stability of the ligand increased (indicated by the stability constant Kc) however I am struggling to explain why. I presume its because the standard potential at the cathode Erhe increases with ligand stability due to greater oxidation.</p>
Answer: <p>The redox potential of Copper does not depend directly on the stability of the complex Cu-ligand. It depends on the concentration of the residual <span class="math-container">$\ce{Cu^{2+}}$</span> ion in solution. And in any solution of <span class="math-container">$\ce{CuSO4}$</span>, adding a ligand produces a decrease in concentration of the free ion <span class="math-container">$\ce{Cu^{2+}}$</span>. So, according to Nernst's law, the potential <span class="math-container">$\ce{Cu^{2+}/Cu}$</span> becomes less positive.</p>
<p>For example, the standard redox potential of <span class="math-container">$\ce{Cu^{2+}/Cu}$</span> is + <span class="math-container">$0.34$</span> V. Let's consider the ligand chloride <span class="math-container">$\ce{Cl-}$</span>. If enough chloride ions <span class="math-container">$\ce{Cl-}$</span> is added to a <span class="math-container">$1$</span> M <span class="math-container">$\ce{Cu^{2+}}$</span> solution so that at the end the excess of <span class="math-container">$\ce{Cl}$</span>- ions is <span class="math-container">$1$</span> M, the complex <span class="math-container">$\ce{CuCl4^{2-}}$</span> will be formed, with a concentration <span class="math-container">$1$</span> M. The measured residual [<span class="math-container">$\ce{Cu^{2+}}$</span>] concentration is <span class="math-container">$2.5 ·10^{-6}$</span> M. In this case, Nernst's law gives the following <span class="math-container">$\ce{Cu^{2+}/Cu}$</span> redox potential : <span class="math-container">$$\ce{E_{Cu^{2+}/Cu} = + 0.34 V + \frac{0.0592}{2}log(2.5·10^{-6}) = + 0.34 V - 0.0296·5.6 = +0.34 V -0.166 V = +0.174 V}$$</span> This +<span class="math-container">$0.174$</span> V is lower than the standard redox potential (+<span class="math-container">$0.34$</span> V).</p>
<p>The stability complex of the <span class="math-container">$\ce{CuCl4^{2-}}$</span> ion is : <span class="math-container">$\ce{K = \frac{[CuCl4^{2-}]}{[Cu^{2+}][Cl^-]^4} = \frac{1}{2.5·10^{-6}·1^4} = 4·10^5 M^{-4}}$</span></p>
<p>If another ligand is used, which makes a stronger complex with copper <span class="math-container">$\ce{Cu^{2+}}$</span>, the stability constant is higher, and the redox potential will be still lower.</p>
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https://chemistry.stackexchange.com/questions/141116/what-is-the-relationship-between-the-stability-constant-of-a-copper-ii-electro
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Question: <p>The general equation for liquid junction potential is given by <span class="math-container">$$E = {-RT\over F}\sum_i\int_a^b{t_i\over z_i}d ln(a_i)$$</span>
Where a and b are the two phases, <span class="math-container">$t_i$</span> is the transport number of the ions and <span class="math-container">$z_i$</span> is the charge on the ions.
When making the additional assumptions that (1) the concentrations are equivalent to the activities everywhere in the junction and that (2) the concentration of each ion follows a linear transition between the two phases, the above equation becomes the Henderson equation:<span class="math-container">$$E = {\sum_i{|Z_i| \over z_i}u_i[c_i(b)-c_i(a)]\over \sum_i|z_i|u_i[c_i(b)-c_i(a)]}{RT\over F}{ln{\sum_iz_iu_ic_i(a)\over \sum_iz_iu_ic_i(b)}}$$</span>But when I tried to evaluate the integral by my self by substituting the equation of transfer number <span class="math-container">$t_i = {z_ic_iu_i\over \sum_iz_ic_iu_i}$</span> and putting <span class="math-container">$a_i = c_i$</span>, I'm stuck with the following: <span class="math-container">$$E = {-RT\over F}\sum_i \int_a^b{{|z_i|\over zi}u_i\over \sum_i|z_i|c_iu_i}dc_i$$</span> The integral is of the form <span class="math-container">$\int{1\over {ax + b}}dx$</span> where <span class="math-container">$a = |z_i|u_i$</span> and <span class="math-container">$b = \sum_{j \ne i}|z_j|c_ju_j$</span> <span class="math-container">$$E = {-RT\over F}\sum_i{1\over z_i}ln\left( {|z_i|u_ic_i(b)+\sum_{j \ne i}|z_j|c_ju_j\over |z_i|u_ic_i(a)+\sum_{j \ne i}|z_j|c_ju_j}\right)$$</span>Which is clearly not the Henderson equation given in Bard and Falkner, and I couldn't figure out where to use the second assumption. Where was I wrong and how to proceed?</p>
Answer:
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https://chemistry.stackexchange.com/questions/142862/how-to-derive-the-henderson-equation-for-liquid-junction-potentials-given-in-bar
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Question: <p>Ordinary Batteries will deplete most of their Power in a 10 year period. What happens to rechargeable Batteries? How long would they still functin properly?</p>
<p>Hope this is the right community for this question. If you need a specific example:</p>
<p>Powertoll Battery:
<a href="https://www.metabo.com/ch/de/zubehoer/zubehoer-akkumaschinen/akkupacks/lihd/akkupack-lihd-18-v-10-0-ah-625549000.html" rel="nofollow noreferrer">https://www.metabo.com/ch/de/zubehoer/zubehoer-akkumaschinen/akkupacks/lihd/akkupack-lihd-18-v-10-0-ah-625549000.html</a></p>
<p>1.5v Battery
<a href="https://www.energizer.com/batteries/energizer-rechargeable-batteries" rel="nofollow noreferrer">https://www.energizer.com/batteries/energizer-rechargeable-batteries</a></p>
<p>Stored in an airsealed container, decent temperature & humidity. And lets say we charge & deplete them 3x per year.</p>
<p>Just to be sure, I'm not asking how many charges they will survive, but after how many decades they are still usable.</p>
Answer: <p>It mostly depend on battery type, and after that current. Batteries usually dislike very high currents, that would make them charge or discharge faster than in 1 hour, because of heating and secondary reactions, and very low currents, that would make them discharge in a year, because of dendrite formation.</p>
<p>Worst type of battery for decade long work are lithium-ion batteries. Their decay is mostly due to time and temperature, rather than cycles used. Exact opposite of your goal. NiCd, NiMH are less energy dense, they also will need additional 3 recharges per year due to self-discharge. All of them lose about half of their maximum capacity in 3 years. Li-ion because of degradation due to temperature. NiCd and NiMH bevause of large self discharge.</p>
<p>LiFePO4 is significantly longer living battery than li-ion. Lead-acid bartery is somewhat similar for your usage. They will lose about half of their capacity in 10 years. They achieve it in a different way. Lead-acid has significant self-discharge, needing an additional recharge every year, but its bulky and degradation takes a long time to affect it all. LiFePO4 is the opposite, very delicate, low self discharge, and degradation is just slower in general. At this timeframe your device will become obsolete.</p>
<p>Lithium titanate, LTO, is probably the best of what we have today for your usage. It is very durable and degrades very slowly. Its hard to put an estimate on this, but i guess you can expect about 25 years till half of it maximum capacity is gone. At this time frame wires will rust away first.</p>
<p>If you need even longer working time, you could check supercapacitors. They have 10 times less energy density, but they will work till their shell is rusted away. About 50 years till half of of them are rusted. They also need 100 recharges per year due to self discharge.</p>
<p>If you want even longer life, check ordinary capacitors and compound filled PCB. Capacity is 100 times less than batteries. Those will work till UV, ozone and moisture will destroy the compound or wires going to the board. At this time frame your smart PCB components are likely to fail first. About 100 years. Will need a few recharges per year due to selfdischarge.</p>
<p>For longer still, will need something exotic, like flywheels and magnetic levitation in a vacuum chamber.</p>
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https://chemistry.stackexchange.com/questions/143664/how-long-do-rechargeable-batteries-last-vs-standard-batteries
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Question: <p>In a <a href="https://en.wikipedia.org/wiki/Concentration_cell" rel="nofollow noreferrer">concentration cell</a>, one can measure a voltage because in one of the half cells (the one with lower concentration electrolyte), the atoms tend to dissipate more into the electrolyte as ions, leaving more electrons behind (compared to the other half cell), thus creating a current.</p>
<p>I am not sure if I understand, <em>why</em> a lower concentration causes the atoms to dissipate more. Is it because the atoms of the metal have more contact to water?</p>
Answer: <p>It is not primarily that the process <span class="math-container">$\ce{M(s) -> M^n+(aq) + n e-}$</span> is faster for more diluted solutions.</p>
<p>The main reason is that the opposite process <span class="math-container">$\ce{M^n+(aq) + n e- -> M(s)}$</span> is slower for more diluted solutions.</p>
<p>As the net effect, there is dissolution of <span class="math-container">$\ce{M^n+(aq)}$</span> ions and the electrode gaining more negative potential, until the rate of both processes gets equal and the electrode reaches its equilibrium potential.</p>
<p>Because the rate of ion dissolution grows with increased potential,
while the rate of ion depositions grows with decreased potential.</p>
<p>The final effect is, the electrode in then more diluted ion solution has the more negative potential.</p>
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https://chemistry.stackexchange.com/questions/145201/why-does-lower-concentration-cause-the-metal-atoms-to-dissipate-more
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