page_content stringlengths 12 2.63M | metadata unknown |
|---|---|
For example, for malonic acid (p $K_{\rm a1}=2.85$ and p $K_{\rm a1}=5.70$ ) we can prepare buffers with pH values of
$$pH = 2.85 + log \frac{C_{HM^-}}{C_{H_2M}}$$
$$pH = 5.70 + log \frac{C_{M^{2-}}}{C_{HM^{-}}}$$
where H<sub>2</sub>M, HM<sup>-</sup> and M<sup>2-</sup> are malonic acid's different acid-base form... | {
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Because of these ionic atmospheres, each ion's apparent charge at the edge of its ionic atmosphere is less than the ion's actual charge.
Adding 10 g of KNO<sub>3</sub> to the solution and stirring to dissolve the solid, produces the result shown in <u>Figure 6.15b</u>. The solution's lighter color suggests that addin... | {
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It is possible to estimate activity coefficients using the EXTENDED DEBYE-HÜCKEL EQUATION
$$\log \gamma_A = \frac{-0.51 \times z_A^2 \times \sqrt{\mu}}{1 + 3.3 \times \alpha_A \times \sqrt{\mu}}$$
6.63
where $z_A$ is the ion's charge, $\alpha_A$ is the hydrated ion's effective diameter in nanometers (Table 6.2)... | {
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$$\gamma_{10\frac{1}{3}} = 0.810$$
Defining the equilibrium concentrations of $Pb^{2+}$ and $IO_3^-$ in terms of the variable x
| Concentrations | $Pb(IO_3)_2$ (s) | = | $Pb^{2+}$ (aq) | + | $2 IO_3^- (aq)$ |
|----------------|------------------|---|----------------|---|-----------------|
| Initial | s... | {
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Select the *Value of:* radio button and enter 0 in the box. Place the cursor in the box for *By Changing Cells:* and click on cell
**Figure 6.17** Spreadsheet demonstrating the use of Excel's Solver function to find the root of a cubic equation. The spreadsheet in (a) shows the cubic equation in cell B2 and the initi... | {
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| (a) | | A | В | С | D |
|-----|---|----------|--------------------|-------------------|--------------------|
| | 1 | pH = | 3.00 | 2.00 | 1.00 |
| | 2 | [H3O+] = | =10^-b1 | =10^-c1 |... | {
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$$> eqn = function(x) \{4*x^3 + 0.4*x^2 - 2.5e-13\}$$
Because our equation is a function, the **uniroot** command can send a value of *x* to *eqn* and receive back the equation's corresponding value. Finally, we use the **uniroot** command to find the root.
$$>$$
**uniroot**(eqn, lower = 0, upper = 0.1, tol = 1e–... | {
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```
A simpler, more compact way to do this is > pH = seq(1,7,1) > eval(pH) or > eval(seq(1,7,1)) where the sequence command has a for-
```
seq(lower limit, upper limit, step size)
| (a) pH<br>error | (b) | pH | error | (c) | pH | error |
|------------------|-----|----|-----------... | {
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Alternatively, we can use a cumulative, or overall formation constant, *b*2, for the metal–ligand complex ML2, in which both ligands are added to M.
In an oxidation–reduction reaction, one of the reactants is oxidized and another reactant is reduced. Instead of using an equilibrium constants to characterize an oxidat... | {
"Header 1": "Equilibrium Chemistry",
"Header 3": "Chapter Overview",
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Be sure to state and justify any assumptions you make in solving the problems.
- 10. Ignoring activity effects, calculate the molar solubility of Hg<sub>2</sub>Br<sub>2</sub> in the following solutions. Be sure to state and justify any assumption you make in solving the problems.
- a. a saturated solution of Hg<sub>2</... | {
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For your convenience, here are hyperlinks to the appendices containing these constants
Appendix 10: Solubility Products
Appendix 11: Acid Dissociation Constants
Appendix 12: Metal-Ligand Formation Constants
Appendix 13: Standard State Reduction Potentials
$$K = \frac{K_4}{(K_1)^2} = \frac{(5.0)}{(0.40)^2} = 3... | {
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**Figure 6.21** Ladder diagram for Practice Exercise 6.5
| Concentrations | $Hg_2Cl_2$ (s) | = | $Hg_2^{2+}$ (aq) | + | 2Cl <sup>-</sup> (aq) |
|----------------|----------------|---|------------------|---|-----------------------|
| Initial | solid | | 0 ... | {
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Because $NH_3$ is a base, our first assumption is
$$[OH^{-}] >> [H_3O^{+}]$$
which simplifies the charge balance equation to
$$[NH_4^+] = [OH^-]$$
Because NH<sub>3</sub> is a weak base, our second assumption is
$$[NH_3] >> [NH_4^+]$$
which simplifies the mass balance equation to
$$C_{\text{NH}_3} = 0.05... | {
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This assumes, of course, that we can ignore the contributions of $Hg_2^{2+}$ and $Cl^-$ from the solubility of $Hg_2Cl_2$ .
Next we use equation 6.63 to calculate the activity coefficients for $Hg_2^{2+}$ and $Cl^-$ .
$$\log \gamma_{\text{Hg}_{2}^{2+}} = \frac{-0.51 \times (+2)^{2} \times \sqrt{0.10}}{1 + 3... | {
"Header 1": "Equilibrium Chemistry",
"Header 3": "Chapter Overview",
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+ error = Ag + AgNH3 + NH4 + H3O-OH-I
+ output = data.frame(pI, error)
+ print(output)
+ }
```
The function accepts an initial guess for pI and calculates the concentrations of each species in solution using the definition of pI to calculate [I<sup>-</sup>], using the $K_{\rm sp}$ to obtain [Ag<sup>+</sup>], usi... | {
"Header 1": "Equilibrium Chemistry",
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- 7A [The Importance of Sampling](#page-295-1)
- 7B [Designing a Sampling Plan](#page-298-1)
- 7C [Implementing the Sampling Plan](#page-310-1)
- 7D [Separating The Analyte From Interferents](#page-320-1)
- 7E [General Theory of Separation Efficiency](#page-320-2)
- 7F [Classifying Separation Techniques](#page-323-1)
-... | {
"Header 1": "Collecting and Preparing Samples",
"Header 3": "Chapter Overview",
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#### Example 7.2
The following data were collected as part of a study to determine the effect of sampling variance on the analysis of drug-animal feed formulations.<sup>2</sup>
| <u>%</u> | Drug (w/v | w) | % Drug (w/w) | | | | |
|----------|-----------|--------|--------------|--------|-------... | {
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If we proceed by moving down the third column, then the 10 samples are as follows:
| Sample | Number | Row | Column | Sample | Number | Row | Column |
|--------|--------|-----|--------|--------|--------|-----|--------|
| 1 | 76831 | 68 | 31 | 6 | 41701 | 17 | 01 |
| 2 | 66558 | 65 | 58 ... | {
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In this case cost, expedience, and accessibility are more important than ensuring a random sample.
#### **7B.2 What Type of Sample to Collect**
Having determined from where to collect samples, the next step in designing a sampling plan is to decide on the type of sample to collect. There are three common methods fo... | {
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Predict the percent relative standard deviation and the absolute standard deviation if we collect 5.00-g samples.
#### SOLUTION
To determine the sampling constant, $K_{\rm s}$ , we need to know the average mass of the cereal samples and the relative standard deviation for the amount of ash in those samples. The av... | {
"Header 1": "Collecting and Preparing Samples",
"Header 3": "Chapter Overview",
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#### **Practice Exercise 7.3**
Olaquindox is a synthetic growth promoter in medicated feeds for pigs. In an analysis of a production lot of feed, five samples with nominal masses of 0.95 g were collected and analyzed, with the results shown in the following table.
What is the value of $K_s$ and what size samples ... | {
"Header 1": "When we use equation 7.7, we must express the standard deviation for sampling, $s_{samp}$ , and the error, e, in the same way. If $s_{samp}$ is reported as a percent relative standard deviation, then the error, e, is reported as a percent relative error. When you use equation 7.7, be sure to check that... |
Propose a sampling strategy that provides a maximum relative error of $\pm 0.50\%$ ( $\alpha = 0.05$ ) and a maximum cost of \$700.
Click here to review your answer to this exercise.
#### **7C Implementing the Sampling Plan**
Implementing a sampling plan usually involves three steps: physically removing the sam... | {
"Header 1": "When we use equation 7.7, we must express the standard deviation for sampling, $s_{samp}$ , and the error, e, in the same way. If $s_{samp}$ is reported as a percent relative standard deviation, then the error, e, is reported as a percent relative error. When you use equation 7.7, be sure to check that... |
Disadvantages include the tendency for some analytes to adsorb to the container's walls, the presence of analytes at concentrations too low to detect with suitable accuracy and precision, and the presence of reactive analytes, such as ozone and nitrogen oxides, that may react with the container or that may otherwise al... | {
"Header 1": "When we use equation 7.7, we must express the standard deviation for sampling, $s_{samp}$ , and the error, e, in the same way. If $s_{samp}$ is reported as a percent relative standard deviation, then the error, e, is reported as a percent relative error. When you use equation 7.7, be sure to check that... |
For example, a Ni-bearing ore with an average particle size of 5 mm may require a sample that weighs one ton to obtain a reasonable *ssamp*. Reducing the sample's average particle size allows us to collect the same number of particles with a smaller, more manageable mass. Second, many analytical techniques require that... | {
"Header 1": "When we use equation 7.7, we must express the standard deviation for sampling, $s_{samp}$ , and the error, e, in the same way. If $s_{samp}$ is reported as a percent relative standard deviation, then the error, e, is reported as a percent relative error. When you use equation 7.7, be sure to check that... |
| Table 7.3 | Melting | | Common Fluxes for Decomposing Inorganic Samples |
|-----------|----------------------|---------------|-------------------------------------------------|
| Flux | Temperature (o<br>C) | Crucible | Typical Samples |
| Na2CO3 ... | {
"Header 1": "When we use equation 7.7, we must express the standard deviation for sampling, $s_{samp}$ , and the error, e, in the same way. If $s_{samp}$ is reported as a percent relative standard deviation, then the error, e, is reported as a percent relative error. When you use equation 7.7, be sure to check that... |
$$S_{samp}^* = k_A(C_A)_{\circ} 7.17$$
Substituting equation 7.12 and equation 7.17 into equation 7.16, and rearranging
$$E = \frac{k_A(C_A + K_{A,I} \times C_I) - k_A(C_A)_{\circ}}{k_A(C_A)_{\circ}}$$
$$E = \frac{C_A + K_{A,I} \times C_I - (C_A)_{\circ}}{(C_A)_{\circ}}$$
$$E = \frac{C_A}{(C_A)_{\circ}} - \fr... | {
"Header 1": "When we use equation 7.7, we must express the standard deviation for sampling, $s_{samp}$ , and the error, e, in the same way. If $s_{samp}$ is reported as a percent relative standard deviation, then the error, e, is reported as a percent relative error. When you use equation 7.7, be sure to check that... |
In the photo on the left, the filter is attached to a syringe. Samples are placed in the syringe and pushed through the filter. The filtrate is collected in a test tube or other suitable container. (c) A disposable filter system with a 0.22 µm cellulose acetate membrane filter. The sample is added to the upper unit and... | {
"Header 1": "When we use equation 7.7, we must express the standard deviation for sampling, $s_{samp}$ , and the error, e, in the same way. If $s_{samp}$ is reported as a percent relative standard deviation, then the error, e, is reported as a percent relative error. When you use equation 7.7, be sure to check that... |
In fact, the equilibrium constant for the reaction in which EDTA displaces the masking agent
Ni(CN)<sub>4</sub><sup>2-</sup>(aq) + Y<sup>4-</sup>(aq)
$$\Rightarrow$$
NiY<sup>2-</sup>(aq) + 4CN<sup>-</sup>(aq)
$$K = \frac{K_1}{\beta_4} = \frac{4.2 \times 10^{18}}{1.7 \times 10^{30}} = 2.5 \times 10^{-12}$$
is suff... | {
"Header 1": "When we use equation 7.7, we must express the standard deviation for sampling, $s_{samp}$ , and the error, e, in the same way. If $s_{samp}$ is reported as a percent relative standard deviation, then the error, e, is reported as a percent relative error. When you use equation 7.7, be sure to check that... |
Those metals that form amphoteric hydroxides, however, do not precipitate because they react to form higher-order hydroxo-complexes.
| Table 7.7 | Examples of Using a Chemical Reaction and a Distillation to Separate an Inorganic Analyte From Interferents | |
|-----------------|-----------------... | {
"Header 1": "When we use equation 7.7, we must express the standard deviation for sampling, $s_{samp}$ , and the error, e, in the same way. If $s_{samp}$ is reported as a percent relative standard deviation, then the error, e, is reported as a percent relative error. When you use equation 7.7, be sure to check that... |
<sup>15</sup> Jeannot, M. A.; Cantwell, F. F. *Anal. Chem.* **1997**, *69*, 235–239.
<sup>16</sup> Alltech Associates *Extract-Clean SPE Sample Preparation Guide*, Bulletin 83.
**Figure 7.22** Selection of solid phase extraction cartridges for liquid samples. The solid adsorbent is... | {
"Header 1": "When we use equation 7.7, we must express the standard deviation for sampling, $s_{samp}$ , and the error, e, in the same way. If $s_{samp}$ is reported as a percent relative standard deviation, then the error, e, is reported as a percent relative error. When you use equation 7.7, be sure to check that... |
Before their analysis by gas chromatography, trihalomethanes are separated from their aqueous matrix using a liquid–liquid extraction with pentane.<sup>20</sup>
In a simple liquid—liquid extraction the solute partitions itself between two immiscible phases. One phase usually is an aqueous solvent and the other phase ... | {
"Header 1": "When we use equation 7.7, we must express the standard deviation for sampling, $s_{samp}$ , and the error, e, in the same way. If $s_{samp}$ is reported as a percent relative standard deviation, then the error, e, is reported as a percent relative error. When you use equation 7.7, be sure to check that... |
Solving equation 7.26 for $V_{org}$ , and making appropriate substitutions for $(q_{aq})_1$ and $V_{aq}$ gives
$$V_{\text{org}} = \frac{V_{\text{aq}} - (q_{\text{aq}})_1 V_{\text{aq}}}{(q_{\text{aq}})_1 D} = \frac{50.00 \text{ mL} - (0.001) (50.00 \text{ mL})}{(0.001) (5.00 \text{ mL})} = 9990 \text{ mL}$$
In ... | {
"Header 1": "When we use equation 7.7, we must express the standard deviation for sampling, $s_{samp}$ , and the error, e, in the same way. If $s_{samp}$ is reported as a percent relative standard deviation, then the error, e, is reported as a percent relative error. When you use equation 7.7, be sure to check that... |
$$K_{a} = \frac{[H_{3}O_{aq}^{+}][A_{aq}^{-}]}{[HA_{aq}]}$$
7.30
Solving equation 7.30 for the concentration of A<sup>-</sup> in the aqueous phase
$$[\mathbf{A}_{aq}^{-}] = \frac{K_{\mathbf{a}} \times [\mathbf{H} \mathbf{A}_{aq}]}{[\mathbf{H}_{\mathbf{3}} \mathbf{O}_{aq}^{+}]}$$
and substituting into equation 7... | {
"Header 1": "When we use equation 7.7, we must express the standard deviation for sampling, $s_{samp}$ , and the error, e, in the same way. If $s_{samp}$ is reported as a percent relative standard deviation, then the error, e, is reported as a percent relative error. When you use equation 7.7, be sure to check that... |
The fraction of metal ion that remains in the aqueous phase is
$$(Q_{aq})_1 = \frac{100.0 \text{ mL}}{(0.0438)(10.00 \text{ mL}) + 100.0 \text{mL}} = 0.996$$
At a pH of 1.00, we extract only 0.40% of the metal into the organic phase. Changing the pH to 3.00, however, increases the extraction efficiency to 97.8%. Fi... | {
"Header 1": "When we use equation 7.7, we must express the standard deviation for sampling, $s_{samp}$ , and the error, e, in the same way. If $s_{samp}$ is reported as a percent relative standard deviation, then the error, e, is reported as a percent relative error. When you use equation 7.7, be sure to check that... |
To analyze a shipment of 100 barrels of an organic solvent, you plan to collect a single sample from each of 10 barrels selected at random.
<sup>22</sup> Corl, W. E. *Spectroscopy* **1991**, *6(8)*, 40–43.
From which barrels should you collect samples if the first barrel is given by the twelfth entry in the random ... | {
"Header 1": "When we use equation 7.7, we must express the standard deviation for sampling, $s_{samp}$ , and the error, e, in the same way. If $s_{samp}$ is reported as a percent relative standard deviation, then the error, e, is reported as a percent relative error. When you use equation 7.7, be sure to check that... |
| Nominal Mass (g) | Actual Mass (g) | % w/w KH2PO4 |
|------------------|-----------------|--------------|
| 0.25 | 0.2515 | 0.847 |
| | 0.2465 | 0.598 |
| | 0.2770 | 0.431 |
| | 0.2460 | 0.842 ... | {
"Header 1": "When we use equation 7.7, we must express the standard deviation for sampling, $s_{samp}$ , and the error, e, in the same way. If $s_{samp}$ is reported as a percent relative standard deviation, then the error, e, is reported as a percent relative error. When you use equation 7.7, be sure to check that... |
To evaluate the procedure's accuracy, a standard sample of ore known to have a Co/Fe ratio of 10.2 is analyzed. When pure samples of Co and Fe are taken through the procedure the following calibration relationships are obtained
$$S_{\text{Co}} = 0.786 \times m_{\text{Co}}$$
and $S_{\text{Fe}} = 0.699 \times m_{\text... | {
"Header 1": "When we use equation 7.7, we must express the standard deviation for sampling, $s_{samp}$ , and the error, e, in the same way. If $s_{samp}$ is reported as a percent relative standard deviation, then the error, e, is reported as a percent relative error. When you use equation 7.7, be sure to check that... |
**Figure 7.32** Extraction scheme for Problem 7.30.
**Figure 7.33** Extraction scheme for Problem 7.31.
| pН | Hg <sup>2+</sup> | Pb <sup>2+</sup> | $Zn^{2+}$ |
|----|------------------|------------------|-----------|
| 1 | 3.3 | 0.0 ... | {
"Header 1": "When we use equation 7.7, we must express the standard deviation for sampling, $s_{samp}$ , and the error, e, in the same way. If $s_{samp}$ is reported as a percent relative standard deviation, then the error, e, is reported as a percent relative error. When you use equation 7.7, be sure to check that... |
For this scenario, the percent relative error is
$$e = t\sqrt{\frac{s_{samp}^2}{n_{samp}} + \frac{s_{meth}^2}{n_{samp}n_{rep}}} = 2.179\sqrt{\frac{0.10}{1} + \frac{0.20}{1 \times 13}} = 0.74$$
where t(0.05, 12) is 2.179. Because this percent relative error is larger than $\pm 0.50\%$ , this is not a suitable sampl... | {
"Header 1": "When we use equation 7.7, we must express the standard deviation for sampling, $s_{samp}$ , and the error, e, in the same way. If $s_{samp}$ is reported as a percent relative standard deviation, then the error, e, is reported as a percent relative error. When you use equation 7.7, be sure to check that... |
- 8A [Overview of Gravimetric Methods](#page-361-1)
- 8B [Precipitation Gravimetry](#page-363-1)
- 8C [Volatilization Gravimetry](#page-385-1)
- 8D [Particulate Gravimetry](#page-394-1)
- 8E [Key Terms](#page-398-1)
- 8F [Chapter Summary](#page-398-2)
- 8G [Problems](#page-398-3)
- 8H [Solutions to Practice Exercises](... | {
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See Chapter 11 on electrochemical methods of analysis for a further discussion of electrogravimetry.
We will return to this concept of applying a conservation of mass later in the chapter when we consider specific examples of gravimetric methods. Other examples of definitive techniques are coulometry and isotope-dilu... | {
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<u>Problem 4</u> in the end-of-chapter problems asks you to show that equation 8.11 is correct by completing the derivation.
chloride is important if we wish to determine the concentration of silver by precipitating AgCl. In particular, we must avoid a large excess of chloride.
Another important parameter that ma... | {
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"Header 3": "Chapter Overview",
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**Figure 8.3** Ball-and-stick model showing the lattice structure of AgCl. Each silver ion in the lattice's interior binds with six chloride ions and each chloride ion in the interior binds with six silver ions. Those ions on the lattice's surface or edges bind to fewer than six ions and carry a partial charge. A silve... | {
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One solution to the latter problem is to generate the precipitant *in situ* as the product of a slow chemical reaction, which effectively maintains a constant *RSS*. Because the precipitate forms under conditions of low *RSS*, initial nucleation produces a small number of particles. As additional precipitant forms, par... | {
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Heating the solution and the precipitate provides a third way to induce coagulation. As the temperature increases, the number of ions in the primary adsorption layer decreases, which lowers the precipitate's surface charge. In addition, heating increases the particles' kinetic energy, allowing them to overcome the el... | {
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With the publication of the 20th Edition in 1998, this method is no longer listed as an approved method.
pletely removes these volatile species. For example, one standard gravimetric method for the determination of magnesium involves its precipitation as $MgNH_4PO_4 \cdot 6H_2O$ . Unfortunately, this precipitate is ... | {
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<u>Table 8.3</u> lists examples of several common organic precipitants.
Precipitation gravimetry continues to be listed as a standard method for the determination of $SO_4^{2^-}$ in water and wastewater analysis.<sup>8</sup> Precipitation is carried out using $BaCl_2$ in an acidic solution (adjusted with HCl to a... | {
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| Table 8.2 | eactions for the Homogeneous Preparation of Selected aorganic Precipitants | | | | |
|-------------------------------|--------------------------------------------------------------------------------|--|--|--|--|
| Precipitan | t Reaction ... | {
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But what if we cannot separately precipitate the two analytes? To find the concentrations of both
#### **Practice Exercise 8.2**
A 0.7336-g sample of an alloy that contains copper and zinc is dissolved in 8 M HCl and diluted to 100 mL in a volumetric flask. In one analysis, the zinc in a 25.00-mL portion of the sol... | {
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For example, a conservation of electrons requires that the electrons released by Na<sub>3</sub>PO<sub>3</sub> end up in the product, Hg<sub>2</sub>Cl<sub>2</sub>, yielding the following stoichiometric conversion factor:
$$\frac{2 \text{ mol } Na_3 PO_3}{\text{mol } Hg_2Cl_2}$$
This conversion factor provides a dire... | {
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A less obvious way to improve a method's sensitivity is indicated by the term of 1/2 in equation 8.14, which accounts for the stoichiometry between the analyte and precipitate. We can also improve sensitivity by forming a precipitate that contains fewer units of the analyte.
#### Practice Exercise 8.5
Suppose you w... | {
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The balance pan and the thermocouple are then lowered into the furnace.
nace. The sample's mass and the mass of the residue are measured using an analytical balance.
Trapping and weighing the volatile products of a thermal decomposition requires specialized equipment. The sample is placed in a closed container and ... | {
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Volatilization gravimetry also is used to determine biomass in waters and wastewaters. Biomass is a water quality index that provides an indication of the total mass of organisms contained within a sample of water. A known volume of the sample is passed through a preweighed 0.45-µm membrane filter or a glass-fiber fi... | {
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#### SOLUTION
The change in the sample's mass is 5.25 mg, which corresponds to
$$5.25 \text{ mg lost} \times \frac{100.0 \text{ mg MgC}_2 \text{O}_4 \cdot \text{H}_2 \text{O}}{69.08 \text{ mg lost}} = 7.60 \text{ mg MgC}_2 \text{O}_4 \cdot \text{H}_2 \text{O}$$
The %w/w MgC<sub>2</sub>O<sub>4</sub>•H<sub>2</sub... | {
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Particulate gravimetry is important in the environmental analysis of water, air, and soil samples. The analysis for suspended solids in water samples, for example, is accomplished by filtering an appropriate volume of a wellmixed sample through a glass fiber filter and drying the filter to constant weight at 103–105<su... | {
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What is the optimum pH for the quantitative precipitation of $Zn(OH)_2$ ? For your solubility diagram, plot log(S) on the y-axis and pH on the x-axis. See the appendices for relevant equilibrium constants.
- 4. Starting with equation 8.10, verify that equation 8.11 is correct.
- 5. For each of the following precipitat... | {
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Sinha and Shome described a gravimetric method for molybdenum in which it is precipitated as MoO<sub>2</sub>(C<sub>13</sub>H<sub>10</sub>NO<sub>2</sub>)<sub>2</sub> using *n*-benzoylphenylhydroxylamine, C<sub>13</sub>H<sub>11</sub>NO<sub>2</sub>, as the precipitant.<sup>12</sup> The precipitate is weighed after ignitin... | {
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While working as a laboratory assistant you prepared 0.4 M solutions of AgNO<sub>3</sub>, Pb(NO<sub>3</sub>)<sub>2</sub>, BaCl<sub>2</sub>, KI and Na<sub>2</sub>SO<sub>4</sub>. Unfortunately, you became distracted and forgot to label the solutions before leaving the
laboratory. Realizing your error, you label the sol... | {
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In the presence of water vapor the surface of zirconia, ZrO<sub>2</sub>, chemically adsorbs H<sub>2</sub>O, forming surface hydroxyls, ZrOH (additional water is physically adsorbed as H<sub>2</sub>O). When heated above 200 °C, the surface hydroxyls convert to H<sub>2</sub>O(g), releasing one molecule of water for every... | {
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| [thiourea] (M) | $\Delta f(Hz)$ | [thiourea] (M) | $\Delta f(Hz)$ | |
|-----------------------|----------------|-----------------------|----------------|--|
| $3.00\times10^{-7}$ | 74.6 | $1.50 \times 10^{-6}$ | 327 | |
| $5.00 \times 10^{-7}$ | 120 | $2.50 \times 1... | {
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#### **Practice Exercise 8.4**
To find the mass of $(NH_4)_3PO_4 \cdot 12MoO_3$ that will produce 0.600 g of PbMoO<sub>3</sub>, we first use a conservation of mass for molybdenum; thus
$$0.600 \text{ g PbMoO}_{3} \times \frac{1 \text{ mol Mo}}{351.2 \text{ g PbMoO}_{3}} \times \frac{1876.59 \text{ g (NH}_{4})_{... | {
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Adapting this equation to a sample that contains $CaC_2O_4$ , $MgC_2O_4$ , and inert materials is easy; thus
$$\begin{split} \text{g CaC}_2\text{O}_4 &= (0.1794\,\text{g} - 0.1294\,\text{g}) \times \\ &\frac{1\,\text{mol CO}_2}{44.01\,\text{g CO}_2} \times \frac{128.10\,\text{g CaC}_2\text{O}_4}{\text{mol CO}_2} = ... | {
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- 9A [Overview of Titrimetry](#page-415-1)
- 9B [Acid–Base Titrations](#page-419-1)
- 9C [Complexation Titrations](#page-460-1)
- 9D [Redox Titrations](#page-478-1)
- 9E [Precipitation Titrations](#page-501-1)
- 9F [Key Terms](#page-509-1)
- 9G [Chapter Summary](#page-509-2)
- 9H [Problems](#page-510-1)
- 9I [Solutions... | {
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Because the titration of $Mg^{2+}$ with EDTA
$$Mg^{2+}(aq) + Y^{4-}(aq) \Rightarrow MgY^{2-}(aq)$$
Depending on how we are detecting the endpoint, we may stop the titration too early or too late. If the end point is a function of the titrant's concentration, then adding the titrant too quickly leads to an early e... | {
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**9B Acid–Base Titrations**
Before 1800, most acid–base titrations used H2SO4, HCl, or HNO3 as acidic titrants, and K2CO3 or Na2CO3 as basic titrants. A titration's end
| Table 9.1<br>Volume (mL) | Class | Specifications for Volumetric Burets<br>Subdivision (mL) | Tolerance (mL) |
|--------------------------|------... | {
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For example, after adding 30.0 mL of titrant the concentration of OH<sup>-</sup> is
$$[OH^{-}] = \frac{(\text{mol NaOH})_{\text{added}} - (\text{mol HCl})_{\text{initial}}}{\text{total volume}} = \frac{M_b V_b - M_a V_a}{V_a + V_b}$$
$$[OH^{-}] = \frac{(0.200 \text{ M}) (30.0 \text{ mL}) - (0.100 \text{ M}) (50.0 \te... | {
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Step 2: Before adding the titrant, the pH is determined by the titrand, which in this case is a weak acid.
Because the equilibrium constant for reaction 9.2 is quite large
$$K = K_a/K_w = 1.75 \times 10^9$$
we can treat the reaction as if it goes to completion.
Step 3: Before the equivalence point, the pH is de... | {
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Alternatively, we can calculate acetate's concentration using the initial moles of acetic acid; thus
$$\begin{split} [CH_3COO^-] &= \frac{(mol\ CH_3COOH)_{initial}}{total\ volume} \\ &= \frac{(0.100\ M)\,(50.0\ mL)}{50.0\ mL + 25.0\ mL} \\ &= 0.0667\ M \end{split}$$
Step 5: Calculate pH values after the equivalen... | {
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#### Practice Exercise 9.3
Sketch a titration curve for the titration of 25.0 mL of 0.125 M NH<sub>3</sub> with 0.0625 M HCl and compare to the result from <u>Practice Exercise</u> 9.2.
Click <u>here</u> to review your answer to this exercise.
**Figure 9.8** Illustrations showing ... | {
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We can use this change in color to indicate the end point of a titration provided that it occurs at or near the titration's equivalence point.
As an example, let's consider an indicator for which the acid form, HIn, is yellow and the base form, In<sup>-</sup>, is red. The color of the indicator's solution
The same ... | {
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The simplest method for finding the end point is to locate the titration curve's inflection point, which is shown by the arrow. This is also the least accurate method, particularly if the titration curve has a shallow slope at the equivalence point.
Another method for locating the end point is to plot the first deriv... | {
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As shown in Figure 9.15, however, a thermometric titration curve usually shows curvature near the equivalence point due to an incomplete neutralization reaction or to the excessive dilution of the titrand and the titrant during the titration. The latter problem is minimized by using a titrant that is 10–100 times more ... | {
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The description here is based on Method 13.86 as published in *Official Methods of Analysis*, 8th Ed., Association of Official Agricultural Chemists: Washington, D. C., 1955.
All other things being equal, the strength of a weak acid increases if we place it in a solvent that is more basic than water, and the strength... | {
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If the end point pH is between 6 and 10, however, the neutralization of $CO_3^{2-}$ requires one proton
$$CO_3^{2-}(aq) + H_3O^+(aq) \longrightarrow H_2O(l) + HCO_3^-(aq)$$
and the net reaction between $CO_2$ and $OH^-$ is
$$CO_2(aq) + OH^-(aq) \longrightarrow HCO_3^-(aq)$$
Under these conditions some ... | {
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Acid–base titrimetry continues to be listed as a standard method for the determination of alkalinity, acidity, and free $CO_2$ in waters and wastewaters. Alkalinity is a measure of a sample's capacity to neutralize acids. The most important sources of alkalinity are $OH^-$ , $HCO_3^-$ , and $CO_3^{2-}$ , althoug... | {
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Including a reducing agent, such as salicylic acid, converts this nitrogen to a –3 oxidation state, eliminating this source of error. Table 9.7 provides additional examples in which an element is converted quantitatively into a titratable acid or base.
4 Ferek, R. J.; Lazrus, A. L.; Haagenson, P. L.; Winchester, J. W... | {
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Because phenolphthalein's end point pH is 8.3–10.0 (see <u>Table 9.4</u>), the titration must proceed to the third equivalence point and the titration reaction is
$$C_6H_8O_7(aq) + 3OH^-(aq) \longrightarrow C_6H_5O_7^{3-}(aq) + 3H_2O(b)$$
To reach the equivalence point, each mole of citric acid consumes three moles... | {
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#### SOLUTION
The HCl in the collection flask reacts with two bases
$$HCl(aq) + NH_3(aq) \longrightarrow NH_4^+(aq) + Cl^-(aq)$$
$$HCl(aq) + OH^{-}(aq) \longrightarrow H_2O(l) + Cl^{-}(aq)$$
The collection flask originally contains
$$(0.1047 \text{ M HCl}) (0.05000 \text{ L HCl}) = 5.235 \times 10^{-3} \tex... | {
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As outlined in Table 9.9, each species or mixture of species has a unique relationship between the volumes of titrant needed to reach these two end points. Note that mixtures containing three or more these species are not possible.
Use a ladder diagram to convince yourself that mixtures containing three or more of th... | {
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Consider, for example, a solution of acetic acid, CH<sub>3</sub>COOH, for which the dissociation constant is
$$K_{\rm a} = \frac{[{\rm H_3O^+}][{\rm CH_3COO^-}]}{[{\rm CH_3COOH}]}$$
When the concentrations of CH<sub>3</sub>COOH and CH<sub>3</sub>COO<sup>-</sup> are equal, the $K_a$ expression reduces to $K_a = [... | {
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The microburet has a 1-2 µm tip filled with an agar gel membrane. The tip of the microburet is placed within a drop of the sample solution, which is suspended in heptane, and the titrant is allowed to diffuse into the sample. The titration's progress is monitored using an acid–base indicator and the time needed to reac... | {
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In 1945, Schwarzenbach introduced aminocarboxylic acids as multidentate ligands. The most widely used of these new ligands—ethylenediaminetetraacetic acid, or EDTA—forms a strong 1:1 complex with many metal ions. The availability of a ligand that gives a single, easily identified end point made complexation titrimetr... | {
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Before adding EDTA, the mass balance on $Cd^{2+}$ , $C_{Cd}$ , is
$$C_{Cd} = [Cd^{2+}] + [Cd(NH_3)^{2+}] + [Cd(NH_3)^{2+}] + [Cd(NH_3)^{2+}] + [Cd(NH_3)^{2+}] + [Cd(NH_3)^{2+}]$$
and the fraction of uncomplexed $Cd^{2+}$ , $\alpha_{Cd}^{2+}$ , is
$$\alpha_{\rm Cd^{2+}} = \frac{\rm [Cd^{2+}]}{C_{\rm Cd}}$$
9.1... | {
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In addition, EDTA will compete with NH<sub>3</sub> for the Cd<sup>2+</sup>. To evaluate the titration curve, therefore, we first need to calculate the conditional formation constant for CdY<sup>2–</sup>. From Table 9.10 and Table 9.12 we find that $\alpha_{\rm Y}$ 4– is 0.367 at a pH of 10, and that $\alpha_{\rm Cd}$... | {
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$$K_{\rm f}'' = \frac{[\text{CdY}^{2-}]}{C_{\rm Cd} C_{\rm EDTA}} = \frac{3.33 \times 10^{-3} - x}{(x)(x)} = 9.5 \times 10^{14}$$
$$x = C_{\rm Cd} = 1.87 \times 10^{-9} \,\text{M}$$
Once again, to find the concentration of uncomplexed Cd<sup>2+</sup> we must account for the presence of NH<sub>3</sub>; thus
$$[Cd^... | {
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**Figure 9.29** Illustrations showing the steps in sketching an approximate titration curve for the titration of 50.0 mL of $5.00 \times 10^{-3} \text{ M Cd}^{2+}$ with 0.0100 M EDTA in the presence of 0.0100 M NH<sub>3</sub>: (a) locating the equivalence point volume; (b) plotting tw... | {
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Even if a suitable indicator does not exist, it often is possible to complete an EDTA titration by introducing a small amount of a secondary metal–EDTA complex if the secondary metal ion forms a stronger complex with the indicator and a
<u>Figure 9.30</u> is essentially a two-variable ladder diagram. The solid lines ... | {
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Because of calmagite's acid–base properties, the range of pMg values over which the indicator changes color depends on the titrand's pH (<u>Figure 9.30</u>). Figure 9.33 shows the titration curve for a 50-mL solution of 10<sup>-3</sup> M Mg<sup>2+</sup> with 10<sup>-2</sup> M EDTA at pHs of 9, 10, and 11. Superimposed ... | {
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$$8.135 \times 10^{-3} \text{ M} \times 0.05000 \text{ L} = 4.068 \times 10^{-4} \text{ mol Ca}^{2+}$$
which means that $4.068 \times 10^{-4}$ moles of EDTA are used in the titration. The molarity of EDTA in the titrant is
$$\frac{4.068 \times 10^{-4} \text{ mol EDTA}}{0.04263 \text{ L}} = 9.543 \times 10^{-3} ... | {
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$$\frac{1.524 \times 10^{-3} \text{ mol Ni}}{50.00 \text{ mL}} \times 250.0 \text{ mL} \times \frac{58.69 \text{ g Ni}}{\text{mol Ni}} = 0.4472 \text{ g Ni}$$
$$\frac{0.4472 \text{ g Ni}}{0.7176 \text{ g sample}} \times 100 = 62.32\% \text{ w/w Ni}$$
$$\frac{5.42 \times 10^{-4} \text{ mol Fe}}{50.00 \text{ mL}} \... | {
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$$E_{\rm rxn} = E_{B_{\rm ox}B_{\rm red}} - E_{A_{\rm ox}/A_{\rm red}}$$
After each addition of titrant the reaction between the titrand and the titrant reaches a state of equilibrium. Because the potential at equilibrium is zero, the titrand's and the titrant's reduction potentials are identical.
$$E_{B_{ox}/B_{... | {
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| Table 9.15 Data for the Titration of 50.0 mL of 0.100 M $Fe^{2+}$ with 0.100 M $Ce^{4+}$ | | | |
|------------------------------------------------------------------------------------------|--------------|------------------------------|--------------|
| Volume o... | {
"Header 1": "Titrimetric Methods",
"Header 3": "Step 1: Calculate the volume of titrant needed to reach the equivalence point.",
"token_count": 1908,
"source_pdf": "datasets/websources/biochem/clairvoyance.ipynb.pdf"
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**Figure 9.37** Illustrations showing the steps in sketching an approximate titration curve for the titration of 50.0 mL of 0.100 M Fe<sup>2+</sup> with 0.100 M Ce<sup>4+</sup> in 1 M HClO<sub>4</sub>: (a) locating the equivalence point volume; (b) plotting two points before the equival... | {
"Header 1": "Titrimetric Methods",
"Header 3": "Step 1: Calculate the volume of titrant needed to reach the equivalence point.",
"token_count": 2015,
"source_pdf": "datasets/websources/biochem/clairvoyance.ipynb.pdf"
} |
**Figure 9.38** Titration curve for the titration of 50.0 mL of 0.100 M Fe<sup>2+</sup> with 0.0200 M $MnO_4^-$ at a fixed pH of 1 (using $H_2SO_4$ ). The equivalence point is shown by the **red** dot.
$$E_{eq} = \frac{E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} + 5E_{\text{MnO}\bar{... | {
"Header 1": "Titrimetric Methods",
"Header 3": "Step 1: Calculate the volume of titrant needed to reach the equivalence point.",
"token_count": 2002,
"source_pdf": "datasets/websources/biochem/clairvoyance.ipynb.pdf"
} |
#### Representative Method 9.3
#### **Determination of Total Chlorine Residual**
#### DESCRIPTION OF THE METHOD
The chlorination of a public water supply produces several chlorine-containing species, the combined concentration of which is called the total chlorine residual. Chlorine is present in a variety of c... | {
"Header 1": "Titrimetric Methods",
"Header 3": "Step 1: Calculate the volume of titrant needed to reach the equivalence point.",
"token_count": 1607,
"source_pdf": "datasets/websources/biochem/clairvoyance.ipynb.pdf"
} |
The sample is placed at the top of the column and moves through the column under the influence of gravity or vacuum suc-
| Table 9.17 Examples of Reactions For Reducing a Titrand's Oxidation State Using a Reduction Column | | ... | {
"Header 1": "Titrimetric Methods",
"Header 3": "Step 1: Calculate the volume of titrant needed to reach the equivalence point.",
"token_count": 1937,
"source_pdf": "datasets/websources/biochem/clairvoyance.ipynb.pdf"
} |
Diphenyl-
The standardization reactions are
$$Ce^{4+}(aq) + Fe^{2+}(aq) \rightarrow$$
$Fe^{3+}(aq) + Ce^{3+}(aq)$
$2Ce^{4+}(aq) + H_2C_2O_4(aq) \rightarrow$
$2Ce^{3+}(aq) + 2CO_2(g) + 2H^+(aq)$
The standardization reactions are
$$MnO_{4}^{-}(aq) + 5Fe^{3+}(aq) + 8H^{+}(aq) \longrightarrow$$
$$Mn^{2+}(aq) + 5Fe^... | {
"Header 1": "Titrimetric Methods",
"Header 3": "Step 1: Calculate the volume of titrant needed to reach the equivalence point.",
"token_count": 2024,
"source_pdf": "datasets/websources/biochem/clairvoyance.ipynb.pdf"
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Having determined the free chlorine residual in the water sample, a small amount of KI is added, which catalyzes the reduction of monochloramine, NH<sub>2</sub>Cl, and oxidizes a portion of the DPD back to its red-colored form. Titrating the oxidized DPD with ferrous ammonium sulfate yields the amount of NH<sub>2</su... | {
"Header 1": "Titrimetric Methods",
"Header 3": "Step 1: Calculate the volume of titrant needed to reach the equivalence point.",
"token_count": 2040,
"source_pdf": "datasets/websources/biochem/clairvoyance.ipynb.pdf"
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When the oxidation is complete, an excess of KI is added, which converts any unreacted $IO_4^-$ to $IO_3^-$ and $I_3^-$ .
$$IO_{4}^{-}(aq) + 3I^{-}(aq) + H_{2}O(l) \longrightarrow IO_{3}^{-}(aq) + I_{3}^{-}(aq) + 2OH^{-}(aq)$$
The $I_3^-$ is then determined by titrating with $S_2\mathrm{O}_3^{2-}$ using st... | {
"Header 1": "Titrimetric Methods",
"Header 3": "Step 1: Calculate the volume of titrant needed to reach the equivalence point.",
"token_count": 1970,
"source_pdf": "datasets/websources/biochem/clairvoyance.ipynb.pdf"
} |
Finally, because each mole of OCl<sup>-</sup> produces one mole of $I_3^-$ , and each mole of $I_3^-$ reacts with two moles of $S_2O_3^{2-}$ , we know that every mole of NaOCl in the sample ultimately results in the consumption of two moles of $Na_2S_2O_3$ .
The moles of $Na_2S_2O_3$ used to reach the titrat... | {
"Header 1": "Titrimetric Methods",
"Header 3": "Step 1: Calculate the volume of titrant needed to reach the equivalence point.",
"token_count": 1850,
"source_pdf": "datasets/websources/biochem/clairvoyance.ipynb.pdf"
} |
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