leemai_all_chunks.json

["Chapter # 03\n\nChemical Equilibria\n\n \n\nHow do substances behave'in various environments? This requires an understanding of chemical reactions and\nequilibrium. When an additional ion is added, the common ion effect predicts solution reactions. The strength of\nacids and bases can be estimated from their dissociation constants, which differentiate strong acids from weak\nones. Buffer solutions maintain pH equilibrium and regulate blood pH. This you will understand in this chapter.\nThe distribution of solutes among solvents is predicted by the partition coefficient, which is useful in estimating\nthe distribution of drugs within the body. So it is very important for drug design. These concepts play a key role in\nindustrial and environmental science.\n\nStudent Learning Outcomes (SLOs)\n\n> Use the extent of ionization and the acid dissociation constant, Ka, to distinguish between strong and\nweak acids. Ee\n\n \n\nSTRENGE ACIDS AND BASES\nThe degree to which different Bronsted acids give off protons is", "tent of ionization and the acid dissociation constant, Ka, to distinguish between strong and\nweak acids. Ee\n\n \n\nSTRENGE ACIDS AND BASES\nThe degree to which different Bronsted acids give off protons is called \"acid strength\". A relatively strong acid\nis an acid that can give off more protons than another acid. Hydrochloric acid, for example, is relatively strong\ncompared to acetic acid, which is relatively strong relative to water. Bases differ in the degree to which they\naccept protons as well. A strong base can accept protons to a greater extent than another base, so ammonia\nis relatively stronger relative to water because it can accept protons to a higher degree than water.\n\n2.ii Differentiate between strong and weak acids using the extent of ionization and Ka. Teen)\n\n3.ji Compare and contrast strong and weak acids based on their extent of ionization and the value of K,.\n\n> Strong and Weak Acids\n\n\u00a2 The extent of ionization and the acid dissociation constant Ka \u2018can be used to disting", "ntrast strong and weak acids based on their extent of ionization and the value of K,.\n\n> Strong and Weak Acids\n\n\u00a2 The extent of ionization and the acid dissociation constant Ka \u2018can be used to distinguish between strong\nand weak acids.\n\ne The strength of an acid is generally expressed in terms of the acid ionization constant, Ks of the acid.\nConsider the case of ionization of a general acid HX in water. In this aqueous solution, the established\nequilibrium may be represented as follows:\n\nHX+H,O,) = Hj0(,,) +X (aq)\ne The equilibrium constant K for this ionization process may be written as follows:\n[H.0\" JEx\"]\n[Hx][H,0]\n\n[Ho >]\n\nOr k[H,0] [Ax]\n\ne Since water is a solvent, it is present in excess and therefore its concentration may be regarded as\nconstant. Thus, K[H20] is another constant and is designated as Ka.\n\nH,0\u00b0 |[ x\" ]\nK[H20] = Ka= be\n\n\u00a9 studyplusplus.com rs\n\n \n\n63\n\f\ne K,is termed as the acid dissociation constant. It is a measure of the extent to which an acid is ionized or\ndisso", " is designated as Ka.\n\nH,0\u00b0 |[ x\" ]\nK[H20] = Ka= be\n\n\u00a9 studyplusplus.com rs\n\n \n\n63\n\f\ne K,is termed as the acid dissociation constant. It is a measure of the extent to which an acid is ionized or\ndissociated at the equilibrium state. It must be kept in mind that the acid dissociation <onstant. \u2018ss, 1s\ndependent on temperature. Therefore, the value of K, should be mentioned along with the temperature\nat which K, was determined. The dissociation constant, Ks, of acetic acid in water at 25\u00b0C is 1.8 x 10\u00b05.\nThe comparison of Ka Values of different acids provides a method to compare their strengths.\n\ne \"The greater the value of Ks, the stronger the acid\u201d.\n\ne The values of K, are usually inconvenient numbers, therefore, for convenience, these values are\nconverted to pK, values. The relationship between Ka, and pK; is as follows:\n\n_ PK, = \u2014log Ka\n\ne Since pK, refers to the negative logarithm of K,, the lower the pK, value, the stronger the acid, since a\nlower pK, value corresponds to a higher ", "etween Ka, and pK; is as follows:\n\n_ PK, = \u2014log Ka\n\ne Since pK, refers to the negative logarithm of K,, the lower the pK, value, the stronger the acid, since a\nlower pK, value corresponds to a higher K, value. Table Jists the ionization constants and pK; values of\nsome common acids in water at 25\u00b0C. Which acid is the strongest? Which acid is the weakest?\n\nTable: lonisation constants and pK, of Acids\n\n    \n\nName of Acid \u2014 Formula\n\nPerchloric acid | HCtO, | 1.0x10% | -10.0 |\nHydroiodic acid | HI | 1.0x10\" | -10.0 |\nHydrobromic acid | HBr =|. 1.0x10\u00b0 | -9.0 |\nHydrochloric acid | Hct | 1.0x10\u00b0 |-6.0 |\nSulphuric acid | HxSOs | 1.0x10? |.-3.0 |\n\n| HF 7.\n\np HCOOH |\n\n       \n          \n\nHydrofluoric acid\nFormic acid HCOOH | 1.8x 10~\nBenzoic acid | CcHsCOOH | 6.3x 10\u00b0 | +4.2 |\n\ne Which acid is stronger HCl or HF?\n\n\u2018Example 3.1 (\n\nCalculate the concentration of H* ions of a solution that contains\n1.0 M HF (K, = 7.2 x 10%)\n\n  \n\n \n\nSolution:\nHF (aq) = = H%faq) + Fo(ea)\nInitial conc. 1.0M 0 0\nEq. c", "nger HCl or HF?\n\n\u2018Example 3.1 (\n\nCalculate the concentration of H* ions of a solution that contains\n1.0 M HF (K, = 7.2 x 10%)\n\n  \n\n \n\nSolution:\nHF (aq) = = H%faq) + Fo(ea)\nInitial conc. 1.0M 0 0\nEq. conc.(mole dm=3) 1.0- x x xX\n[Ht |[F\nHF\n7ax104 \u00ab\n1.0-x\n\nSince x is very small as compared to 1.0, the term in the denominator can be approximated as follows:\n1.0-x = 1.0\n\nK, =\n\n7.2 x 104\n\n1\nx = 0.0268M\n{H*] = 0.0268M\n\n\u00a9 __ studyplusplus.com o \u00b0\n\f\n    \n \n\n   \n\nStudent Learning Outcomes (SLOs)\n\n> ___Use the extent of ionization and the base dissociation constant, Kp, to distinguish between strong and\nweak bases.\n\n2.lii Differentiate between strong and weak bases using the extent of ionization and Kp. (Cael\n\n3.iii \u201cElaborate on the factors that distinguish strong and weak bases, incorporating the concept of ionization\nand the base dissociation constant (Ks).\n\n> Strong and Weak Bases\n\ne The strength of a base is the ability to accept a proton from a solvent. Hydroxides of alkali metals such as\n", "concept of ionization\nand the base dissociation constant (Ks).\n\n> Strong and Weak Bases\n\ne The strength of a base is the ability to accept a proton from a solvent. Hydroxides of alkali metals such as\nsodium hydroxide and potassium hydroxide are strong bases and ionize completely in aqueous solution.\n\n+\n\nNaOH.) > Nay a) * OH.)\n\nKOH,,.) \u2014>Kj,,) + OH(,.)\n\ne The OH\u201cion thus formed is a Bronsted base because it can accept proton H\u2019.\nOH\" + H* > H20\ne The ability of a base to accept a proton from an acid, usually water, is termed as the strength of the\nbase. For a base B, an equilibrium reaction with water can be represented by the following equation:\nB+H,0(,.) ==BH/,.) +OHi.)\n\n   \n \n\ne The equilibrium constant Ky is called the base ionization constant and can be derived. As;\n\nos [BH ][or ]\n\n[8]\ne Ky value will be large if the degree of ionization of\nthe base B is high i.e. if the base B is strong. The\n\nValue of Ky. will be small for a weak base B. Again, [NameofBase | Formula | K\u00bb | pK\u2019 |\n\nf", "e Ky value will be large if the degree of ionization of\nthe base B is high i.e. if the base B is strong. The\n\nValue of Ky. will be small for a weak base B. Again, [NameofBase | Formula | K\u00bb | pK\u2019 |\n\nfor convenience, a parameter pK\u00bb has been devised | Diethlyamine | (CzHs)2NH_ | 9.6x10~ | 3.02 |\n\nto express the Ky value in convenient numbers. | Ethylamine | (CaHs)2NH2 | 5.6 x 10~* | 3.25 |\nThus, pK, is defined as the negative logarithm of Ky. | | Methylamine | CHsNH2 \"| 4.5x10~* | 3.34 |\npk, = -logks | [Ammonia | Nis 17x20 | 476,\n\n* According to these values, ammonia is a stronger | [Pyridine | CsHsN _| 5.6x 107 | 8.25 |\nbase than pyridine and aniline but weaker than TAniline | CeHsNH2__| 4.3x 107\u00b0 | 9.37 |\n\nmethylamine and ethylamine. Also, diethyl amine is\nthe strongest base among all those listed in Table.\n\nTable: Ky and pK\u00bb Values of Some Common Bases\n\n \n\nStudent Learning Outcomes (SLOs)\nExplain what is meant by a chemical buffer and how a buffer system works. (For context this shou", "se listed in Table.\n\nTable: Ky and pK\u00bb Values of Some Common Bases\n\n \n\nStudent Learning Outcomes (SLOs)\nExplain what is meant by a chemical buffer and how a buffer system works. (For context this should\n\ninclude: a. defining what is a buffer solution. b. Explain how a buffer solution can be made. c.\nexplaining how buffer solutions control pH: use chemical equations in these explanations. d. Describe\n_and explain the uses of buffer solutions, including the role of HCO; in controlling pH in blood.\n\n \n\nBUFFER SOLUTIONS AND THEIR APPLICATIONS\n\n.3.iv_ Provide a comprehensive explanation of buffer solutions, including how they are prepared, how the:\ncontrol pH, and their uses in different scenarios.\n\n2.iv. Define a buffer solution and provide an example of how it can be made.\nA buffer solution is a solution, the pH of which does not change significantly when a small amount of acid or base\nis added to it. Such a solution has a constant pH which does not change keeping it constant for a long t", "s a solution, the pH of which does not change significantly when a small amount of acid or base\nis added to it. Such a solution has a constant pH which does not change keeping it constant for a long time.\n\n \n   \n \n\noS studyplusplus.com e\n\n65\n\f\n> \u00a9 Types of Buffer Solutions:\n\nA buffer solution can be made in two ways:\n\n(1) By mixing a weak acid and.a salt of it with a strong base. Such solutions give acidic buffers with pH less\nthan 7. e.g. CHs3COOH + CHsCOONa.\n\n(2) By mixing a weak base and a salt of it with a strong acid. Such solutions will give basic buffers with a pH\n\nof more than 7. e.g. NH\u00abOH + NH,Cl.\n2.v__How does a buffer solution control pH? Include chemical equations in your explanation. | (Exercise) |\n\n\u00bb Buffer Action:\nConsider a buffer solution of CH;3COOH and CH3COONa. The common ion effect helps us to understand how\nthe buffer will work. CH;COOH being a weak electrolyte undergoes very little dissociation. When CH3COONa,\na strong electralyte is added to the CHiCCO solution", "mon ion effect helps us to understand how\nthe buffer will work. CH;COOH being a weak electrolyte undergoes very little dissociation. When CH3COONa,\na strong electralyte is added to the CHiCCO solution, the dissociation of CHaCOOH is suppressed due to the\ncommon ion effect of CHs;COO\u201d\u201d\nCH,COOH,,) +H,0,,) == CH,COO;,,) +H 0/2)\nCH,COONa,, \u2014 ==CH,COO/,, +Naz,,)\n{i) Suppose we add a few drops of HCL to it. Its H* ions are used up by CH3COO7 ions.\nThus the addition of HCZ will not change the pH of the buffer solution. |\n#,O\nCH,COOH,.) ==CH,COO(,.) +Hi,<)\nHCI 9) \u2014> Hig + Clay\nCH,COO7,,) +H/,,) ==CH,COOH,,.)\n(ii) In the same buffer solution, if a strong base is ern, it is neutralised by the acid.\n\nCH,COOH,, =CH,c00;,, +H;,.)\nNaOH.) \u2014> Naj,,) # OH.)\n\nHag) + OH(oq) == 420)\nThus the addition of NaOH will not change the value of pH. iM\n\n> Calculation of pH of Buffer Solution\n\ne The concentration of conjugate base in the reaction mixture is predominately supplied by the salt which :\n\nis a strong el", "will not change the value of pH. iM\n\n> Calculation of pH of Buffer Solution\n\ne The concentration of conjugate base in the reaction mixture is predominately supplied by the salt which :\n\nis a strong electrolyte. Therefore, assuming the concentration of conjugate base is equal to that of salt\nand the original concentration of acid as equilibrium concentrations, the pH of a buffer can\u2019be calculated.\ne The following example explains the calculations associated with buffer solutions.\n\n(a) Calculate the pH of an acetic acid-sodium acetate buffer solution containing 1.0 moles of each component.\n\n(b) What will be the pH of this solution after the addition of 0.01 mole of hydrochloric acid to 1dm? of the\nsolution? Assume that the volume of the solution remains unchanged with the addition of hydrochloric\nacid. (K, for acetic acid is 1.8 x 10\u00b0).\n\nSoiution:\n\n(a) The pH of the buffer solutions can be calculated by assuming the equilibrium concentration of both the\n\nacid and its conjugate base as th", "loric\nacid. (K, for acetic acid is 1.8 x 10\u00b0).\n\nSoiution:\n\n(a) The pH of the buffer solutions can be calculated by assuming the equilibrium concentration of both the\n\nacid and its conjugate base as the starting concentration.\n\nThus [CH3CO2H] = 1.0M [CH3COO7}=1.0M\n\nFor acetic acid dissociation; :\n\n       \n        \n    \n       \n       \n    \n    \n \n \n\n[cr,coo-][H\" ]\nKs = =1.8x10\"\n\n    \n\n18x10> =\n\nThus, the pH of the buffer solution is 4.745.\n\nSo studyplusplus.com oe\n\nain\n\n66\n\f\n(b) After. HCl addition:\nHCl.) )\n0.01 mole 0.01 mole0.01 mole\n\nInitially, there were 1.0 mole of CH3COOH and 1.0 mole of CH3COO present per dm? of the solution. After\nthe addition of hydrochloric acid 0.01 mole of CH3COO ions is combined with the H* ions formed from the\ndissociation of 0.01 mole of added hydrochloric acid. This can be written as:\n\nCH,COOj,,) +H, \u2014> CH,CO,H,.)\n\nha 1\na +Cr (00)\n\n(eq)\n0.01 mole0.01 mole 0.01 mole\n\nThus, the numbers of moles of acetic acid and acetate ions, after the addition of hydroch", "id. This can be written as:\n\nCH,COOj,,) +H, \u2014> CH,CO,H,.)\n\nha 1\na +Cr (00)\n\n(eq)\n0.01 mole0.01 mole 0.01 mole\n\nThus, the numbers of moles of acetic acid and acetate ions, after the addition of hydrochloric acid are:\nCH,CO,H=(1.0+0.01)mole =1.01 mole\n\nCH,COO\u2122 =(1.0--0.01)mole =0.99 mole \u00b0\nThe equilibrium equation for this new situation can be written as;\nCH,COO\u2122 |[H* }\n[CH,CO,H]\n\nit) = Ka*[CH,CO,H\n\"| CH, -coo\u2122\n. 1.8x107 (1.01\n{H*] & ( )\n0.99\npH = \u2014log(1.83 x 1075)\n= 4.736\n\nNotice there is a slight cones) in pH from 4.745 to 4.736 which is only a difference of 0.009. Thus, a buffer\np small amount.\n\n2.vi. Describe the uses of buffer solutions in various applications. (nes)\n\n> licati r Solutions\nBuffers play a pain role in various everyday applications, helping to maintain stability and prevent abrupt\nchanges in different systems. Here are some important applications of buffers in everyday life.\n\n2.vii_ Explain how HCO;\" plays a role in controlling pH in the blood.\n\n \n\n \n\n1. The buffer sys", "t abrupt\nchanges in different systems. Here are some important applications of buffers in everyday life.\n\n2.vii_ Explain how HCO;\" plays a role in controlling pH in the blood.\n\n \n\n \n\n1. The buffer system in blood plasma\ne The bicarbonate ion (HCO3\u00b0) plays a crucial role in maintaining blood pH through the bicarbonate buffer\nsystem. Blood pH is a measure of its acidity or alkalinity, and it must be tightly regulated for various\nphysiological processes to function properly. The normal pH range of blood is approximately 7.35 \u2014 7.45,\na value higher than 7.8 or lower than 6.8 can lead to death.\ne Inthe bicarbonate buffer system, there is a reversible exchange of carbon dioxide (CO2) and bicarbonate\nions in the blood. The reaction can be summarized as follows:\n(a) Formation of carbonic acid (H2COs):\nCO,,)\n(b) Decomposition of carbonic acid into bicarbonate (HCO;-) and hydrogen ion (H*):\nH,CO,,,) HCO},.) +H.)\ne This system is catalyzed by an enzyme found in red blood cells called carbonic anh", "Os):\nCO,,)\n(b) Decomposition of carbonic acid into bicarbonate (HCO;-) and hydrogen ion (H*):\nH,CO,,,) HCO},.) +H.)\ne This system is catalyzed by an enzyme found in red blood cells called carbonic anhydrase. CO2, H2CO3,\nHCO; and H\u2019 levels are related, and any imbalance can affect the pH of the blood. Thus HCO;~ helps\nregulate blood pH:\n\n+H,0,) =H,CO,,,)\n\n\u00a9 __ studyplusplus.com \u00b0 67\n\f\n3.\n\n6.\n\n   \n\n>\n\n- Student Learning Outcomes (SLOs)\n> Calculate concentration of slightly soluble salts.\n\nNeutralization of acids:\n\ne When there are many acids (H\u2019* ions) in the blood, bicarbonate acts as a base, binding to these hydrogen\nions and forming carbonic acid. This reaction helps to neutralize and eliminate excess acidity:\n\nHCO a0) + Hag) HCO aia)\n\nNeutralization of bases: .\n\ne Conversely, if there are many bases in the blood, such as hydroxide ions (OH\"), the carbonic acid\ndissociates, releasing hydrogen ions that can combine with excess bases:\n\nOH(,.) +H,CO,,,) =SHCO},,,) +H,0;\n\n\u00a2 Participating ", "e many bases in the blood, such as hydroxide ions (OH\"), the carbonic acid\ndissociates, releasing hydrogen ions that can combine with excess bases:\n\nOH(,.) +H,CO,,,) =SHCO},,,) +H,0;\n\n\u00a2 Participating in these reversible reactions, bicarbonate ions help maintain the acid-base balance in the\nblood. This buffer system is crucial in preventing rapid changes in blood pH and provides a stable\nenvironment for enzymes and other biochemical processes to function optimally. The respiratory and\nrenal systems also.play an important role in regulating CO2 and bicarbonate levels to maintain blood pH\nin a normal range.\n\nHousehold Cleaning Products:\n\ne Detergents: Many cleaning agents and detergents contain buffers to maintain a stable pH, ensuring\neffective cleaning without causing harm to the surface or skin.\n\nPersonal Care Products:\n\ne Shampoos and Soaps: Buffers are often used to maintain the pH of shampoos and soaps, preventing\nskin irritation and ensuring the products are gentle.\n\nSwimming Pools", " skin.\n\nPersonal Care Products:\n\ne Shampoos and Soaps: Buffers are often used to maintain the pH of shampoos and soaps, preventing\nskin irritation and ensuring the products are gentle.\n\nSwimming Pools:\n\ne Water Treatment: Buffers are employed in swimming pools to help maintain a stable pH level,\npreventing corrosion of pool equipment and ensuring a comfortable environment for swimmers.\n\nFood and Beverage Industry:\n\ne Food Preservation: Buffers are used in food processing to control acidity, preserve flavours, and\nmaintain the stability of certain food products.\n\ne Beverage Production: Buffers help control the pH in beverages, ensuring consistency in taste and\npreventing spoilage.\n\nPhotography:\n\n\u00a2 Developer Solutions: Buffers are used in photographic developer solutions to maintain a stable pH,\nallowing for controlled and consistent film or print development.\n\nr\n\n' SOLUBILITY PRODUCT AND PRECIPITATION REACTIONS\n\nNow we will discuss some of the important equilibria which have some analyt", " pH,\nallowing for controlled and consistent film or print development.\n\nr\n\n' SOLUBILITY PRODUCT AND PRECIPITATION REACTIONS\n\nNow we will discuss some of the important equilibria which have some analytical importance.\n\nSolubility Product\n\nWhen an excess of slightly soluble ionic compound is mixed with\n\nwater. Some of it dissolves and the remaining compound settles at | K,, is used for solutes which are only\nthe bottom. Dynamic equilibrium is established between an | Slightly soluble and do not completely\nundissolved solid compound and its ions in the saturated solution. | %!s0Wve in solution. The higher the Ksp\nF le, when CaF. is mixed with water. The followi Sa CE 2 COMNRONEG, Se Mane\nor JeNP. , p 2 ater. \u20ac TOHOWINE | soluble in water.\n\nequilibrium is established. \u2019\n\n2.\nCaF 4) = Cag +2 Fry\n\n \n\noS studyplusplus.com e\n\f\nKe for this equilibrium can be written as:\n2\n\nCa\u00ae || F\n[caF,]\nSince CaF is a slightly soluble salt its concentration almost remains constant.\n\nTherefore,\n\nKe =\n\nK.[car] =", "\n\n \n\noS studyplusplus.com e\n\f\nKe for this equilibrium can be written as:\n2\n\nCa\u00ae || F\n[caF,]\nSince CaF is a slightly soluble salt its concentration almost remains constant.\n\nTherefore,\n\nKe =\n\nK.[car] = [ca*][e]\nKos = (Ca'I[F*)\nWhere K;, is a constant known as the solubility product constant? It,is defined as the product of the molar\nconcentrations of ions, each raised to a power which is equal to the coefficient of the ion in the balanced\nchemical equation.\n\nIn general, Ks, expression of any slightly soluble ionic compound, AmBn can be written as\nAn B(x) = mA) + nBiet)\n\nco[eT oT\n\nThis means that the solubility product constant is equal to the product of the equilibrium concentration of\nions each raised toa power equal to the number of such ions in the formula unit of the compound.\n\nConcept Assessment Exercise 3.1 |\n\n1) Write the dissociation many and the solubility product constant expression for each of the\nfollowing solids.\n\nro OZ mo 3, BaS0,\n\nRana PHSO4. = Phi) +SOZi.q)- |, ANOH) 5) ", "Assessment Exercise 3.1 |\n\n1) Write the dissociation many and the solubility product constant expression for each of the\nfollowing solids.\n\nro OZ mo 3, BaS0,\n\nRana PHSO4. = Phi) +SOZi.q)- |, ANOH) 5) Alina +30H,,4, | BASO,,) \u2014=Baj,., +SO%,,.\n\nKsp K,, =[Pb\"*][SOz] K,, =[AP*][OH7} K,, =[Ba\u2019*}[SO%]\nExpression\n\n> Calculating Concentrations of ions\n\u00a9 Given the Ksp for AgBr(Ksp = 1.60 x 107) we can use the Ksp expression to calculate the concentrations\nof each ion in the solution.\ne If we represent the x moles of AgBr dissolved in one dm? of water\nAgBris) = == ABlag\u201d \u2014 + BF faq)\nx x x\nKsp = [Ag*] [Br]\n1.60107 = x.x\n, x = 16x10\nx = 1.26x10\u00b0M\nSo, [Ag*) = [Br] =1.26x 10M\n\n2.vili Calculate the concentration of a slightly soluble salt given its solubility product constant (Ksp). | (Exercise)\n$$\n\n\u00a9 Slightly soluble salts dissociate only to.a small extent in water, leading to low ion concentrations. To\ncalculate the concentration of these salts in solution, we use their solubility product constant ", "ightly soluble salts dissociate only to.a small extent in water, leading to low ion concentrations. To\ncalculate the concentration of these salts in solution, we use their solubility product constant (Ksp).\n\u00a2 The Ks, expression relates the concentrations of the ions in equilibrium with the undissolved salt.\n> Steps to calculate the concentration:\n1. Write the dissociation equation for. the salt.\n2. Set up the Kp expression.\n3. Solve for ion concentrations based on stoichiometry.\n\n    \n     \n\n} studyplusplus.com }\n\n69\n\f\nCalculate the concentration of Silver chloride (AgCt) in its aqueous solution.\n1. Dissociation:\n\nAgcl,,) = Ag* (2) +cr (2)\n\ncy =[ ee\" [or]\n\n2. Ksp expression:\n\nFor AgCl, K,, =1.8x107\" at25\u00b0C.\n3. Concentration Calculation:\nFisiio each mole of AgCt gives one mole of Ag* and Ct ions, let the solubility of AgCL be xmol/dm\u2019. Then,\n\n1.8 x 107\u00b0\n1.34 x 1075 mol/dm?\n\nThus, the solubili in water is approximately 1.34 x 10\u00b0> mol/dm?\n\n! 7. Concept Assessment Exercise 3.2\n\n1) \u2014 Ksp f", " ions, let the solubility of AgCL be xmol/dm\u2019. Then,\n\n1.8 x 107\u00b0\n1.34 x 1075 mol/dm?\n\nThus, the solubili in water is approximately 1.34 x 10\u00b0> mol/dm?\n\n! 7. Concept Assessment Exercise 3.2\n\n1) \u2014 Ksp for PbF2 is 4 x 10~. Calculate the concentration Pb* and F-ions in solution.\nAns.\n\n   \n\nPbE,,. == Phi.) +2F,\n\nKs, expression: .\n\nK,, =[Pb\u2122* IF?\nFor PbF 2s), Ksp = 4 x 10%\nConcentration Calculation:\n\nSince each mole of PbF2 gives one mole of Pb?* and F- ions, let the solubility of PbF2 be x mol/dm?. Then,\n\nKsp = [x] [x?\n4x10% = = x.(2x)?\n4x10% = 4\n43 = 4x10%\nee Axt0t\n4\n* = 1x10\u00b0*\nxX = 2.15x 10 mol/dm?\nThus, the solubility/concentration of Pb?* = - 2.15 x 10? mol/dm?\nF = 22.15 x10 mol/dm?\n\n2) Kp for BaSO, is 1 x 10\u00b0. Write the equation and the equilibrium expression for the dissolving of barium\n\n4.30 x 10\u00b0 mol/dm?\n\nsulphate. Also calculate the concentration of each ion.\n\nAns.\n\nBaSO,, \u2014=Bar\u2019, +SOz,\n\nKy expression:\n\n(99)\n\nK,, =[Ba**}[SO2\"]\n\nFor BaSO,,\n\nstudyplusplus.com\n\nKsp = 1 x 10\u00b0\n\no\n\nOe ee", "4.30 x 10\u00b0 mol/dm?\n\nsulphate. Also calculate the concentration of each ion.\n\nAns.\n\nBaSO,, \u2014=Bar\u2019, +SOz,\n\nKy expression:\n\n(99)\n\nK,, =[Ba**}[SO2\"]\n\nFor BaSO,,\n\nstudyplusplus.com\n\nKsp = 1 x 10\u00b0\n\no\n\nOe ee i es ww we ww me ow ee ww we ow we ww ww we nw ow on ow ow w= ow = on = oe\n\nN\noO\n\f\n' | Concentration Calculation: '\n| Since each mole of BaSO, gives one mole of Ba\u201d* and SO, ions, let the solubility of BaSO, be x mol/dm\u2019. l\n' Then,\n| Ke = dX]\n| 1x10\" = x |\n1 <= 1x10\" 1\n| x = 1x10%mol/dm? |\n+ Thus, the solubility (or concentration of Ba\u201d and SO.*) in water is approximately 1x 10\u00b0 mol/dm*_ ______\n\n   \n\nStudent Learning Outcomes (SLOs) .\n> Explain common ion effects giving suitable examples.\nCOMMON ION EFFECT\n\n2.1 Explain the common ion effect with a suitable example. Ca)\n\n3.1 Discuss the principles behind the common ion effect, providing two examples to illustrate its application.\n\nAn interesting situation arises when a weak electrolyte and a salt containing a common ion are present\nsimulta", "ples behind the common ion effect, providing two examples to illustrate its application.\n\nAn interesting situation arises when a weak electrolyte and a salt containing a common ion are present\nsimultaneously in an aqueous solution. For example, in a solution of ae acid, hydrofluoric acid Ka = 7.2 x 10\u00b0,\nits salt sodium fluoride produces the common ion.\n\nKCIO,, F2K9\n\nKCI.) Kah (29)\n\nSince HF is a weak electrolyte, it slightly dissociates. NaF being a strong\nelectrolyte breaks up completely into its ions, The common ion F- produced by .\nNaF will upset its equilibrium. This will increase the concentration of Frions.\nAccording, to Le Chatelier's principle, the equilibrium will shift to the left to use\nsome of the Fions. This will decrease the dissociation of HF. Thus dissociation\nof HF will decrease in the presence of dissolved NaF. This means as a result of\nthe equilibrium shift, the concentration of HF.will increase.\nSimilarly, when a highly soluble salt is added to the saturated solutio", "crease in the presence of dissolved NaF. This means as a result of\nthe equilibrium shift, the concentration of HF.will increase.\nSimilarly, when a highly soluble salt is added to the saturated solution of a less soluble salt containing a common\nion. The degree of dissociation of the less soluble.salt decreases. Therefore, it causes a decrease in its solubility.\nThe term common ion effect is used to describe the behaviour of a solution in which the same ion is produced by\ntwo different compounds. \"The phenomenon in which the degree of ionization or solubility of an electrolyte is\nsuppressed by the addition of highly soluble electrolyte containing a common ion is called common ion effect\".\n> Examples\n\ni. Potassium per chlorate KCIO, is moderately soluble in water. When highly soluble KCI is added to the\n\nsaturated solution of KCIO\u00ab. It causes an increase in the concentration of K* ion.\nKCIO,;,) \u2014Kj,.) + ClO;\n\nKC} PK) Tag\nAccording to Le Chatelier's principle K* ions will react with ClO, ", "added to the\n\nsaturated solution of KCIO\u00ab. It causes an increase in the concentration of K* ion.\nKCIO,;,) \u2014Kj,.) + ClO;\n\nKC} PK) Tag\nAccording to Le Chatelier's principle K* ions will react with ClO, ions to form KCIOas), This will suppress,\nthe ionization of KCIO\u00ab. Thus it will precipitate out.\ni. When HCl gas is passed through the saturated solution of NaCl (Brine), it increases the concentration of\n\nCrion,\n\n+ ClO;\\.4)\n\nDO YOU KNOW\n\nThe separation and identification of\ncations Into analytical groups is based on\nsolubility product principle and common\njon effect. In general, any procedure\nthat involves precipitation follows these\nprinciples.\n\n       \n   \n   \n     \n\n4(e9)\n\nNaCl.) \u2014=Naj,,) +Cl,\n\nrl 1\n\nAcce\n\noS studyplusplus.com e\n\n71\n\f\nSelf-Check Exercise 3.3\n\ni. Ammonium Chloride, NH,CI is a water-soluble salt. What will happen if this salt is added to a solution\ncontaining ammonium hydroxide?\nNH,OH.q) == NHji.9) + OH)\n\nAns. Adding ammonium chloride to the solution of ammonium hydroxid", " a water-soluble salt. What will happen if this salt is added to a solution\ncontaining ammonium hydroxide?\nNH,OH.q) == NHji.9) + OH)\n\nAns. Adding ammonium chloride to the solution of ammonium hydroxide will decrease the concentration of\nhydroxide ion and shifts the equilibrium to the left. This is due to common ion effect, where the presence\nof acommon ion from added NH,Cl suppresses the dissociation of NH4OH.\n\nii. Carbonic acid is a weak acid. It ionizes in water as follows\n\nH,COy,,) == 2H, +COZ,.\n\nWhat will happen if a strong electrolyte such as Na2CO; is added to a solution containing carbonic acid?\n\nAns. When Na2CO; is added to a solution containing carbonic acid a neutralization reaction will occur. The\ncarbonate ions from Na2CO3 will react with the hydrogen ions from H2CO3 forming bicarbonate ions and\nultimately water and carbon dioxide. This will cause the pH of the solution to increase as the carbonic acid\nis neutralized,\n\nStudent Learning Outcomes (SLOs)\n\nState what is meant b", "carbonate ions and\nultimately water and carbon dioxide. This will cause the pH of the solution to increase as the carbonic acid\nis neutralized,\n\nStudent Learning Outcomes (SLOs)\n\nState what is meant by the term partition coefficient, Kpee\n\nCalculate a partition coefficient for a system in which the sohute is in the same physical state in the\ntwo solvents.\n\nExplain the factors affecting the value of a partition coefficient in terms of the polarities of the solute\nand the solvent used.\n\nTHE PARTITION COEFFICIENT\nIf you put two immiscible liquids like ether and water in a separatory funnel and shake. On standing, you can see\ntwo layers. Ether is less dense, so it forms the upper layer. Assume that solute X dissolves in both liquids.\nIf you prepare a solution of known concentration of X.in water and mix it with ether in a separatory funnel and\nshake well. On standing, the two layers separate again.\nNow determine the concentration of the water layer. You will notice a decrease in concentrat", "er and mix it with ether in a separatory funnel and\nshake well. On standing, the two layers separate again.\nNow determine the concentration of the water layer. You will notice a decrease in concentration.\nWhy? Where did the solute go?\nAnalysis of the ether layer shows that there is also a solute in the ether layer. This means that the solute is\ndistributed between the two solvents. A dynamic equilibrium is created between these two solutions. At\nequilibrium, the concentration of X becomes constant in both layers.\nX(in water) =\u2014 X{(in ether)\nX in ether\noi [x in water]\n\nThis equilibrium constant is called the partition coefficient and is represented by the symbol Kc. It has a constant\nvalue at a constant temperature. .\n\nThe distribution law, or Nernst distribution law, gives a generalization that governs the distribution \u201ca a solute\nbetween\u2018two immiscible solvents. This law was first given by Nernst, who studied the partitioning of several\nsolutes between various suitable pairs of solven", " that governs the distribution \u201ca a solute\nbetween\u2018two immiscible solvents. This law was first given by Nernst, who studied the partitioning of several\nsolutes between various suitable pairs of solvents.\n\nThe partition coefficient (Kp<) is defined as the ratio of the concentrations of a solute in two different immiscible\nsolvents in contact with each other when equilibrium has been established at a particular temperature. The\nPartition coefficient is a ratio of two concentrations, so it has no units.\n\noS studyplusplus.com e\n\n \n\n72\n\f\nCalculation of Partition Coefficient\n\n \n\nThe partition coefficient (Kp<) for a system in which the solute is in the same physical state in the two solvents\ncan be calculated using the equilibrium expression.\n\nACTIVITY 3.1: Calculating the partition coefficient\nProcedure\n\ne\ne\ne\n\nMeasure 100 cm? of a 0.150 M solution of aqueous methylamine and add it into a separatory funnel.\nAdd 75 cm? of an organic solvent in the separatory funnel.\n\nShake the separatory fun", "Procedure\n\ne\ne\ne\n\nMeasure 100 cm? of a 0.150 M solution of aqueous methylamine and add it into a separatory funnel.\nAdd 75 cm? of an organic solvent in the separatory funnel.\n\nShake the separatory funnel gently but thoroughly to allow the solute to distribute between the two\nsolvents. Allow the system to reach equilibrium (about 5-10 minutes).\n\nLet the layers settle and separate the two solvents into different beakers.\n\nMeasure the concentration of the solute in each solvent.\n\nCalculations\n\nData for calculations\n\n1.\n2.\n3.\n4.\n5.\n6.\n\nCalculate the partition coefficient of methylamine in the\norganic solvent and water.\n\nSolution\n\nWhen 100 cm? of a 0.150 moldm\u2122 solution of aqueous methylamine (CH3NH2) is shaken with 75.0 cm?\nof an organic solvent in the separating funnel to allowed to come to equilibrium. Only 25 cm? of the\naqueous layer is run off and titrated against 0.225MHCL 7.05 cm? of HCL was used.\n\nVolume of aqueous methylamine solution: 100 cm?\nInitial concentration of methylamine: ", "quilibrium. Only 25 cm? of the\naqueous layer is run off and titrated against 0.225MHCL 7.05 cm? of HCL was used.\n\nVolume of aqueous methylamine solution: 100 cm?\nInitial concentration of methylamine: 0.150 moldm?\nVolume of organic solvent: 75.0 cm?\n\nVolume of aqueous layer titrated: 25.0 cm?\n\nVolume of HCL used for titration: 7.05 cm?\nConcentration of HCL: 0.225 moldm=\n\nICH,NH, Lane sexe\n[CH,NH, Jansen\n\nStep 1: Write down the equilibrium equation:\nCH,NH, (aq) == CH,NH, (organic solvent)\n\n \n\nStep 2: Write down the Kpc expression:\n\ny [CH,NH, (organiclayer)\n1 =\nCH,NH,,,9)\n\nStep 3: Determine the total moles of methylamine in the original solution.\nDetermine the total moles of CH3NHz in the original solution.\n0.100x 150\n1000\n= 0.015 mol\nThese moles were distributed between the two layers.\nStep 4: Determine the no. of moles methylamine in the aqueous layer.\n\nCH,NH,j,9) +HCl,q) > CH,NH,Cl,,.)\n\n25 cm? of aqueous layer reacted with 7.05 cm? of 0.225 MHCI\n1 mol of CH3Nh:2 is: 1 mol of HCI\n\nTotal", "4: Determine the no. of moles methylamine in the aqueous layer.\n\nCH,NH,j,9) +HCl,q) > CH,NH,Cl,,.)\n\n25 cm? of aqueous layer reacted with 7.05 cm? of 0.225 MHCI\n1 mol of CH3Nh:2 is: 1 mol of HCI\n\nTotal moles CH3NH2\n\nn = MxVdm3\nmole of CH3NH2 reacted with HCI solution = 0.225x\n\n \n\n7.05\n= 0.001586 mol\nmo\n\nAs 25 cm? ofthe aqueous layer was titrated, the 0.001586 mol of methylamine was present in 25 cm? of\nthe aqueous layer.\n\nSo, the moles of methylamine present in 100 cm? of aqueous layer = ee = 0.00634 mol\n\n23\n\n\u00a9 studyplusplus.com o es\n\f\n| 1) When 2 grams of a solute is shaken with a mixture of 150 cm\u2019 of water and 20 cm? of chloroform. After\nshaking, 1.5 grams of the solute is found in the chloroform layer. Calculate the partition coefficient.\n1 Solution:\n_ Conc. Of solute in Organic Layer\nma Conc. of solute in H,O\nTotal solute = 2.0g\nAfter Shaking\nIn CHCl layer = 1.5\u00a2\nH20 layer = 0.58\n\nStep 5: Determine the number of moles of CH3NHz2 present in the organic layer.\n\nMol CH3NH2 (organic la", "a Conc. of solute in H,O\nTotal solute = 2.0g\nAfter Shaking\nIn CHCl layer = 1.5\u00a2\nH20 layer = 0.58\n\nStep 5: Determine the number of moles of CH3NHz2 present in the organic layer.\n\nMol CH3NH2 (organic layer) = mol CH3NH2 (total) - mol CH3NH2 (aqueous layer)\n= 0.015-\u2014 0.00634\n= 0.00867 mol\n\nStep 6: Change the number of moles into concentrations:\n\n1000\n\nConcentration (CH3NHz2) in aqueous layer) = 0.0063 x 300 = 0.063 moldm=3\n\n \n\n= = 0.116 moldm=?\n\n \n\nConcentration (CH3NHz) in organic layer) = 0.0866 x\n\nStep 7: Substitute the values into the Kpc expression:\n\n0.116\n\u2014 0.0364\nKpc = 1.83\n\nSince the value of Ky. is larger than 1, methylamine is more soluble in \u2018iu organic solvent than in water.\n\n\u2018 Concept Assessment Exercise 3.3\n\nVol. of HO = 150cm?=1.15 dm?\nVol. of CHCl; = 20cm?=0.02 dm?.\n1.5g\na. 3\nOozdm ~ 75 8/dm\n\n0.5g\n= 3,33 g/dm?\n0.15dm* \u201c\n\nm solutegia,\n=\" [ solute, |\n\n75g /dm?*\n- Balan pasa\n\n3.33g/dm\n\nConc. of solute in CHCl; =\n\nConc. of solute in HO =\n\n \n\n> Factors Affecting the Numerical v", "Oozdm ~ 75 8/dm\n\n0.5g\n= 3,33 g/dm?\n0.15dm* \u201c\n\nm solutegia,\n=\" [ solute, |\n\n75g /dm?*\n- Balan pasa\n\n3.33g/dm\n\nConc. of solute in CHCl; =\n\nConc. of solute in HO =\n\n \n\n> Factors Affecting the Numerical value of Partition coefficient\n\nThe partition coefficient (Kp) indicates how a solute distributes itself between two immiscible phases,\n\nusually a non-polar solvent and a polar solvent.\nFactors that affect the numerical value-of the partition coefficient are:\n\n4. The Polarity of the solute:\nA polar solute has a greater affinity for a polar phase (e.g., water). This is because polar solutes interact\n\nwith polar solvents through dipole-dipole or hydrogen bonding interactions.\n\n \n\nFor example: A polar solute like ethanol (a polar molecule because of the hydroxyl group) between\noctanol and water. Ethanol is more soluble in water than octanol, resulting in a higher partition\n\ncoefficient in favour of the polar phase.\n\n} studyplusplus.com }\n\n74\n\f\nThe Solvent Polarity:\n\nThe nature of the solvents ", "thanol is more soluble in water than octanol, resulting in a higher partition\n\ncoefficient in favour of the polar phase.\n\n} studyplusplus.com }\n\n74\n\f\nThe Solvent Polarity:\n\nThe nature of the solvents used in the partition plays a decisive role. Non-polar solvents promote the\n\npartitioning of non-polar solutes, while polar solvents promote the partitioning of polar solutes.\n\nFor example: When a non-polar solute such as benzene is added, its partition coefficient is higher in\n\noctanol than in water, because benzene is non-polar and interacts more with non-polar octanol.\n\nTemperature:\n\nTemperature can affect the partition coefficient by affecting the solubility and kinetic energy of the\nsolute in both. solvents. In general, an increase in temperature can increase the solubility of solutes in\n\npolar solvents.\n\nFor instant: The partitioning of a polar solute such as aspirin between octanol and water can show\ndifferent values at different temperatures. At higher temperatures, aspirin is more", "\npolar solvents.\n\nFor instant: The partitioning of a polar solute such as aspirin between octanol and water can show\ndifferent values at different temperatures. At higher temperatures, aspirin is more soluble in water than\n\nin octanol.\n\nMolecular structure and size\n\nLarge solute molecules can interact with solvents differently than smaller molecules. Molecular\n\nstructure, including functional groups, can also affect the distribution coefficient.\n\nConsider the partitioning of octanol and water between two molecules of similar structure but different\nsizes, such as ethyl acetate and butyl acetate. A larger butyl acetate molecule may have a lower partition\n\ncoefficient in favour of the non-polar phase due to increased steric hindrance.\n\nThe degree to which different Bronsted acids give off protons is called \"acid strength\"\n\nThe greater the value of Ka, the stronger the acid\".\n\nThe strength of a base is the ability to accept a proton from a solvent\n\nA buffer solution is a solution, the pH ", "otons is called \"acid strength\"\n\nThe greater the value of Ka, the stronger the acid\".\n\nThe strength of a base is the ability to accept a proton from a solvent\n\nA buffer solution is a solution, the pH of which does not change significantly when a small amount of\nacid or base is added to it.\n\nThe bicarbonate ion (HCO;-) plays a crucial role in maintaining blood pH through the bicarbonate\nbuffer system.\n\nThe normal pH range of blood is approximately 7.35-7.45.\n\nThe solubility product is defined as the product of the equilibrium concentrations of ions, each raised\nto a power which is the coefficient of the ion in the balance chemical equation.\n\nThe phenomenon in which the degree of ionization or solubility of an electrolyte is suppressed by the\naddition of a highly soluble electrolyte containing a common ion is called the common ion effect.\n\nThe partition coefficient (Kp<) is defined as the ratio of the concentrations of a solute in two different\nimmiscible solvents in contact with each ot", "ng a common ion is called the common ion effect.\n\nThe partition coefficient (Kp<) is defined as the ratio of the concentrations of a solute in two different\nimmiscible solvents in contact with each other when equilibrium has been established at a particular\ntemperature.\n\nNernst'\u2019s law states that a solute is distributed between two layers of immiscible solvents so that the\nratio of its concentration in each solvent is equal to its solubility.\n\n \n\nSo studyplusplus.com oe\n\n75\n\f\nExercise\n\n \n\ni. Which phenomenon describes a shift in equilibrium due to the addition of anion already involved in the\n\nequilibrium?\n\na), Le Chatelier's Principle b) Common lon Effect \u2014_c) Nernst's Law d) Avogadro's Principle\nii. What property is used to distinguish between strong and weak acids?\n\na) Molar mass b) Extent of ionization c) Density d) Melting point\niii. Which factor is considered when distinguishing between strong and weak bases?\n\na) Colour b) Odor c) Extent of ionization d) Solubility\n\niv. What defi", "ent of ionization c) Density d) Melting point\niii. Which factor is considered when distinguishing between strong and weak bases?\n\na) Colour b) Odor c) Extent of ionization d) Solubility\n\niv. What defines a buffer solution?\na) High concentration of ions\nb) Presence of a weak acid and its conjugate base (or a weak base and its conjugate acid)\nc) Low pH d) Absence of ions\n\nv. Howcana buffer solution be made?\n\na) Mixing strong acids and bases\nb) Mixing a weak acid and its conjugate base (or a weak bas2 and its conjugate acid)\n\nc) Diluting a strong acid d) Adding water to a strong base\nvi. What role does HCO; play in controlling pH in blood?\na) Neutralization of acids b) Buffering against changes in pH\nc) Catalysing reactions d) Enhancing oxygen transport\nvii. How is the concentration of a slightly soluble salt calculated?\na) Using the solubility product constant (Ksp) b) Titration with a strong base\nc) Measuring density d) Conductivity measurement\nviii. Which term is used to describe the s", "htly soluble salt calculated?\na) Using the solubility product constant (Ksp) b) Titration with a strong base\nc) Measuring density d) Conductivity measurement\nviii. Which term is used to describe the strength of an acid in terms of its ionization in water?\na) Acid Concentration b) Acid Dissociation Constant (Ka)\nc) Acid Molarity . d) Acid Equilibrium Constant\nix. Which type of solvent would favour the partitioning of a polar solute?\na) Non-polar solvent _\u2014_b) Hydrophobic solvent _c) Polar solvent d) Aprotic solvent\n\nx. | What is the partition coefficient defined as?\na) Ratio of solute concentration in one solvent to the other\nb) Ratio of solute concentration in a single solvent\nc) Ratio of solute mass in one solvent to the other\nd) Ratio of solute mass in a single solvent\n\nPp leo teal vy [ey [eis [oro Te alee\n\n> 2) Short Answer Questions\n\n   \n\ni. Explain the common ion effect with a suitable example.\n\n \n\nAns. See Page No. (71)\nii. Differentiate between strong and weak acids using the ex", "y [eis [oro Te alee\n\n> 2) Short Answer Questions\n\n   \n\ni. Explain the common ion effect with a suitable example.\n\n \n\nAns. See Page No. (71)\nii. Differentiate between strong and weak acids using the extent of ionization and Ky.\nAns. See Page No. (63)\niii. Differentiate between strong and weak bases using the extent of ionization and Kp.\nAns. See Page No. (65)\n\n \n\n \n\n \n\n \n\noS studyplusplus.com e\n\n76\n\f\n \n\niv. Define a buffer solution and provide an example of how it can be made.\n\nAns. See Page No. (65)\n\nv. How does a buffer solution control pH? Include chemical equations in your explanatian.\n\nAns. See Page No. (66)\n\nvi. Describe the uses of buffer solutions in various applications.\n\nAns. See Page No. (67)\n\nvii. Explain how HCO; plays a role in controlling pH in the blood.\n\nAns. See Page No. (67) .\n\nviii, Calculate the concentration of a slightly soluble salt given its solubility product constant (Ks).\nAns. See Page No. (69)\n\ni. Discuss the principles behind the common ion effect, providin", "67) .\n\nviii, Calculate the concentration of a slightly soluble salt given its solubility product constant (Ks).\nAns. See Page No. (69)\n\ni. Discuss the principles behind the common ion effect, providing two examples to illustrate its application.\n\nAns. See Page No. (71)\n\nji. Compare and contrast: strong and weak acids based on their extent of ionization and the value of Ka.\n\nAns. See Page No. (63)\n\nji. Elaborate on the factors that distinguish strong and weak bases, incorporating the concept of ionization\nand the base dissociation constant (Ky).\n\nAns. See Page No. (65) ;\n\niv. Provide a comprehensive explanation of buffer solutions, including how they are\u2019prepared, how they\ncontrol pH, and their uses in different scenarios.\n\nAns. See Page No. (65)\nv. Calculate the pH of an NH,OH and NH,CL buffer solution containing one mole of each. Kp of NH,OH is 1.8 x 105,\nAns. [NH,OH] 1.0 moles.dm=?\n\n \n\n \n\n \n\n \n\n \n\n \n\n \n\n \n\n \n\n   \n\n \n\n \n\n \n\n \n\n \n\n \n\n \n\n \n\n\"\n\nand (NH.Cl] = 1.0 moles.dm-3\nkK = 1.8x10\u00b0\n-", "ffer solution containing one mole of each. Kp of NH,OH is 1.8 x 105,\nAns. [NH,OH] 1.0 moles.dm=?\n\n \n\n \n\n \n\n \n\n \n\n \n\n \n\n \n\n \n\n   \n\n \n\n \n\n \n\n \n\n \n\n \n\n \n\n \n\n\"\n\nand (NH.Cl] = 1.0 moles.dm-3\nkK = 1.8x10\u00b0\n-logk\u00bb = -log1.8x 10-5\nPK, = 4.74\n: \u201c [salt]\nusing, POH = PK + 108 Those]\n\nsalt\npH = 4.74 + log Sd\n\npH = 4.74 + log = 4.74 + log!\n\nlog1 = 0\nPH = 4.74+0=4.74\npH = 4.74\nAs pH+pOH = 14\n~ So, pOH = 14-pH\n= 14-4.74\npOH = 9.26\n\n} studyplusplus.com }\n\n77\n\f\nvi. A solute has a partition coefficient Kpenzenr/water = 4,100 cm? of water is shaken with 100 cm\u2019 of benzene\ncontaining 0.20 mol of the solute. Find the amount of solute that remains in water at equilibrium.\n\nAns. Partition coefficient K = Coensene =4\nCyrater\nVolume of benzene = 100cm?>\nVolume of water = 100cm?\nTotal amount of solute = 0.20 mol\nAmount of solute that remains in water at equilibrium = ?\n\nLet Amount of solute that remains in water at equilibrium = x\n\nSo, 0.20 \u2014 x = amount of solute in benzene\n\n \n\nUse the partition coefficient:\n0.", "ount of solute that remains in water at equilibrium = ?\n\nLet Amount of solute that remains in water at equilibrium = x\n\nSo, 0.20 \u2014 x = amount of solute in benzene\n\n \n\nUse the partition coefficient:\n0.20-x)/100 _ 0.20-x A\nx/100 ate he\n\nSolve the equation:\n\n0.20-x 4\n\nx\n0.20-x = 4x\n0.20 = Sx\nx = = =0.04mol\n\n \n\n1. By adding NH,CL to a solution of NH,OH in water, the ionization of NH,OH:\n\n(A)Increases (B) Decreases (C) Remains Same (D) Cannot be predicted\n2. When HClis passed in saturated solution of NaCl, the solubility of NaCl is:\n(A) Increased | (B) Decreased (C) Not affected (D) None of all\n\n3. When ionic product of a solution is greater than the solubility product at a particular temperature then\nthe solution is said to be:\n\n(A) Unsaturated (B) Saturated (C) Very dilute (D) Super saturated\n4. The solution which resists change in its pH either an acid or base is added in It Is called:\n\n(A) Buffer solution (B) Acid (C) Base (D) Alkali\n5. The pH of mixture of CH;\\COONa and CH;COOH is:\n\n(A", "\n4. The solution which resists change in its pH either an acid or base is added in It Is called:\n\n(A) Buffer solution (B) Acid (C) Base (D) Alkali\n5. The pH of mixture of CH;\\COONa and CH;COOH is:\n\n(A)7 *  (B)>7 (C)<7 (D) 1\n6. Purification of NaCl by passing HC! gas Is an example of:\n\n(A) Filtration (B) Sublimation (C) lonic product (D) Common ion effect\n7. | Mixture of NH,OH and NH,Cl makes a buffer whose pH is:\n\n(A) Less than seven (B)7 (C) More than seven (D)4\n8. Whenis HCl added to H2S aqueous solution, the ionization of H2S:\n\n(A) Increase (B) Remain constant (C) Decreases (D) Increase rapidly\n9:, Sum of pK, and pK; Is equal to:\n\n(A) 14 (B) 7 (c)o (D)1\n10. The'strongest acid among Halogen acids is:\n\n(A) HCI (B) HBr (C) HI (D) HF\n11. Ionization of hydrogen sulphide gas is suppressed by: '\n\n_ (A) KCl (B) NaCl (C) HCI (D) NH,CI\n\n\u00a9 __ studyplusplus.com o 78\n\f\n12.\n\n13.\n\n14.\n\n15.\n\n16.\n\n17.\n\n18.\n\n19.\n\n20,\n\n21.\n\n22.\n\n23.\n\n24,\n\n25.\n\n26.\n\n27.\n\n28.\n\n29.\n\n30.\n\n31.\n\n32.\n\n33.\n\nThe value of pKy a", " '\n\n_ (A) KCl (B) NaCl (C) HCI (D) NH,CI\n\n\u00a9 __ studyplusplus.com o 78\n\f\n12.\n\n13.\n\n14.\n\n15.\n\n16.\n\n17.\n\n18.\n\n19.\n\n20,\n\n21.\n\n22.\n\n23.\n\n24,\n\n25.\n\n26.\n\n27.\n\n28.\n\n29.\n\n30.\n\n31.\n\n32.\n\n33.\n\nThe value of pKy at 25\u00b0C for water Is:\n\n(A) 10\u00b0\u201d (B) 7 (c).10-\"\nAcid having K, > 1 will be:\n(A) Weak (B) Very weak (C) Moderate\n\nThe ionic product of water will increase if:\n(A) H* ions are added\n(C) Temperature is increased\n\nThe units of K, for the reaction of ammonia synthesis are:\n\n(B) OH\u201d ions are added\n\n(A) mol? dm\u00ae (B) mol dm\u00ae (C) mol? dm?\nThe value of K,, at 252C is:\n(A) 0.11 x 10-4 (B) 0.30 x 1074 (Cc) 1 1074\nThe units for K,, of H2O are:\n\nmole\n(A) Came (B) mol? dm* (C) mol? dm\u00ae\nThe term pH was introduced by:\n(A) Henderson (B) Sorenson (C) Goldsmith\npH of a solution is 9, the solution is a:\n(A) Weak acid (B) Weak base (C) Strong acid\nThe pH of 10 mol dm= of an aqueous solution of NaOH is:\n(A) 2 (B) 12 (C)3\nThe value of pH of pure water at 25\u00b0C is:\n(A) 14 (B)7 (C) 1x 10-*\npH of Bananas is:\n(A) 2.1. (", "d (B) Weak base (C) Strong acid\nThe pH of 10 mol dm= of an aqueous solution of NaOH is:\n(A) 2 (B) 12 (C)3\nThe value of pH of pure water at 25\u00b0C is:\n(A) 14 (B)7 (C) 1x 10-*\npH of Bananas is:\n(A) 2.1. (B) 4.6 (C) 9.4\nPH of milk is:\n(A) 6.5 (B) 8.5 (C) 10.5\n\nThe pH of 107 mol dm\" of an aqueous solution of HCl is:\n\n(A) 2.0 (B) 2.7 (C).3.0\nWhich aqueous solution has highest pH?\n\n(A) 0.1M NaOH (B) 0.1M H2SO,4 (C) 0.1 M HCI\nThe nature of milk is:\n\n(A) Acidic (B) Basic (C) Neutral\n\nThe ionization constant of pure water at 25\u00b0C is:\n(A) 1.8 x 10726 mol dm? (B) 1.6 x 10\u2122 mol dm=3\n\nThe pH of 10~ mol / dm? of Ba(OH), is:\n\n(C) 1.0 x 10-* mol? dm-*\n\n(D) 14\n\n(D) Strong\n\n(D) H* and OH\" ions are added in equal amount\n\n(D) mol? dm?\n\n(D) 3 x 10-**\n\n(D) mol-? dm\n\n(D) Thomson\n(D) Strong base\n(D) 10\n\n(D) 1x 10\"\n\n(D) 9.6\n\n(D) 12.5\n\n(D) 1.5\n\n(D) 0.2M HNO;\n(D) Normal\n\n(D) 1.8 x 10\u00b0 mol? dm*\n(D) 10.3\n\n(D) 9.35\n\n(D) 4.5\n\n(D) Highly basic\n(D) 3.5\n\n(D) 10000\n\n79\n\n(A) 4.5 (B) 6.4 (C) 7.5\npH of human blood is:\n(A) 6.", "\n(D) 9.6\n\n(D) 12.5\n\n(D) 1.5\n\n(D) 0.2M HNO;\n(D) Normal\n\n(D) 1.8 x 10\u00b0 mol? dm*\n(D) 10.3\n\n(D) 9.35\n\n(D) 4.5\n\n(D) Highly basic\n(D) 3.5\n\n(D) 10000\n\n79\n\n(A) 4.5 (B) 6.4 (C) 7.5\npH of human blood is:\n(A) 6.35 (B) 7.35 (C) 8.35\nApproximate pH of apple is:\n(A).2.7 (B) 3.1 (C) 4.2\nRain water is:\n(A) Slightly acidic (B) Slightly basic (C) Neutral\npH value of vinegar is:\n(A) 1.1 (B) 2.0 (C) 2.8\nA solution with pH = 2 is more acidic than a solution with pH = 6 by a factor of:\n(A) 4 (B)8 (C) 1000\nSo studyplusplus.com oe\n\f\n34. pH value for 1.0 M HCI solution Is:\n\n(A) 0.0 (B) 0.5 (C) 0.7 (D) 0.8\n35. pH of soft drinks is approximately:\n(A) 1.5 (B) 1.0 (C) 2.0 (D) 3.0\n36. The solution having zero pH will be:\n(A) Acidic (B) Highly acidic (C) Neutral (D) Basic\n37. The pH of tomato is: ; ;\n(A) 12 (B) 4.2 (C) 7.2 (D) 9.2\n38. AgCl dissolved with conc. (2x107). Ksp will be:\n(A) 3.6x10\u00b06 (B) 3.6x105 (C) 7.2x10% \u00bb (D) 4x104\n39. If Ksp = [M*?}?[X3]\u2019, the chemical formula of the compound is:\n(A) M2X2 (B) M2X3 (C", "7.2 (D) 9.2\n38. AgCl dissolved with conc. (2x107). Ksp will be:\n(A) 3.6x10\u00b06 (B) 3.6x105 (C) 7.2x10% \u00bb (D) 4x104\n39. If Ksp = [M*?}?[X3]\u2019, the chemical formula of the compound is:\n(A) M2X2 (B) M2X3 (C) MaX2 (D) MX2\n40. Precipitation occurs when:\n(A) Ksp <ionic.product (B) Ksp > ionic product (C) Ksp = ionic product (D)~/Ksp < ionic product\n41. HCl when added to H2S solution: 5\n(A) Suppresses the ionization of H2S (B) Enhances the ionization\n(C) Solution becomes coloured (D) Does not affect\n42. When HCl is added to the aqueous solution of NaCl; the ionization of NaCl:\n(A) Increases (B) Decreases (C) Remains same _(D) First decreases and then increases\n43. Asolution with a pK, value of 9, suggest that it is a:\n(A) Strong acid (B) Weak acid (C) Weak base (D) Strong base\n44. K, is acid association constant, large pK, value for an acid indicates that acid Is:\n(A) Moderate (B) Water soluble (C) Strong (D) Weak\n45. The solution in which pH is maintained, when a small amount of acid or base is", "tion constant, large pK, value for an acid indicates that acid Is:\n(A) Moderate (B) Water soluble (C) Strong (D) Weak\n45. The solution in which pH is maintained, when a small amount of acid or base is added to it, is known as:\n(A) Aqueous solution \u2014_(B) Dilute solution (C) Concentrated solution (D) Buffer solution\n46. Which one is the example of buffer? \u2018\n(A) HCI/NaCl (B) NaOH/H2COs3 (C) NHaOH/NH,Cl (D) NaOH/NaCl\n47. Anacidic buffer solution can be prepared by mixing: ; :\n(A) Weak acid and its salt with strong base (B) Strong acid and its salt with weak base\n(C) Weak base and its salt with strong acid (D) Strong base and its salt with weak acid\n48. pH of a buffer solution containing 1M CH3COOH and 1M CH3COONa (pK; = 4.74) is:\n(A) 4.74 (B) 3.74 (C) 5.74 (D) 6.74\n\n  \n\nAdditional Short Answer Questions\n\nQ-1 Sum of pK, and pK\u00bb at 25\u00b0C.\n\nAns. pK, + pK,=pK, pK, =\u2014log K,\nKw =10\u2122 at 25\u00b0C\nPK, + pK, =\u2014logk,,\npK, + pK, =\u2014log10\u2122*\npK, +pK, =14 logy, =1\nIt is true at 25\u00b0C.\n\n \n\noS studyplusplus.com e", " Questions\n\nQ-1 Sum of pK, and pK\u00bb at 25\u00b0C.\n\nAns. pK, + pK,=pK, pK, =\u2014log K,\nKw =10\u2122 at 25\u00b0C\nPK, + pK, =\u2014logk,,\npK, + pK, =\u2014log10\u2122*\npK, +pK, =14 logy, =1\nIt is true at 25\u00b0C.\n\n \n\noS studyplusplus.com e\n\f\n \n\nQ.2 Sum of pK, and pK, at 100\u00b0C.\n\nAns. pK, + pK, =pK,\npK, +pK, =-logk,\npK, + pk, =\u2014log(7.5x10\u2122)K, =7.5x10\u2122 at100\u00b0C.\npK, +pK, =13.125\n\nHence at 100\u00b0C, the sum of pK, and py is less than 14.\nThe reason for this is that the value of ionization constant (Kw) of water has greater value and its \u2014log(pKw)\n\nhas lesser value than that at 25\u00b0C.\nQ.3 Calculate pH of a buffer solution in which 0.11 molar CH;COONa and 0.09 molar acetic acid solutions are\npresent. K, for CHsCOOH is 1.85 x 10\u00b0\n\n \n\n.\n\nGiven Data: [CH; COONa)] = 0.11M\n. [CH;COOH] = 0.09M\nK, Of CH3COOH = 1.85x 10%\nRequired: pH of Buffersolution = ?\nSolution: pK, = \u2014logk, \u00a9\n\n= \u2014log 1.8 x 10% = 4.74\n\nSalt\npH = pKa+ log\n\n0.11\npH = 4.74 + 1085 o9\n\npH = 4.74+0.087\npH = 4.83\n\nQ.4 The solubility of PbF2 at 25\u00b0C is 0.64gdm\". Calculate K,, of P", "ution = ?\nSolution: pK, = \u2014logk, \u00a9\n\n= \u2014log 1.8 x 10% = 4.74\n\nSalt\npH = pKa+ log\n\n0.11\npH = 4.74 + 1085 o9\n\npH = 4.74+0.087\npH = 4.83\n\nQ.4 The solubility of PbF2 at 25\u00b0C is 0.64gdm\". Calculate K,, of PbF2.\nGiven Data: Solubility of PbF. = 0.64 g dm at 25\u00b0C\nRequired: Ksp = ?\nSolution: First of all convert the concentration from g dm\u2122 to moles dm?\nMass of PbF, dissolved dm== 0.64g\nMolecular mass of PbF2 = 207.2+19x2\n\n= 207.2+38\n= 245.2g/mole\n-3\nNumber of moles of PbF2= Ses\n245.2gmol\"\n= 2.6x 10? mole dm?\nBalance Equation:\nPbF 2s) asc Phra + 2Fica)\n2.6 x 10 mole dm= 0 \u201c0 at t=0Sec\n0 mole dm? 2.610 mole dm= 2x2.6 x 10 moles dm\u2122 t=eq\nThe expression of Ksp is\nes 2 ert ye\nKee = [Pb(aq)] [F(aq)) .\n\nPutting values of concentration\nKsp = (2.6 x 107) (2x 2.6 x 10\u00b03)2\n\nKy, = 7.0x 10%\nemi ice i a\nQ5 Ca(OH)2 is sparingly soluble compound. Its solubility constant Is 6.5 x 10, Calculate the solubility of Ca(OH) 2.\nGiven Data: Solubility constant=Ksp = 6.510%\nRequired: Solubility of Ca(OH): = ?\n\n\u00a9 studyp", "H)2 is sparingly soluble compound. Its solubility constant Is 6.5 x 10, Calculate the solubility of Ca(OH) 2.\nGiven Data: Solubility constant=Ksp = 6.510%\nRequired: Solubility of Ca(OH): = ?\n\n\u00a9 studyplusplus.com res 81\n\f\nSolution:\n\n \n\n \n\n \n\n \n\nBalance Equation: Ca(OH) 2 Caiva + 20H)\nLet us suppose\nSolubility of Ca(OH)2 = S\nCa(OH): Cais) + 2OH (ea)\nSs 0 QO at t=Osec\nSs Ss 2s t= equilibrium\n\nThe concentration of OH ~ is double than the concentration of Ca, so\nKsp = [Cajeay) [OHjeq)*\n65x10 = $x (2s)\u201d\n\n6.5x10% = $x4s\u2019\n3\n\n \n\n6.5x10\u00b0 = 4S\n4s\u2019 = 65x10\ng) 2\u00bb 8:5x10\n4\n(seae\")\"\nS$ =\n4\nS = (1.625 x 10-)\u00a53\n= (1.625) \u00a53 x 10-6\u00ab\u00a53\nS = 1175x107\n\nHence At equilibrium stage\nS = [Ca*]=1.175 x 10? mol dm?\n[OH~] 2S =2.x 1.175 x 107\n= 2.350 x 107 mol dm?\nConcentration of Ca** ions = 1.175 x 10 mol dm?\nConcentration of OH\u2122 ions = 2.350 x 10 mol dm?\n\n \n\nQ.6 Why the K,, of water increases with the increase of temperature?\nAns. As you know that ionic product of water (Kw) is\nKw = [H*][OH7]\nAs the temperature i", "ation of OH\u2122 ions = 2.350 x 10 mol dm?\n\n \n\nQ.6 Why the K,, of water increases with the increase of temperature?\nAns. As you know that ionic product of water (Kw) is\nKw = [H*][OH7]\nAs the temperature increases, more water molecules dissociate. As a result the value of Kw increases with\nincreasing temperature. But this increase in K,, is not regular. The value of Ky, increases almost 75 times when\nthe temperature is increased from 0\u00b0C to 100\u00b0C.\n\n \n\n \n\nQ.7 Why water is a weak electrolyte?\n\nAns. Whenever some quantity of acid or base is added in water, then K, remains the same, but [H*] and [OH\u2122]\nare no more equal. -\n\n \n\nIn neutral water:\n[H*] = [OH]\n{H*][H*] = 10-*\n[HP = 10\n[H*) = 10\u00b0 moles dm?\nSimilarly [OH] = 10\u00b0? moldm?\n\nThis means that out of 55.5 moles of pure water in one dm\u2019 of it, only 10-\u201d moles of it are dissociated into\nions. This shows that water is a very weak electrolyte.\n\nSo studyplusplus.com oe\n\f\n \n\nQ.8 What is Henderson equation?\nAns. i. Henderson\u2019s equation for acidic bu", "y 10-\u201d moles of it are dissociated into\nions. This shows that water is a very weak electrolyte.\n\nSo studyplusplus.com oe\n\f\n \n\nQ.8 What is Henderson equation?\nAns. i. Henderson\u2019s equation for acidic buffers\n\nsalt!\npH = pKa + lossy\n\nli. Henderson\u2019s equation for basic buffers\n[salt]\n[base]\n\n \n\n \n\npOH = pK, + log\n\n \n\nQ.9 How NaCl can be purified by common ion effect?\nAns. NaCt Purification: NaCl can be purified by common ion effect. NaCl is purified by passing hydrogen chloride\ngas through the saturated brine. NaCL is fully ionized in the solution.\n\n \n\n \n\nNaCl) = Najaq) + Clog\n[Na*} (Cr)\nKe = \u201c\" [NaCl]\nHCl ionizes as HCL (g) == Hag + Chea)\n\n \n\nOn passing HCL gas, concentration of CL ions is increased. Therefore, NaCt crystallizes out of the solution to\nmaintain the constant value of equilibrium constant. This type of effect is called common ion effect.\nNa(aq) + Cliaq) === NaC\n\n \n\nQ.10 Write two uses of buffer solution.\nAns. Uses of Buffer Solution:\n\ni. If we wants to study a reaction under", "m constant. This type of effect is called common ion effect.\nNa(aq) + Cliaq) === NaC\n\n \n\nQ.10 Write two uses of buffer solution.\nAns. Uses of Buffer Solution:\n\ni. If we wants to study a reaction under conditions that would suffer any associated change in the pH of\nreaction mixture. So, by a suitable choice of the solutes, a chemist can ensure that a solution will not\nexperience more than a very small change in pH, even if small amount of a strong acid or a strong base\nis added. ;\n\nii. Buffers are important in many areas of chemistry and allied sciences like molecular biology,\nmicrobiology, cell biology, soil sciences, nutrition and the clinical analysis.\n\nQ.11 What is the partition coefficient?\n\nAns. It's the ratio of the concentration of a solute in two immiscible phases at equilibrium, representing its\ndistribution between those phases.\n\nQ.12 How is the partition coefficient calculated?\n\nAns. It's calculated by dividing the concentration of the solute in the stationary phase by the c", "epresenting its\ndistribution between those phases.\n\nQ.12 How is the partition coefficient calculated?\n\nAns. It's calculated by dividing the concentration of the solute in the stationary phase by the concentration of\nthe solute in the mobile phase (Kg = Cx/Cm).\n\nQ.13_ What does a high partition coefficient indicate?\n\nAns. A high partition coefficient suggests that the solute is more soluble in the organic or stationary phase\ncompared to the aqueous or mobile phase.\n\nQ.14 What does a low partition coefficient indicate?\n\nAns. A low partition coefficient indicates that the solute is more soluble in the aqueous or mobile phase.\n\nQ.15 What are the units of the partition coefficient?\n\nAns. The units of the partition coefficient are dimensionless as they are a ratio of concentrations.\nQ.16 Is the partition coefficient dependent on temperature?\n\nAns. Yes, as an equilibrium constant, the partition coefficient is dependent on temperature.\nQ.17 How Is the partition coefficient relevant in drug des", "partition coefficient dependent on temperature?\n\nAns. Yes, as an equilibrium constant, the partition coefficient is dependent on temperature.\nQ.17 How Is the partition coefficient relevant in drug design?\n\nAns. In drug design, the partition coefficient (often expressed as log P) is used as a measure of a drug's\nlipophilicity and its potential for membrane permeability, including crossing cell membranes.\n\n \n\n \n\n \n\n \n\n \n\n \n\n \n\n \n\n \n\n \n\n \n\n \n\n \n\n \n\n \n\noS studyplusplus.com e\n\n83\n\f\n \n\nQ.18 Can the partition coefficient be used to determine the mass of a solute extracted?\n\nAns. Yes, by. knowing the partition coefficient and initial concentrations, one can calculate the mass of a solute\nextracted into a different phase, even with successive extractions.\n\nQ.19 What is the difference between the partition coefficient (K) and the distribution coefficient (D)?\n\nAns. While both relate to the distribution of a solute, the partition coefficient (K) refers specifically to the un-\nionized form of the ", "rtition coefficient (K) and the distribution coefficient (D)?\n\nAns. While both relate to the distribution of a solute, the partition coefficient (K) refers specifically to the un-\nionized form of the solute, whereas the distribution coefficient (D) considers all species (ionized and un-\nivaized) present at a given pH.\n\n \n\n \n\n \n\n \n\nQ.20 Caii the partition coefficient change with temperature, and if so, how?\n\nns. artition coefficient (Kpc) can indeed change with temperature, reflecting the temperature dependence\nc. -ubility. Solubility, a key factor in determining Kpc, is influenced by temperature because it affects the\ninte vactions between solute and solvent molecules. For instance, an increase in temperature generally\nincreases the solubility of solids and liquids in a solvent, leading to a change in the Kpc value. Conversely, for\ngases, solubility typically decreases with an increase in temperature, altering the Kpc accordingly. The extent\nof this change depends on the nature of the ", " a change in the Kpc value. Conversely, for\ngases, solubility typically decreases with an increase in temperature, altering the Kpc accordingly. The extent\nof this change depends on the nature of the solute and the solvent. For example, if a solute becomes\n\n \n\n>\n\nsignificantly more soluble in one solvent compared to another with a rise in temperature, the Kpc will change\n\naccordingly. This change is important in practical applications, like in pharmaceuticals, where temperature-\ncontrolled environments are crucial for maintaining drug efficacy. .\n\nQ.21 Give two examples of a buffer solution.\n\nAns. Examples:\n\n(i) CH3COOH (Acetic acid) +CH3COONa (Sodium acetate)\n\n \n\n \n\n(Weakacid) - (Salt of acetic acid) ;\n(ii) NH,OH (Ammonium hydroxide) + NH\u00abCl(Ammonium chloride)\n(Weak base) (Salt of weak base)\n\n \n\nQ.22 Calculate the pH of formic acid-sodium formate buffer solution containing 1.0 mole of each component.\n(K, for formic acid is 1.8 x 10) \u00a9\nData: [HCOOH] = 1.0 mole dm?\n{HCOONa] = 1.0 mole d", "se)\n\n \n\nQ.22 Calculate the pH of formic acid-sodium formate buffer solution containing 1.0 mole of each component.\n(K, for formic acid is 1.8 x 10) \u00a9\nData: [HCOOH] = 1.0 mole dm?\n{HCOONa] = 1.0 mole dm?\nK, for formic acid = 1.8x 10%\n\n \n\nPK, = \u2014logKs = \u2014log (1.8 x 10~)\npK, = 3.745\nRequired: - pH of buffer = ?\n\n~ Solucion:According to Henderson\u2019s equation;\n\n-pH\n\nsalt\nPK, + log ar\nH = 37483 oe ean\nPH = 3.745 +1087 9 mole dm\nPH = 3.745+log1 _ ee log 1=0\npH = 3.745\nFe\nQ.23 What will be the pH of the solution after addition of 0.10 mole of hydrochloric acid gas to 1.01 dm? volume\nof the buffer solution in part (a) Assume that the volume of solution remains unchanged on addition of\nhydrochtoric acid (K, for formic acid is 1.8 x 10-*).\nAns. Since HCl is a strong acid so it completely ionizes. It means it produces 0.1 mole H\u2019* ions, These H* ions react\n\nwith 0.1 mole of HCOO\u2122 ions. Hence, out of 1.0 moles of salt (HCOO\u2122 Na*), 0.9 moles (1.0 \u2014 0.1 = 0.9) of salt\nare left behind.\n\nOn the other ha", "eans it produces 0.1 mole H\u2019* ions, These H* ions react\n\nwith 0.1 mole of HCOO\u2122 ions. Hence, out of 1.0 moles of salt (HCOO\u2122 Na*), 0.9 moles (1.0 \u2014 0.1 = 0.9) of salt\nare left behind.\n\nOn the other hand, concentration of acid (HCOOH) is increased from 1.0 moles to 1.1 moles (1.0 + 0.15 1.1).\nHence, new.concentrations will be:\n\noS studyplusplus.com e\n\n84\n\f\nNumber of moles of HCOONa = 0.9 moles\nNumber of moles of HCOOH = 1.1 moles\nSince volume of buffer solution is 1.01 dm\u2019, therefore, the molar concentrations will be\n\n0. 2.\n[HCOONa] =\n\n \n\n1.01 = 0-891 mole dm?\nan\n[HCOOH] Loi = 1.089 mole dm-3\n\nK, for formic acid = 1.8x 10\n\npK, = \u2014logK, = \u2014log (1.8 x 10\u201c)\npK, = 3.745\nAccording to Henderson\u2019s equation;\nsalt\u2019\n\u201cpH = pKat+ los at\n0.891 mole dm?\npH = 3.745 +1087 089 mole dm>\npH = 3.745-0.0872\npH = 3.658\n\nQ.24 How does the solubility of a sparingly-soluble salt get affected due to the presence of a soluble salt having\n\none common ion?\n\nAns. This phenomenon of common ion effect can be discussed", "= 3.658\n\nQ.24 How does the solubility of a sparingly-soluble salt get affected due to the presence of a soluble salt having\n\none common ion?\n\nAns. This phenomenon of common ion effect can be discussed under Le-Chatlier principle. It says that if\n\n. concentration of reactants is increased, then reaction goes to forward direction to attain same K. or K, value.\nIn case of saturated solution of NaCt, Na,\u00b0, and CL, are produced. The addition CL, from HCL gas will push\nthe reaction to backward direction and NaCt will be precipitated in that saturated solution.\n\nNaCl === Na,2, + Cte,\nHL <== HE, + C42,\nNa\u00ae, + Cue, NaCl) Reverse process\nExcess ions ppt.\n\n \n\n \n\n \n\nQ.25 Why are the cations of group (II) of basic radicals in salt analysis precipitated as sulphides from acidic\n\nsolutions while those of groups (IV) are precipitated from ammonical solution?\n\nAns. In these two groups, the comparison is in suppression and enhancing ionization nee 1 ont! least .? oe are\n\non a conc. of -, ions is suffici", " groups (IV) are precipitated from ammonical solution?\n\nAns. In these two groups, the comparison is in suppression and enhancing ionization nee 1 ont! least .? oe are\n\non a conc. of -, ions is sufficient to do the precipitation of Pb\u00ae, cu\u00ae ; Hg\u201d ca\u00ae, Bi eas,\n\nsb\u201d: and sn\u00ae / sn\u00ae . The \u2014 oo. of sulphides of these radicals are very low. In group (IV) of basic\nradical analysis is i.e., of zn ; Mn\u201d. ; Co\u201d i\u2018 nie , we replace HCl byNH, OH.\n\n \n\nNH@ + OH)\n\nH2S 2H + e.\n\nOH\u00ae consumes H\u00ae and helps H:S to dissociate more. Sufficient conc. of s\u00ae ion is necessary to do\nprecipitation of these radicals.\n\n \n\nNH, OH\n\n \n\n \n\nne er SSS\nQ.26 Why is an excess of ammonium chloride added to the solution before adding NH,OH for\n\ndetecting/identifying cationsduring salt analysis of group III?\n\nAns. NH,CLis a stronger electrolyte than NH,OH. First of all the addition of NH,CI increases the conc. of Na Bey\n\nions which suppress the ionization of NH,OH. The common ion effect decreases the conc. of OH\u00ae ions eo\n\nFe\u2019, ", "r electrolyte than NH,OH. First of all the addition of NH,CI increases the conc. of Na Bey\n\nions which suppress the ionization of NH,OH. The common ion effect decreases the conc. of OH\u00ae ions eo\n\nFe\u2019, Mn\u00b0, and zn\u201d, are precipitated. If excessofO = ivete the \u00ab basic radicals of \u00bb ma\nprecipitated. In this way confusion arises.\n\n\u00a9 studyplusplus.com o 8\u00b0\n\f\n \n\nQ.27 The concentration of silver and chloride ions in saturated solution of silver chloride at 298 K is 1.0x 10\u00b0M\nindividually. Calculate the solubility product of AgCL.\n\nAns. The solubility product of AgCL is written from its equation of dissociation.\nAgChs, ABs) * Chay\nKsp = [AB say) (Che)\n. The molar concentrations of Ag\u00ae and CI\u00ae ions are 1 x 10\u00b0 M each at 298 K.\n=. Ksp =.(1 x 1075 M) (1 x 10-5 M) = 107\u00b0 Mm? , -\nOr Ksp = 10\u00b0\u00b0mol\u2019dm~* \u201c. 1M=moldm>\n\n \n\n \n\n \n\n \n\nQ.28 A particular saturated solution of silver chromate, Ag2CrOs, has [Ag\u00ae] = 5.0 x 10\u00b0 M and\n{cro} = 4.4 x 10\u201c M. What is the value of K;, for Ag2CrO.?\n\nAns. Ag2CrOsg is spari", " 1M=moldm>\n\n \n\n \n\n \n\n \n\nQ.28 A particular saturated solution of silver chromate, Ag2CrOs, has [Ag\u00ae] = 5.0 x 10\u00b0 M and\n{cro} = 4.4 x 10\u201c M. What is the value of K;, for Ag2CrO.?\n\nAns. Ag2CrOsg is sparingly soluble salt. The conc. of ions are given in terms of molarity or moldm\u2122>.\nConc. of Ag, = 5.0x10\u00b0M\n\nConc. of COS = 44x 104M\nThe equation for dissociation is\n\n \n\n \n\nAg2CrOais) Agi&) + CrO,2,\n\nKp = [AgS,) [Cr0,2) \u2018\n= (5x 10M)? (4.4 104M)\n= (5)? x (4.4).107 x 104 M3\n= 25x4.4x 10M?\nKsp = 110x104 = 1.1 x 107 M?= 1.1 x 10%moPdm?\nQ.29 The K;, of AgBr is 5.4 x 10\u00b0 mol*dm* at 298 K. Calculate solubility of AgBr.\nAns. Ksp of AgBr = 5.4x 10-3mol*dm*\nTemperature 298 K\n\nSolubility (s) of AgBr = ?\nThe formula of solubility for uni-valent electrolyte is\n4 Ss = VK\u00bb\nPutting values\n\nS$ = V Keo = /5.4x108 mol? dm*\nS = 54x10\" =4/54 x 107 moldm=\n\nSolubility of AgBr 7.348 x 107\u201d moldm=?\n\nQ.30 Define K, and pK, and their applications.\nAns. lonization constant of an acid (Ks):\nThe quantitative measure of the", "ol? dm*\nS = 54x10\" =4/54 x 107 moldm=\n\nSolubility of AgBr 7.348 x 107\u201d moldm=?\n\nQ.30 Define K, and pK, and their applications.\nAns. lonization constant of an acid (Ks):\nThe quantitative measure of the strength of an acid is called ionization constant of an acid (Ks). Greater the\nvalue of K, stronger will be the acid and vice versa.\nPK}\n\u2018The negative log of dissociation constant of acid (Ka) is called pK;:\npK, = \u2014log K,\n\n(wd\n\nLarger the value of pK., weaker will be the acid and vice versa.\nThe value of K; is used to determine the percentage ionization of an acid.\n\nRelationship between K, and pK:\n\n\u00a9 studyplusplus.com o 8\u00b0\n\f", "In a chemical reaction, chemical equilibrium is the state in which both the reactants and products are present in concentrations which have no further tendency to change with time, so that there is no observable change in the properties of the system. This state results when the forward reaction proceeds at the same rate as the reverse reaction. The reaction rates of the forward and backward reactions are generally not zero, but they are equal. Thus, there are no net changes in the concentrations of the reactants and products. Such a state is known as dynamic equilibrium.\nIt is the subject of study of equilibrium chemistry.\n\n\n== Historical introduction ==\nThe concept of chemical equilibrium was developed in 1803, after Berthollet found that some chemical reactions are reversible. For any reaction mixture to exist at equilibrium, the rates of the forward and backward (reverse) reactions must be equal. In the following chemical equation, arrows point both ways to indicate equilibrium. A ", "eaction mixture to exist at equilibrium, the rates of the forward and backward (reverse) reactions must be equal. In the following chemical equation, arrows point both ways to indicate equilibrium. A and B are reactant chemical species, S and T are product species, and \u03b1, \u03b2, \u03c3, and \u03c4 are the stoichiometric coefficients of the respective reactants and products:\n\n\u03b1 A + \u03b2 B \u21cc \u03c3 S + \u03c4 T\nThe equilibrium concentration position of a reaction is said to lie \"far to the right\" if, at equilibrium, nearly all the reactants are consumed. Conversely the equilibrium position is said to be \"far to the left\" if hardly any product is formed from the reactants.\nGuldberg and Waage (1865), building on Berthollet's ideas, proposed the law of mass action:\n\n  \n    \n      \n        \n          \n            \n              \n                \n                  forward reaction rate\n                \n              \n              \n                \n                =\n                \n                  k\n                ", "       \n                \n                  forward reaction rate\n                \n              \n              \n                \n                =\n                \n                  k\n                  \n                    +\n                  \n                \n                \n                  \n                    A\n                  \n                  \n                    \u03b1\n                  \n                \n                \n                  \n                    B\n                  \n                  \n                    \u03b2\n                  \n                \n              \n            \n            \n              \n                \n                  backward reaction rate\n                \n              \n              \n                \n                =\n                \n                  k\n                  \n                    \u2212\n                  \n                \n                \n                  \n                    S\n                  \n                  \n                    \u03c3\n   ", "k\n                  \n                    \u2212\n                  \n                \n                \n                  \n                    S\n                  \n                  \n                    \u03c3\n                  \n                \n                \n                  \n                    T\n                  \n                  \n                    \u03c4\n                  \n                \n              \n            \n          \n        \n      \n    \n    {\\displaystyle {\\begin{aligned}{\\text{forward reaction rate}}&=k_{+}{\\ce {A}}^{\\alpha }{\\ce {B}}^{\\beta }\\\\{\\text{backward reaction rate}}&=k_{-}{\\ce {S}}^{\\sigma }{\\ce {T}}^{\\tau }\\end{aligned}}}\n  \n\nwhere A, B, S and T are active masses and k+ and k\u2212 are rate constants. Since at equilibrium forward and backward rates are equal:\n\n  \n    \n      \n        \n          k\n          \n            +\n          \n        \n        \n          \n            {\n            \n              A\n            \n            }\n          \n          \n            \u03b1\n         ", "       \n          k\n          \n            +\n          \n        \n        \n          \n            {\n            \n              A\n            \n            }\n          \n          \n            \u03b1\n          \n        \n        \n          \n            {\n            \n              B\n            \n            }\n          \n          \n            \u03b2\n          \n        \n        =\n        \n          k\n          \n            \u2212\n          \n        \n        \n          \n            {\n            \n              S\n            \n            }\n          \n          \n            \u03c3\n          \n        \n        \n          \n            {\n            \n              T\n            \n            }\n          \n          \n            \u03c4\n          \n        \n      \n    \n    {\\displaystyle k_{+}\\left\\{{\\ce {A}}\\right\\}^{\\alpha }\\left\\{{\\ce {B}}\\right\\}^{\\beta }=k_{-}\\left\\{{\\ce {S}}\\right\\}^{\\sigma }\\left\\{{\\ce {T}}\\right\\}^{\\tau }}\n  \n\nand the ratio of the rate constants is also a constant, now known as an equilibrium constant.\n", "\\{{\\ce {B}}\\right\\}^{\\beta }=k_{-}\\left\\{{\\ce {S}}\\right\\}^{\\sigma }\\left\\{{\\ce {T}}\\right\\}^{\\tau }}\n  \n\nand the ratio of the rate constants is also a constant, now known as an equilibrium constant.\n\n  \n    \n      \n        \n          K\n          \n            c\n          \n        \n        =\n        \n          \n            \n              k\n              \n                +\n              \n            \n            \n              k\n              \n                \u2212\n              \n            \n          \n        \n        =\n        \n          \n            \n              {\n              \n                S\n              \n              \n                }\n                \n                  \u03c3\n                \n              \n              {\n              \n                T\n              \n              \n                }\n                \n                  \u03c4\n                \n              \n            \n            \n              {\n              \n                A\n              \n              \n        ", "                }\n                \n                  \u03c4\n                \n              \n            \n            \n              {\n              \n                A\n              \n              \n                }\n                \n                  \u03b1\n                \n              \n              {\n              \n                B\n              \n              \n                }\n                \n                  \u03b2\n                \n              \n            \n          \n        \n      \n    \n    {\\displaystyle K_{c}={\\frac {k_{+}}{k_{-}}}={\\frac {\\{{\\ce {S}}\\}^{\\sigma }\\{{\\ce {T}}\\}^{\\tau }}{\\{{\\ce {A}}\\}^{\\alpha }\\{{\\ce {B}}\\}^{\\beta }}}}\n  \n\nBy convention, the products form the numerator.\nHowever, the law of mass action is valid only for concerted one-step reactions that proceed through a single transition state and is not valid in general because rate equations do not, in general, follow the stoichiometry of the reaction as Guldberg and Waage had proposed (see, for example, nucleophilic al", "e transition state and is not valid in general because rate equations do not, in general, follow the stoichiometry of the reaction as Guldberg and Waage had proposed (see, for example, nucleophilic aliphatic substitution by SN1 or reaction of hydrogen and bromine to form hydrogen bromide). Equality of forward and backward reaction rates, however, is a necessary condition for chemical equilibrium, though it is not sufficient to explain why equilibrium occurs.\nDespite the limitations of this derivation, the equilibrium constant for a reaction is indeed a constant, independent of the activities of the various species involved, though it does depend on temperature as observed by the van 't Hoff equation. Adding a catalyst will affect both the forward reaction and the reverse reaction in the same way and will not have an effect on the equilibrium constant. The catalyst will speed up both reactions thereby increasing the speed at which equilibrium is reached.\nAlthough the macroscopic equilib", "me way and will not have an effect on the equilibrium constant. The catalyst will speed up both reactions thereby increasing the speed at which equilibrium is reached.\nAlthough the macroscopic equilibrium concentrations are constant in time, reactions do occur at the molecular level. For example, in the case of acetic acid dissolved in water and forming acetate and hydronium ions,\n\nCH3CO2H + H2O  \u21cc  CH3CO\u22122 + H3O+\na proton may hop from one molecule of acetic acid onto a water molecule and then onto an acetate anion to form another molecule of acetic acid and leaving the number of acetic acid molecules unchanged. This is an example of dynamic equilibrium. Equilibria, like the rest of thermodynamics, are statistical phenomena, averages of microscopic behavior.\nLe Ch\u00e2telier's principle (1884) predicts the behavior of an equilibrium system when changes to its reaction conditions occur. If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to pa", " predicts the behavior of an equilibrium system when changes to its reaction conditions occur. If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to partially reverse the change. For example, adding more S (to the chemical reaction above) from the outside will cause an excess of products, and the system will try to counteract this by increasing the reverse reaction and pushing the equilibrium point backward (though the equilibrium constant will stay the same).\nIf mineral acid is added to the acetic acid mixture, increasing the concentration of hydronium ion, the amount of dissociation must decrease as the reaction is driven to the left in accordance with this principle. This can also be deduced from the equilibrium constant expression for the reaction:\n\n  \n    \n      \n        K\n        =\n        \n          \n            \n              {\n              \n                \n                  CH\n                  \n                    3\n         ", "e reaction:\n\n  \n    \n      \n        K\n        =\n        \n          \n            \n              {\n              \n                \n                  CH\n                  \n                    3\n                  \n                  \n                    \n                  \n                \n                \n                  CO\n                  \n                    2\n                  \n                  \n                    \u2212\n                  \n                \n              \n              }\n              {\n              \n                \n                  H\n                  \n                    3\n                  \n                  \n                    \n                  \n                \n                \n                  O\n                  \n                    +\n                  \n                \n              \n              }\n            \n            \n              \n                {\n                \n                  CH\n                  \n                    3\n                  \n  ", "         \n              \n              }\n            \n            \n              \n                {\n                \n                  CH\n                  \n                    3\n                  \n                  \n                    \n                  \n                \n                \n                  CO\n                  \n                    2\n                  \n                  \n                    \n                  \n                \n                H\n                }\n              \n            \n          \n        \n      \n    \n    {\\displaystyle K={\\frac {\\{{\\ce {CH3CO2-}}\\}\\{{\\ce {H3O+}}\\}}{{\\ce {\\{CH3CO2H\\}}}}}}\n  \n\nIf {H3O+} increases {CH3CO2H} must increase and CH3CO\u22122 must decrease. The H2O is left out, as it is the solvent and its concentration remains high and nearly constant.\nJ. W. Gibbs suggested in 1873 that equilibrium is attained when the \"available energy\" (now known as Gibbs free energy or Gibbs energy) of the system is at its minimum value, assuming the reacti", "tant.\nJ. W. Gibbs suggested in 1873 that equilibrium is attained when the \"available energy\" (now known as Gibbs free energy or Gibbs energy) of the system is at its minimum value, assuming the reaction is carried out at a constant temperature and pressure. What this means is that the derivative of the Gibbs energy with respect to reaction coordinate (a measure of the extent of reaction that has occurred, ranging from zero for all reactants to a maximum for all products) vanishes (because dG = 0), signaling a stationary point. This derivative is called the reaction Gibbs energy (or energy change) and corresponds to the difference between the chemical potentials of reactants and products at the composition of the reaction mixture. This criterion is both necessary and sufficient. If a mixture is not at equilibrium, the liberation of the excess Gibbs energy (or Helmholtz energy at constant volume reactions) is the \"driving force\" for the composition of the mixture to change until equilibr", "e is not at equilibrium, the liberation of the excess Gibbs energy (or Helmholtz energy at constant volume reactions) is the \"driving force\" for the composition of the mixture to change until equilibrium is reached. The equilibrium constant can be related to the standard Gibbs free energy change for the reaction by the equation\n\n  \n    \n      \n        \n          \u0394\n          \n            r\n          \n        \n        \n          G\n          \n            \u2296\n          \n        \n        =\n        \u2212\n        R\n        T\n        ln\n        \u2061\n        \n          K\n          \n            \n              e\n              q\n            \n          \n        \n      \n    \n    {\\displaystyle \\Delta _{r}G^{\\ominus }=-RT\\ln K_{\\mathrm {eq} }}\n  \n\nwhere R is the universal gas constant and T the temperature (In Kelvin).\nWhen the reactants are dissolved in a medium of high ionic strength the quotient of activity coefficients may be taken to be constant. In that case the concentration quotient, Kc,\n\n  \n    \n    ", "lvin).\nWhen the reactants are dissolved in a medium of high ionic strength the quotient of activity coefficients may be taken to be constant. In that case the concentration quotient, Kc,\n\n  \n    \n      \n        \n          K\n          \n            c\n          \n        \n        =\n        \n          \n            \n              [\n              \n                S\n              \n              \n                ]\n                \n                  \u03c3\n                \n              \n              [\n              \n                T\n              \n              \n                ]\n                \n                  \u03c4\n                \n              \n            \n            \n              [\n              \n                A\n              \n              \n                ]\n                \n                  \u03b1\n                \n              \n              [\n              \n                B\n              \n              \n                ]\n                \n                  \u03b2\n                \n             ", "  \u03b1\n                \n              \n              [\n              \n                B\n              \n              \n                ]\n                \n                  \u03b2\n                \n              \n            \n          \n        \n      \n    \n    {\\displaystyle K_{\\ce {c}}={\\frac {[{\\ce {S}}]^{\\sigma }[{\\ce {T}}]^{\\tau }}{[{\\ce {A}}]^{\\alpha }[{\\ce {B}}]^{\\beta }}}}\n  \n\nwhere [A] is the concentration of A, etc., is independent of the analytical concentration of the reactants. For this reason, equilibrium constants for solutions are usually determined in media of high ionic strength. Kc varies with ionic strength, temperature and pressure (or volume). Likewise Kp for gases depends on partial pressure. These constants are easier to measure and encountered in high-school chemistry courses.\n\n\n== Thermodynamics ==\nAt constant temperature and pressure, one must consider the Gibbs free energy, G, while at constant temperature and volume, one must consider the Helmholtz free energy, A, for", ".\n\n\n== Thermodynamics ==\nAt constant temperature and pressure, one must consider the Gibbs free energy, G, while at constant temperature and volume, one must consider the Helmholtz free energy, A, for the reaction; and at constant internal energy and volume, one must consider the entropy, S, for the reaction.\nThe constant volume case is important in geochemistry and atmospheric chemistry where pressure variations are significant. Note that, if reactants and products were in standard state (completely pure), then there would be no reversibility and no equilibrium. Indeed, they would necessarily occupy disjoint volumes of space. The mixing of the products and reactants contributes a large entropy increase (known as entropy of mixing) to states containing equal mixture of products and reactants and gives rise to a distinctive minimum in the Gibbs energy as a function of the extent of reaction. The standard Gibbs energy change, together with the Gibbs energy of mixing, determine the equili", "ts and gives rise to a distinctive minimum in the Gibbs energy as a function of the extent of reaction. The standard Gibbs energy change, together with the Gibbs energy of mixing, determine the equilibrium state.\nIn this article only the constant pressure case is considered. The relation between the Gibbs free energy and the equilibrium constant can be found by considering chemical potentials.\nAt constant temperature and pressure in the absence of an applied voltage, the Gibbs free energy, G, for the reaction depends only on the extent of reaction: \u03be (Greek letter xi), and can only decrease according to the second law of thermodynamics. It means that the derivative of G with respect to \u03be must be negative if the reaction happens; at the equilibrium this derivative is equal to zero.\n\n  \n    \n      \n        \n          \n            (\n            \n              \n                \n                  d\n                  G\n                \n                \n                  d\n                  \u03be", "\n      \n        \n          \n            (\n            \n              \n                \n                  d\n                  G\n                \n                \n                  d\n                  \u03be\n                \n              \n            \n            )\n          \n          \n            T\n            ,\n            p\n          \n        \n        =\n        0\n         \n      \n    \n    {\\displaystyle \\left({\\frac {dG}{d\\xi }}\\right)_{T,p}=0~}\n  \n:     equilibrium\nIn order to meet the thermodynamic condition for equilibrium, the Gibbs energy must be stationary, meaning that the derivative of G with respect to the extent of reaction, \u03be, must be zero. It can be shown that in this case, the sum of chemical potentials times the stoichiometric coefficients of the products is equal to the sum of those corresponding to the reactants. Therefore, the sum of the Gibbs energies of the reactants must be the equal to the sum of the Gibbs energies of the products.\n\n  \n    \n      \n        \u03b1\n        \n", " those corresponding to the reactants. Therefore, the sum of the Gibbs energies of the reactants must be the equal to the sum of the Gibbs energies of the products.\n\n  \n    \n      \n        \u03b1\n        \n          \u03bc\n          \n            \n              A\n            \n          \n        \n        +\n        \u03b2\n        \n          \u03bc\n          \n            \n              B\n            \n          \n        \n        =\n        \u03c3\n        \n          \u03bc\n          \n            \n              S\n            \n          \n        \n        +\n        \u03c4\n        \n          \u03bc\n          \n            \n              T\n            \n          \n        \n        \n      \n    \n    {\\displaystyle \\alpha \\mu _{\\mathrm {A} }+\\beta \\mu _{\\mathrm {B} }=\\sigma \\mu _{\\mathrm {S} }+\\tau \\mu _{\\mathrm {T} }\\,}\n  \n\nwhere \u03bc is in this case a partial molar Gibbs energy, a chemical potential. The chemical potential of a reagent A is a function of the activity, {A} of that reagent.\n\n  \n    \n      \n        \n          \u03bc\n          \n       ", "se a partial molar Gibbs energy, a chemical potential. The chemical potential of a reagent A is a function of the activity, {A} of that reagent.\n\n  \n    \n      \n        \n          \u03bc\n          \n            \n              A\n            \n          \n        \n        =\n        \n          \u03bc\n          \n            A\n          \n          \n            \u2296\n          \n        \n        +\n        R\n        T\n        ln\n        \u2061\n        {\n        \n          A\n        \n        }\n        \n      \n    \n    {\\displaystyle \\mu _{\\mathrm {A} }=\\mu _{A}^{\\ominus }+RT\\ln\\{\\mathrm {A} \\}\\,}\n  \n\n(where \u03bcoA is the standard chemical potential).\nThe definition of the Gibbs energy equation interacts with the fundamental thermodynamic relation to produce\n\n  \n    \n      \n        d\n        G\n        =\n        V\n        d\n        p\n        \u2212\n        S\n        d\n        T\n        +\n        \n          \u2211\n          \n            i\n            =\n            1\n          \n          \n            k\n          \n        \n        \n ", "        p\n        \u2212\n        S\n        d\n        T\n        +\n        \n          \u2211\n          \n            i\n            =\n            1\n          \n          \n            k\n          \n        \n        \n          \u03bc\n          \n            i\n          \n        \n        d\n        \n          N\n          \n            i\n          \n        \n      \n    \n    {\\displaystyle dG=Vdp-SdT+\\sum _{i=1}^{k}\\mu _{i}dN_{i}}\n  \n.\nInserting dNi = \u03bdi d\u03be into the above equation gives a stoichiometric coefficient (\n  \n    \n      \n        \n          \u03bd\n          \n            i\n          \n        \n         \n      \n    \n    {\\displaystyle \\nu _{i}~}\n  \n) and a differential that denotes the reaction occurring to an infinitesimal extent (d\u03be). At constant pressure and temperature the above equations can be written as\n\n  \n    \n      \n        \n          \n            (\n            \n              \n                \n                  d\n                  G\n                \n                \n                  d\n                 ", "  \n      \n        \n          \n            (\n            \n              \n                \n                  d\n                  G\n                \n                \n                  d\n                  \u03be\n                \n              \n            \n            )\n          \n          \n            T\n            ,\n            p\n          \n        \n        =\n        \n          \u2211\n          \n            i\n            =\n            1\n          \n          \n            k\n          \n        \n        \n          \u03bc\n          \n            i\n          \n        \n        \n          \u03bd\n          \n            i\n          \n        \n        =\n        \n          \u0394\n          \n            \n              r\n            \n          \n        \n        \n          G\n          \n            T\n            ,\n            p\n          \n        \n      \n    \n    {\\displaystyle \\left({\\frac {dG}{d\\xi }}\\right)_{T,p}=\\sum _{i=1}^{k}\\mu _{i}\\nu _{i}=\\Delta _{\\mathrm {r} }G_{T,p}}\n  \n\nwhich is the Gibbs free energy change for the r", "      \n        \n      \n    \n    {\\displaystyle \\left({\\frac {dG}{d\\xi }}\\right)_{T,p}=\\sum _{i=1}^{k}\\mu _{i}\\nu _{i}=\\Delta _{\\mathrm {r} }G_{T,p}}\n  \n\nwhich is the Gibbs free energy change for the reaction. This results in:\n\n  \n    \n      \n        \n          \u0394\n          \n            \n              r\n            \n          \n        \n        \n          G\n          \n            T\n            ,\n            p\n          \n        \n        =\n        \u03c3\n        \n          \u03bc\n          \n            \n              S\n            \n          \n        \n        +\n        \u03c4\n        \n          \u03bc\n          \n            \n              T\n            \n          \n        \n        \u2212\n        \u03b1\n        \n          \u03bc\n          \n            \n              A\n            \n          \n        \n        \u2212\n        \u03b2\n        \n          \u03bc\n          \n            \n              B\n            \n          \n        \n        \n      \n    \n    {\\displaystyle \\Delta _{\\mathrm {r} }G_{T,p}=\\sigma \\mu _{\\mathrm {S} }+\\tau \\mu _{\\mathr", "\n          \u03bc\n          \n            \n              B\n            \n          \n        \n        \n      \n    \n    {\\displaystyle \\Delta _{\\mathrm {r} }G_{T,p}=\\sigma \\mu _{\\mathrm {S} }+\\tau \\mu _{\\mathrm {T} }-\\alpha \\mu _{\\mathrm {A} }-\\beta \\mu _{\\mathrm {B} }\\,}\n  \n.\nBy substituting the chemical potentials:\n\n  \n    \n      \n        \n          \u0394\n          \n            \n              r\n            \n          \n        \n        \n          G\n          \n            T\n            ,\n            p\n          \n        \n        =\n        (\n        \u03c3\n        \n          \u03bc\n          \n            \n              S\n            \n          \n          \n            \u2296\n          \n        \n        +\n        \u03c4\n        \n          \u03bc\n          \n            \n              T\n            \n          \n          \n            \u2296\n          \n        \n        )\n        \u2212\n        (\n        \u03b1\n        \n          \u03bc\n          \n            \n              A\n            \n          \n          \n            \u2296\n          \n        \n      ", "  \u2296\n          \n        \n        )\n        \u2212\n        (\n        \u03b1\n        \n          \u03bc\n          \n            \n              A\n            \n          \n          \n            \u2296\n          \n        \n        +\n        \u03b2\n        \n          \u03bc\n          \n            \n              B\n            \n          \n          \n            \u2296\n          \n        \n        )\n        +\n        (\n        \u03c3\n        R\n        T\n        ln\n        \u2061\n        {\n        \n          S\n        \n        }\n        +\n        \u03c4\n        R\n        T\n        ln\n        \u2061\n        {\n        \n          T\n        \n        }\n        )\n        \u2212\n        (\n        \u03b1\n        R\n        T\n        ln\n        \u2061\n        {\n        \n          A\n        \n        }\n        +\n        \u03b2\n        R\n        T\n        ln\n        \u2061\n        {\n        \n          B\n        \n        }\n        )\n      \n    \n    {\\displaystyle \\Delta _{\\mathrm {r} }G_{T,p}=(\\sigma \\mu _{\\mathrm {S} }^{\\ominus }+\\tau \\mu _{\\mathrm {T} }^{\\ominus })-(\\alpha \\mu _{\\mathrm {A}", "       B\n        \n        }\n        )\n      \n    \n    {\\displaystyle \\Delta _{\\mathrm {r} }G_{T,p}=(\\sigma \\mu _{\\mathrm {S} }^{\\ominus }+\\tau \\mu _{\\mathrm {T} }^{\\ominus })-(\\alpha \\mu _{\\mathrm {A} }^{\\ominus }+\\beta \\mu _{\\mathrm {B} }^{\\ominus })+(\\sigma RT\\ln\\{\\mathrm {S} \\}+\\tau RT\\ln\\{\\mathrm {T} \\})-(\\alpha RT\\ln\\{\\mathrm {A} \\}+\\beta RT\\ln\\{\\mathrm {B} \\})}\n  \n,\nthe relationship becomes:\n\n  \n    \n      \n        \n          \u0394\n          \n            \n              r\n            \n          \n        \n        \n          G\n          \n            T\n            ,\n            p\n          \n        \n        =\n        \n          \u2211\n          \n            i\n            =\n            1\n          \n          \n            k\n          \n        \n        \n          \u03bc\n          \n            i\n          \n          \n            \u2296\n          \n        \n        \n          \u03bd\n          \n            i\n          \n        \n        +\n        R\n        T\n        ln\n        \u2061\n        \n          \n            \n   ", " \n          \n            \u2296\n          \n        \n        \n          \u03bd\n          \n            i\n          \n        \n        +\n        R\n        T\n        ln\n        \u2061\n        \n          \n            \n              {\n              \n                S\n              \n              \n                }\n                \n                  \u03c3\n                \n              \n              {\n              \n                T\n              \n              \n                }\n                \n                  \u03c4\n                \n              \n            \n            \n              {\n              \n                A\n              \n              \n                }\n                \n                  \u03b1\n                \n              \n              {\n              \n                B\n              \n              \n                }\n                \n                  \u03b2\n                \n              \n            \n          \n        \n      \n    \n    {\\displaystyle \\Delta _{\\mathrm {r} }G_{T,p}=\\sum _{i=1}^{k}\\mu ", "                }\n                \n                  \u03b2\n                \n              \n            \n          \n        \n      \n    \n    {\\displaystyle \\Delta _{\\mathrm {r} }G_{T,p}=\\sum _{i=1}^{k}\\mu _{i}^{\\ominus }\\nu _{i}+RT\\ln {\\frac {\\{\\mathrm {S} \\}^{\\sigma }\\{\\mathrm {T} \\}^{\\tau }}{\\{\\mathrm {A} \\}^{\\alpha }\\{\\mathrm {B} \\}^{\\beta }}}}\n  \n\n  \n    \n      \n        \n          \u2211\n          \n            i\n            =\n            1\n          \n          \n            k\n          \n        \n        \n          \u03bc\n          \n            i\n          \n          \n            \u2296\n          \n        \n        \n          \u03bd\n          \n            i\n          \n        \n        =\n        \n          \u0394\n          \n            \n              r\n            \n          \n        \n        \n          G\n          \n            \u2296\n          \n        \n      \n    \n    {\\displaystyle \\sum _{i=1}^{k}\\mu _{i}^{\\ominus }\\nu _{i}=\\Delta _{\\mathrm {r} }G^{\\ominus }}\n  \n:\nwhich is the standard Gibbs energy change for the rea", "          \u2296\n          \n        \n      \n    \n    {\\displaystyle \\sum _{i=1}^{k}\\mu _{i}^{\\ominus }\\nu _{i}=\\Delta _{\\mathrm {r} }G^{\\ominus }}\n  \n:\nwhich is the standard Gibbs energy change for the reaction that can be calculated using thermodynamical tables.\nThe reaction quotient is defined as:\n\n  \n    \n      \n        \n          Q\n          \n            \n              r\n            \n          \n        \n        =\n        \n          \n            \n              {\n              \n                S\n              \n              \n                }\n                \n                  \u03c3\n                \n              \n              {\n              \n                T\n              \n              \n                }\n                \n                  \u03c4\n                \n              \n            \n            \n              {\n              \n                A\n              \n              \n                }\n                \n                  \u03b1\n                \n              \n              {\n           ", "      \n              {\n              \n                A\n              \n              \n                }\n                \n                  \u03b1\n                \n              \n              {\n              \n                B\n              \n              \n                }\n                \n                  \u03b2\n                \n              \n            \n          \n        \n      \n    \n    {\\displaystyle Q_{\\mathrm {r} }={\\frac {\\{\\mathrm {S} \\}^{\\sigma }\\{\\mathrm {T} \\}^{\\tau }}{\\{\\mathrm {A} \\}^{\\alpha }\\{\\mathrm {B} \\}^{\\beta }}}}\n  \n\nTherefore,\n\n  \n    \n      \n        \n          \n            (\n            \n              \n                \n                  d\n                  G\n                \n                \n                  d\n                  \u03be\n                \n              \n            \n            )\n          \n          \n            T\n            ,\n            p\n          \n        \n        =\n        \n          \u0394\n          \n            \n              r\n            \n          \n   ", "   \n            )\n          \n          \n            T\n            ,\n            p\n          \n        \n        =\n        \n          \u0394\n          \n            \n              r\n            \n          \n        \n        \n          G\n          \n            T\n            ,\n            p\n          \n        \n        =\n        \n          \u0394\n          \n            \n              r\n            \n          \n        \n        \n          G\n          \n            \u2296\n          \n        \n        +\n        R\n        T\n        ln\n        \u2061\n        \n          Q\n          \n            \n              r\n            \n          \n        \n      \n    \n    {\\displaystyle \\left({\\frac {dG}{d\\xi }}\\right)_{T,p}=\\Delta _{\\mathrm {r} }G_{T,p}=\\Delta _{\\mathrm {r} }G^{\\ominus }+RT\\ln Q_{\\mathrm {r} }}\n  \n\nAt equilibrium:\n\n  \n    \n      \n        \n          \n            (\n            \n              \n                \n                  d\n                  G\n                \n                \n                  d\n                 ", "  \n      \n        \n          \n            (\n            \n              \n                \n                  d\n                  G\n                \n                \n                  d\n                  \u03be\n                \n              \n            \n            )\n          \n          \n            T\n            ,\n            p\n          \n        \n        =\n        \n          \u0394\n          \n            \n              r\n            \n          \n        \n        \n          G\n          \n            T\n            ,\n            p\n          \n        \n        =\n        0\n      \n    \n    {\\displaystyle \\left({\\frac {dG}{d\\xi }}\\right)_{T,p}=\\Delta _{\\mathrm {r} }G_{T,p}=0}\n  \n\nleading to:\n\n  \n    \n      \n        0\n        =\n        \n          \u0394\n          \n            \n              r\n            \n          \n        \n        \n          G\n          \n            \u2296\n          \n        \n        +\n        R\n        T\n        ln\n        \u2061\n        \n          K\n          \n            \n              e\n         ", "   \n        \n        \n          G\n          \n            \u2296\n          \n        \n        +\n        R\n        T\n        ln\n        \u2061\n        \n          K\n          \n            \n              e\n              q\n            \n          \n        \n      \n    \n    {\\displaystyle 0=\\Delta _{\\mathrm {r} }G^{\\ominus }+RT\\ln K_{\\mathrm {eq} }}\n  \n\nand\n\n  \n    \n      \n        \n          \u0394\n          \n            \n              r\n            \n          \n        \n        \n          G\n          \n            \u2296\n          \n        \n        =\n        \u2212\n        R\n        T\n        ln\n        \u2061\n        \n          K\n          \n            \n              e\n              q\n            \n          \n        \n      \n    \n    {\\displaystyle \\Delta _{\\mathrm {r} }G^{\\ominus }=-RT\\ln K_{\\mathrm {eq} }}\n  \n\nObtaining the value of the standard Gibbs energy change, allows the calculation of the equilibrium constant.\n\n\n=== Addition of reactants or products ===\nFor a reactional system at equilibrium: Qr = Keq; \u03be = \u03beeq.\n\nIf ", "lue of the standard Gibbs energy change, allows the calculation of the equilibrium constant.\n\n\n=== Addition of reactants or products ===\nFor a reactional system at equilibrium: Qr = Keq; \u03be = \u03beeq.\n\nIf the activities of constituents are modified, the value of the reaction quotient changes and becomes different from the equilibrium constant: Qr \u2260 Keq \n  \n    \n      \n        \n          \n            (\n            \n              \n                \n                  d\n                  G\n                \n                \n                  d\n                  \u03be\n                \n              \n            \n            )\n          \n          \n            T\n            ,\n            p\n          \n        \n        =\n        \n          \u0394\n          \n            \n              r\n            \n          \n        \n        \n          G\n          \n            \u2296\n          \n        \n        +\n        R\n        T\n        ln\n        \u2061\n        \n          Q\n          \n            \n              r\n            \n   ", "     \n        \n          G\n          \n            \u2296\n          \n        \n        +\n        R\n        T\n        ln\n        \u2061\n        \n          Q\n          \n            \n              r\n            \n          \n        \n         \n      \n    \n    {\\displaystyle \\left({\\frac {dG}{d\\xi }}\\right)_{T,p}=\\Delta _{\\mathrm {r} }G^{\\ominus }+RT\\ln Q_{\\mathrm {r} }~}\n  \n and \n  \n    \n      \n        \n          \u0394\n          \n            \n              r\n            \n          \n        \n        \n          G\n          \n            \u2296\n          \n        \n        =\n        \u2212\n        R\n        T\n        ln\n        \u2061\n        \n          K\n          \n            e\n            q\n          \n        \n         \n      \n    \n    {\\displaystyle \\Delta _{\\mathrm {r} }G^{\\ominus }=-RT\\ln K_{eq}~}\n  \n then \n  \n    \n      \n        \n          \n            (\n            \n              \n                \n                  d\n                  G\n                \n                \n                  d\n                  \u03be\n        ", "       \n          \n            (\n            \n              \n                \n                  d\n                  G\n                \n                \n                  d\n                  \u03be\n                \n              \n            \n            )\n          \n          \n            T\n            ,\n            p\n          \n        \n        =\n        R\n        T\n        ln\n        \u2061\n        \n          (\n          \n            \n              \n                Q\n                \n                  \n                    r\n                  \n                \n              \n              \n                K\n                \n                  \n                    e\n                    q\n                  \n                \n              \n            \n          \n          )\n        \n         \n      \n    \n    {\\displaystyle \\left({\\frac {dG}{d\\xi }}\\right)_{T,p}=RT\\ln \\left({\\frac {Q_{\\mathrm {r} }}{K_{\\mathrm {eq} }}}\\right)~}\n  \n\nIn simplifications where the change in reaction quotient is solely ", "   \n    {\\displaystyle \\left({\\frac {dG}{d\\xi }}\\right)_{T,p}=RT\\ln \\left({\\frac {Q_{\\mathrm {r} }}{K_{\\mathrm {eq} }}}\\right)~}\n  \n\nIn simplifications where the change in reaction quotient is solely due to the concentration changes, Qr is referred to as the mass-action ratio, and the ratio Qr/Keq is referred to as the disequilibrium ratio.\n\nIf activity of a reagent i increases \n  \n    \n      \n        \n          Q\n          \n            \n              r\n            \n          \n        \n        =\n        \n          \n            \n              \u220f\n              (\n              \n                a\n                \n                  j\n                \n              \n              \n                )\n                \n                  \n                    \u03bd\n                    \n                      j\n                    \n                  \n                \n              \n            \n            \n              \u220f\n              (\n              \n                a\n                \n                ", "  j\n                    \n                  \n                \n              \n            \n            \n              \u220f\n              (\n              \n                a\n                \n                  i\n                \n              \n              \n                )\n                \n                  \n                    \u03bd\n                    \n                      i\n                    \n                  \n                \n              \n            \n          \n        \n         \n        ,\n      \n    \n    {\\displaystyle Q_{\\mathrm {r} }={\\frac {\\prod (a_{j})^{\\nu _{j}}}{\\prod (a_{i})^{\\nu _{i}}}}~,}\n  \n the reaction quotient decreases. Then \n  \n    \n      \n        \n          Q\n          \n            \n              r\n            \n          \n        \n        <\n        \n          K\n          \n            \n              e\n              q\n            \n          \n        \n         \n      \n    \n    {\\displaystyle Q_{\\mathrm {r} }<K_{\\mathrm {eq} }~}\n  \n and \n  \n    \n      \n        \n        ", "  \n            \n              e\n              q\n            \n          \n        \n         \n      \n    \n    {\\displaystyle Q_{\\mathrm {r} }<K_{\\mathrm {eq} }~}\n  \n and \n  \n    \n      \n        \n          \n            (\n            \n              \n                \n                  d\n                  G\n                \n                \n                  d\n                  \u03be\n                \n              \n            \n            )\n          \n          \n            T\n            ,\n            p\n          \n        \n        <\n        0\n         \n      \n    \n    {\\displaystyle \\left({\\frac {dG}{d\\xi }}\\right)_{T,p}<0~}\n  \n The reaction will shift to the right (i.e. in the forward direction, and thus more products will form).\nIf activity of a product j increases, then \n  \n    \n      \n        \n          Q\n          \n            \n              r\n            \n          \n        \n        >\n        \n          K\n          \n            \n              e\n              q\n            \n          \n     ", "         Q\n          \n            \n              r\n            \n          \n        \n        >\n        \n          K\n          \n            \n              e\n              q\n            \n          \n        \n         \n      \n    \n    {\\displaystyle Q_{\\mathrm {r} }>K_{\\mathrm {eq} }~}\n  \n and \n  \n    \n      \n        \n          \n            (\n            \n              \n                \n                  d\n                  G\n                \n                \n                  d\n                  \u03be\n                \n              \n            \n            )\n          \n          \n            T\n            ,\n            p\n          \n        \n        >\n        0\n         \n      \n    \n    {\\displaystyle \\left({\\frac {dG}{d\\xi }}\\right)_{T,p}>0~}\n  \n The reaction will shift to the left (i.e. in the reverse direction, and thus less products will form).\nNote that activities and equilibrium constants are dimensionless numbers.\n\n\n=== Treatment of activity ===\nThe expression for the equilibrium consta", "everse direction, and thus less products will form).\nNote that activities and equilibrium constants are dimensionless numbers.\n\n\n=== Treatment of activity ===\nThe expression for the equilibrium constant can be rewritten as the product of a concentration quotient, Kc and an activity coefficient quotient, \u0393.\n\n  \n    \n      \n        K\n        =\n        \n          \n            \n              [\n              \n                S\n              \n              \n                ]\n                \n                  \u03c3\n                \n              \n              [\n              \n                T\n              \n              \n                ]\n                \n                  \u03c4\n                \n              \n              .\n              .\n              .\n            \n            \n              [\n              \n                A\n              \n              \n                ]\n                \n                  \u03b1\n                \n              \n              [\n              \n                B\n  ", "             \n                A\n              \n              \n                ]\n                \n                  \u03b1\n                \n              \n              [\n              \n                B\n              \n              \n                ]\n                \n                  \u03b2\n                \n              \n              .\n              .\n              .\n            \n          \n        \n        \u00d7\n        \n          \n            \n              \n                \n                  \n                    \u03b3\n                    \n                      \n                        S\n                      \n                    \n                  \n                \n                \n                  \u03c3\n                \n              \n              \n                \n                  \n                    \u03b3\n                    \n                      \n                        T\n                      \n                    \n                  \n                \n                \n                  \u03c4\n         ", "  \u03b3\n                    \n                      \n                        T\n                      \n                    \n                  \n                \n                \n                  \u03c4\n                \n              \n              .\n              .\n              .\n            \n            \n              \n                \n                  \n                    \u03b3\n                    \n                      \n                        A\n                      \n                    \n                  \n                \n                \n                  \u03b1\n                \n              \n              \n                \n                  \n                    \u03b3\n                    \n                      \n                        B\n                      \n                    \n                  \n                \n                \n                  \u03b2\n                \n              \n              .\n              .\n              .\n            \n          \n        \n        =\n        \n          K\n       ", "           \n                \n                  \u03b2\n                \n              \n              .\n              .\n              .\n            \n          \n        \n        =\n        \n          K\n          \n            \n              c\n            \n          \n        \n        \u0393\n      \n    \n    {\\displaystyle K={\\frac {[\\mathrm {S} ]^{\\sigma }[\\mathrm {T} ]^{\\tau }...}{[\\mathrm {A} ]^{\\alpha }[\\mathrm {B} ]^{\\beta }...}}\\times {\\frac {{\\gamma _{\\mathrm {S} }}^{\\sigma }{\\gamma _{\\mathrm {T} }}^{\\tau }...}{{\\gamma _{\\mathrm {A} }}^{\\alpha }{\\gamma _{\\mathrm {B} }}^{\\beta }...}}=K_{\\mathrm {c} }\\Gamma }\n  \n\n[A] is the concentration of reagent A, etc. It is possible in principle to obtain values of the activity coefficients, \u03b3. For solutions, equations such as the Debye\u2013H\u00fcckel equation or extensions such as Davies equation Specific ion interaction theory or Pitzer equations may be used. However this is not always possible. It is common practice to assume that \u0393 is a constant, and to use the co", "ns such as Davies equation Specific ion interaction theory or Pitzer equations may be used. However this is not always possible. It is common practice to assume that \u0393 is a constant, and to use the concentration quotient in place of the thermodynamic equilibrium constant. It is also general practice to use the term equilibrium constant instead of the more accurate concentration quotient. This practice will be followed here.\nFor reactions in the gas phase partial pressure is used in place of concentration and fugacity coefficient in place of activity coefficient. In the real world, for example, when making ammonia in industry, fugacity coefficients must be taken into account. Fugacity, f, is the product of partial pressure and fugacity coefficient. The chemical potential of a species in the real gas phase is given by\n\n  \n    \n      \n        \u03bc\n        =\n        \n          \u03bc\n          \n            \u2296\n          \n        \n        +\n        R\n        T\n        ln\n        \u2061\n        \n          ", " real gas phase is given by\n\n  \n    \n      \n        \u03bc\n        =\n        \n          \u03bc\n          \n            \u2296\n          \n        \n        +\n        R\n        T\n        ln\n        \u2061\n        \n          (\n          \n            \n              f\n              \n                b\n                a\n                r\n              \n            \n          \n          )\n        \n        =\n        \n          \u03bc\n          \n            \u2296\n          \n        \n        +\n        R\n        T\n        ln\n        \u2061\n        \n          (\n          \n            \n              p\n              \n                b\n                a\n                r\n              \n            \n          \n          )\n        \n        +\n        R\n        T\n        ln\n        \u2061\n        \u03b3\n      \n    \n    {\\displaystyle \\mu =\\mu ^{\\ominus }+RT\\ln \\left({\\frac {f}{\\mathrm {bar} }}\\right)=\\mu ^{\\ominus }+RT\\ln \\left({\\frac {p}{\\mathrm {bar} }}\\right)+RT\\ln \\gamma }\n  \n\nso the general expression defining an equilibrium constant is valid fo", "+RT\\ln \\left({\\frac {f}{\\mathrm {bar} }}\\right)=\\mu ^{\\ominus }+RT\\ln \\left({\\frac {p}{\\mathrm {bar} }}\\right)+RT\\ln \\gamma }\n  \n\nso the general expression defining an equilibrium constant is valid for both solution and gas phases.\n\n\n=== Concentration quotients ===\nIn aqueous solution, equilibrium constants are usually determined in the presence of an \"inert\" electrolyte such as sodium nitrate, NaNO3, or potassium perchlorate, KClO4. The ionic strength of a solution is given by\n\n  \n    \n      \n        I\n        =\n        \n          \n            1\n            2\n          \n        \n        \n          \u2211\n          \n            i\n            =\n            1\n          \n          \n            N\n          \n        \n        \n          c\n          \n            i\n          \n        \n        \n          z\n          \n            i\n          \n          \n            2\n          \n        \n      \n    \n    {\\displaystyle I={\\frac {1}{2}}\\sum _{i=1}^{N}c_{i}z_{i}^{2}}\n  \n\nwhere ci and zi stand for the con", "  z\n          \n            i\n          \n          \n            2\n          \n        \n      \n    \n    {\\displaystyle I={\\frac {1}{2}}\\sum _{i=1}^{N}c_{i}z_{i}^{2}}\n  \n\nwhere ci and zi stand for the concentration and ionic charge of ion type i, and the sum is taken over all the N types of charged species in solution. When the concentration of dissolved salt is much higher than the analytical concentrations of the reagents, the ions originating from the dissolved salt determine the ionic strength, and the ionic strength is effectively constant. Since activity coefficients depend on ionic strength, the activity coefficients of the species are effectively independent of concentration. Thus, the assumption that \u0393 is constant is justified. The concentration quotient is a simple multiple of the equilibrium constant.\n\n  \n    \n      \n        \n          K\n          \n            \n              c\n            \n          \n        \n        =\n        \n          \n            K\n            \u0393\n          \n ", "uilibrium constant.\n\n  \n    \n      \n        \n          K\n          \n            \n              c\n            \n          \n        \n        =\n        \n          \n            K\n            \u0393\n          \n        \n      \n    \n    {\\displaystyle K_{\\mathrm {c} }={\\frac {K}{\\Gamma }}}\n  \n\nHowever, Kc will vary with ionic strength. If it is measured at a series of different ionic strengths, the value can be extrapolated to zero ionic strength. The concentration quotient obtained in this manner is known, paradoxically, as a thermodynamic equilibrium constant.\nBefore using a published value of an equilibrium constant in conditions of ionic strength different from the conditions used in its determination, the value should be adjusted.\n\n\n=== Metastable mixtures ===\nA mixture may appear to have no tendency to change, though it is not at equilibrium. For example, a mixture of SO2 and O2 is metastable as there is a kinetic barrier to formation of the product, SO3.\n\n2 SO2 + O2 \u21cc 2 SO3\nThe barrier can b", "ncy to change, though it is not at equilibrium. For example, a mixture of SO2 and O2 is metastable as there is a kinetic barrier to formation of the product, SO3.\n\n2 SO2 + O2 \u21cc 2 SO3\nThe barrier can be overcome when a catalyst is also present in the mixture as in the contact process, but the catalyst does not affect the equilibrium concentrations.\nLikewise, the formation of bicarbonate from carbon dioxide and water is very slow under normal conditions\n\nCO2 + 2 H2O  \u21cc  HCO\u22123 + H3O+\nbut almost instantaneous in the presence of the catalytic enzyme carbonic anhydrase.\n\n\n== Pure substances ==\nWhen pure substances (liquids or solids) are involved in equilibria their activities do not appear in the equilibrium constant because their numerical values are considered one.\nApplying the general formula for an equilibrium constant to the specific case of a dilute solution of acetic acid in water one obtains\n\nCH3CO2H + H2O \u21cc CH3CO2\u2212 + H3O+\n\n  \n    \n      \n        \n          K\n          \n            ", "a for an equilibrium constant to the specific case of a dilute solution of acetic acid in water one obtains\n\nCH3CO2H + H2O \u21cc CH3CO2\u2212 + H3O+\n\n  \n    \n      \n        \n          K\n          \n            \n              c\n            \n          \n        \n        =\n        \n          \n            \n              [\n              \n                \n                  C\n                  \n                    H\n                    \n                      3\n                    \n                  \n                  C\n                  \n                    O\n                    \n                      2\n                    \n                  \n                \n                \n                  \u2212\n                \n              \n              ]\n              [\n              \n                \n                  \n                    H\n                    \n                      3\n                    \n                  \n                  O\n                \n                \n                  +\n                \n", " \n                    H\n                    \n                      3\n                    \n                  \n                  O\n                \n                \n                  +\n                \n              \n              ]\n            \n            \n              [\n              \n                C\n                \n                  H\n                  \n                    3\n                  \n                \n                C\n                \n                  O\n                  \n                    2\n                  \n                \n                H\n              \n              ]\n              [\n              \n                \n                  H\n                  \n                    2\n                  \n                \n                O\n              \n              ]\n            \n          \n        \n      \n    \n    {\\displaystyle K_{\\mathrm {c} }={\\frac {\\mathrm {[{CH_{3}CO_{2}}^{-}][{H_{3}O}^{+}]} }{\\mathrm {[{CH_{3}CO_{2}H}][{H_{2}O}]} }}}\n  \n\nFor all but very concen", "       \n          \n        \n      \n    \n    {\\displaystyle K_{\\mathrm {c} }={\\frac {\\mathrm {[{CH_{3}CO_{2}}^{-}][{H_{3}O}^{+}]} }{\\mathrm {[{CH_{3}CO_{2}H}][{H_{2}O}]} }}}\n  \n\nFor all but very concentrated solutions, the water can be considered a \"pure\" liquid, and therefore it has an activity of one. The equilibrium constant expression is therefore usually written as\n\n  \n    \n      \n        K\n        =\n        \n          \n            \n              [\n              \n                \n                  C\n                  \n                    H\n                    \n                      3\n                    \n                  \n                  C\n                  \n                    O\n                    \n                      2\n                    \n                  \n                \n                \n                  \u2212\n                \n              \n              ]\n              [\n              \n                \n                  \n                    H\n                    \n       ", "              \n                  \u2212\n                \n              \n              ]\n              [\n              \n                \n                  \n                    H\n                    \n                      3\n                    \n                  \n                  O\n                \n                \n                  +\n                \n              \n              ]\n            \n            \n              [\n              \n                C\n                \n                  H\n                  \n                    3\n                  \n                \n                C\n                \n                  O\n                  \n                    2\n                  \n                \n                H\n              \n              ]\n            \n          \n        \n        =\n        \n          K\n          \n            \n              c\n            \n          \n        \n      \n    \n    {\\displaystyle K={\\frac {\\mathrm {[{CH_{3}CO_{2}}^{-}][{H_{3}O}^{+}]} }{\\mathrm {[{CH_{3}CO_{2}H}]", "\n          K\n          \n            \n              c\n            \n          \n        \n      \n    \n    {\\displaystyle K={\\frac {\\mathrm {[{CH_{3}CO_{2}}^{-}][{H_{3}O}^{+}]} }{\\mathrm {[{CH_{3}CO_{2}H}]} }}=K_{\\mathrm {c} }}\n  \n.\nA particular case is the self-ionization of water\n\n2 H2O \u21cc H3O+ + OH\u2212\nBecause water is the solvent, and has an activity of one, the self-ionization constant of water is defined as\n\n  \n    \n      \n        \n          K\n          \n            \n              w\n            \n          \n        \n        =\n        \n          [\n          \n            H\n            \n              +\n            \n          \n          ]\n          [\n          O\n          \n            H\n            \n              \u2212\n            \n          \n          ]\n        \n      \n    \n    {\\displaystyle K_{\\mathrm {w} }=\\mathrm {[H^{+}][OH^{-}]} }\n  \n\nIt is perfectly legitimate to write [H+] for the hydronium ion concentration, since the state of solvation of the proton is constant (in dilute solutions) and", "hrm {w} }=\\mathrm {[H^{+}][OH^{-}]} }\n  \n\nIt is perfectly legitimate to write [H+] for the hydronium ion concentration, since the state of solvation of the proton is constant (in dilute solutions) and so does not affect the equilibrium concentrations. Kw varies with variation in ionic strength and/or temperature.\nThe concentrations of H+ and OH\u2212 are not independent quantities. Most commonly [OH\u2212] is replaced by Kw[H+]\u22121 in equilibrium constant expressions which would otherwise include hydroxide ion.\nSolids also do not appear in the equilibrium constant expression, if they are considered to be pure and thus their activities taken to be one. An example is the Boudouard reaction:\n\n2 CO \u21cc CO2 + C\nfor which the equation (without solid carbon) is written as:\n\n  \n    \n      \n        \n          K\n          \n            \n              c\n            \n          \n        \n        =\n        \n          \n            \n              [\n              C\n              \n                O\n                \n  ", "          \n            \n              c\n            \n          \n        \n        =\n        \n          \n            \n              [\n              C\n              \n                O\n                \n                  2\n                \n              \n              ]\n            \n            \n              [\n              C\n              O\n              \n                ]\n                \n                  2\n                \n              \n            \n          \n        \n      \n    \n    {\\displaystyle K_{\\mathrm {c} }={\\frac {\\mathrm {[CO_{2}]} }{\\mathrm {[CO]^{2}} }}}\n  \n\n\n== Equilibria Among Multiple Reactions ==\nConsider the case of a dibasic acid H2A. When dissolved in water, the mixture will contain H2A, HA\u2212 and A2\u2212. This equilibrium can be split into two steps in each of which one proton is liberated.\n\n  \n    \n      \n        \n          \n            \n              \n                \n                  \n                    H\n                    \n                      2\n               ", "ton is liberated.\n\n  \n    \n      \n        \n          \n            \n              \n                \n                  \n                    H\n                    \n                      2\n                    \n                    \n                      \n                    \n                  \n                  A\n                  \n                    \n                      \n                        \n                          \n                            \n                              \u21bd\n                            \n                            \n                            \n                            \n                              \u2212\n                            \n                          \n                        \n                      \n                      \n                        \n                          \n                            \n                              \u2212\n                            \n                            \n                            \n                            \n          ", "            \n                            \n                              \u2212\n                            \n                            \n                            \n                            \n                              \u21c0\n                            \n                          \n                        \n                      \n                    \n                  \n                  \n                    HA\n                    \n                      \u2212\n                    \n                  \n                  +\n                  \n                    H\n                    \n                      +\n                    \n                  \n                \n                :\n              \n              \n                \n                  K\n                  \n                    1\n                  \n                \n                =\n                \n                  \n                    \n                      \n                        [\n                        HA\n                        \n      ", "\n                \n                =\n                \n                  \n                    \n                      \n                        [\n                        HA\n                        \n                          \u2212\n                        \n                        ]\n                      \n                      \n                        [\n                        \n                          H\n                          \n                            +\n                          \n                        \n                        ]\n                      \n                    \n                    \n                      \n                        [\n                        \n                          H\n                          \n                            2\n                          \n                          \n                            \n                          \n                        \n                        A\n                        ]\n                      \n                    \n            ", "          \n                            \n                          \n                        \n                        A\n                        ]\n                      \n                    \n                  \n                \n              \n            \n            \n              \n                \n                  \n                    HA\n                    \n                      \u2212\n                    \n                  \n                  \n                    \n                      \n                        \n                          \n                            \n                              \u21bd\n                            \n                            \n                            \n                            \n                              \u2212\n                            \n                          \n                        \n                      \n                      \n                        \n                          \n                            \n                              \u2212\n            ", "   \n                        \n                      \n                      \n                        \n                          \n                            \n                              \u2212\n                            \n                            \n                            \n                            \n                              \u21c0\n                            \n                          \n                        \n                      \n                    \n                  \n                  \n                    A\n                    \n                      2\n                      \u2212\n                    \n                  \n                  +\n                  \n                    H\n                    \n                      +\n                    \n                  \n                \n                :\n              \n              \n                \n                  K\n                  \n                    2\n                  \n                \n                =\n                \n          ", "         :\n              \n              \n                \n                  K\n                  \n                    2\n                  \n                \n                =\n                \n                  \n                    \n                      \n                        [\n                        \n                          A\n                          \n                            2\n                            \n                            \n                              \u2212\n                            \n                          \n                        \n                        ]\n                      \n                      \n                        [\n                        \n                          H\n                          \n                            +\n                          \n                        \n                        ]\n                      \n                    \n                    \n                      \n                        [\n                        HA\n             ", "                   \n                        ]\n                      \n                    \n                    \n                      \n                        [\n                        HA\n                        \n                          \u2212\n                        \n                        ]\n                      \n                    \n                  \n                \n              \n            \n          \n        \n      \n    \n    {\\displaystyle {\\begin{array}{rl}{\\ce {H2A <=> HA^- + H+}}:&K_{1}={\\frac {{\\ce {[HA-] [H+]}}}{{\\ce {[H2A]}}}}\\\\{\\ce {HA- <=> A^2- + H+}}:&K_{2}={\\frac {{\\ce {[A^{2-}] [H+]}}}{{\\ce {[HA-]}}}}\\end{array}}}\n  \n\nK1 and K2 are examples of stepwise equilibrium constants. The overall equilibrium constant, \u03b2D, is product of the stepwise constants.\n\n     \n  \n    \n      \n        \n          \u03b2\n          \n            \n              D\n            \n          \n        \n        =\n        \n          \n            \n              \n                [\n                \n              ", "       \n          \u03b2\n          \n            \n              D\n            \n          \n        \n        =\n        \n          \n            \n              \n                [\n                \n                  A\n                  \n                    2\n                    \n                    \n                      \u2212\n                    \n                  \n                \n                ]\n              \n              \n                \n                  [\n                  \n                    H\n                    \n                      +\n                    \n                  \n                  ]\n                \n                \n                  2\n                \n              \n            \n            \n              \n                [\n                \n                  H\n                  \n                    2\n                  \n                  \n                    \n                  \n                \n                A\n                ]\n              \n            \n          \n      ", " \n                    2\n                  \n                  \n                    \n                  \n                \n                A\n                ]\n              \n            \n          \n        \n        =\n        \n          K\n          \n            1\n          \n        \n        \n          K\n          \n            2\n          \n        \n      \n    \n    {\\displaystyle \\beta _{{\\ce {D}}}={\\frac {{\\ce {[A^{2-}] [H^+]^2}}}{{\\ce {[H_2A]}}}}=K_{1}K_{2}}\n  \n\nNote that these constants are dissociation constants because the products on the right hand side of the equilibrium expression are dissociation products. In many systems, it is preferable to use association constants.\n\n  \n    \n      \n        \n          \n            \n              \n                \n                  \n                    A\n                    \n                      2\n                      \u2212\n                    \n                  \n                  +\n                  \n                    H\n                    \n       ", "A\n                    \n                      2\n                      \u2212\n                    \n                  \n                  +\n                  \n                    H\n                    \n                      +\n                    \n                  \n                  \n                    \n                      \n                        \n                          \n                            \n                              \u21bd\n                            \n                            \n                            \n                            \n                              \u2212\n                            \n                          \n                        \n                      \n                      \n                        \n                          \n                            \n                              \u2212\n                            \n                            \n                            \n                            \n                              \u21c0\n                            \n  ", "                   \u2212\n                            \n                            \n                            \n                            \n                              \u21c0\n                            \n                          \n                        \n                      \n                    \n                  \n                  \n                    HA\n                    \n                      \u2212\n                    \n                  \n                \n                :\n              \n              \n                \n                  \u03b2\n                  \n                    1\n                  \n                \n                =\n                \n                  \n                    \n                      \n                        [\n                        HA\n                        \n                          \u2212\n                        \n                        ]\n                      \n                    \n                    \n                      \n                        [\n            ", "                     \u2212\n                        \n                        ]\n                      \n                    \n                    \n                      \n                        [\n                        \n                          A\n                          \n                            2\n                            \n                            \n                              \u2212\n                            \n                          \n                        \n                        ]\n                      \n                      \n                        [\n                        \n                          H\n                          \n                            +\n                          \n                        \n                        ]\n                      \n                    \n                  \n                \n              \n            \n            \n              \n                \n                  \n                    A\n                    \n                      2\n      ", "                 \n                \n              \n            \n            \n              \n                \n                  \n                    A\n                    \n                      2\n                      \u2212\n                    \n                  \n                  +\n                  2\n                  \n                  \n                    H\n                    \n                      +\n                    \n                  \n                  \n                    \n                      \n                        \n                          \n                            \n                              \u21bd\n                            \n                            \n                            \n                            \n                              \u2212\n                            \n                          \n                        \n                      \n                      \n                        \n                          \n                            \n                         ", "                      \n                        \n                      \n                      \n                        \n                          \n                            \n                              \u2212\n                            \n                            \n                            \n                            \n                              \u21c0\n                            \n                          \n                        \n                      \n                    \n                  \n                  \n                    H\n                    \n                      2\n                    \n                    \n                      \n                    \n                  \n                  A\n                \n                :\n              \n              \n                \n                  \u03b2\n                  \n                    2\n                  \n                \n                =\n                \n                  \n                    \n                      \n             ", "          \u03b2\n                  \n                    2\n                  \n                \n                =\n                \n                  \n                    \n                      \n                        [\n                        \n                          H\n                          \n                            2\n                          \n                          \n                            \n                          \n                        \n                        A\n                        ]\n                      \n                    \n                    \n                      \n                        [\n                        \n                          A\n                          \n                            2\n                            \n                            \n                              \u2212\n                            \n                          \n                        \n                        ]\n                      \n                      \n                       ", "                      \u2212\n                            \n                          \n                        \n                        ]\n                      \n                      \n                        \n                          [\n                          \n                            H\n                            \n                              +\n                            \n                          \n                          ]\n                        \n                        \n                          2\n                        \n                      \n                    \n                  \n                \n              \n            \n          \n        \n      \n    \n    {\\displaystyle {\\begin{array}{ll}{\\ce {A^2- + H+ <=> HA-}}:&\\beta _{1}={\\frac {{\\ce {[HA^-]}}}{{\\ce {[A^{2-}] [H+]}}}}\\\\{\\ce {A^2- + 2H+ <=> H2A}}:&\\beta _{2}={\\frac {{\\ce {[H2A]}}}{{\\ce {[A^{2-}] [H+]^2}}}}\\end{array}}}\n  \n\n\u03b21 and \u03b22 are examples of association constants. Clearly \u03b21 = \u20601/K2\u2060 and \u03b22 = \u20601/\u03b2D\u2060; log \u03b21 = p", "\\ce {A^2- + 2H+ <=> H2A}}:&\\beta _{2}={\\frac {{\\ce {[H2A]}}}{{\\ce {[A^{2-}] [H+]^2}}}}\\end{array}}}\n  \n\n\u03b21 and \u03b22 are examples of association constants. Clearly \u03b21 = \u20601/K2\u2060 and \u03b22 = \u20601/\u03b2D\u2060; log \u03b21 = pK2 and log \u03b22 = pK2 + pK1\nFor multiple equilibrium systems, also see: theory of Response reactions.\n\n\n== Effect of temperature ==\nThe effect of changing temperature on an equilibrium constant is given by the van 't Hoff equation\n\n  \n    \n      \n        \n          \n            \n              d\n              ln\n              \u2061\n              K\n            \n            \n              d\n              T\n            \n          \n        \n        =\n        \n          \n            \n              \u0394\n              \n                H\n                \n                  \n                    m\n                  \n                \n                \n                  \u2296\n                \n              \n            \n            \n              R\n              \n                T\n                \n                  2", "  \n                \n                \n                  \u2296\n                \n              \n            \n            \n              R\n              \n                T\n                \n                  2\n                \n              \n            \n          \n        \n      \n    \n    {\\displaystyle {\\frac {d\\ln K}{dT}}={\\frac {\\Delta H_{\\mathrm {m} }^{\\ominus }}{RT^{2}}}}\n  \n\nThus, for exothermic reactions (\u0394H is negative), K decreases with an increase in temperature, but, for endothermic reactions, (\u0394H is positive) K increases with an increase in temperature. An alternative formulation is\n\n  \n    \n      \n        \n          \n            \n              d\n              ln\n              \u2061\n              K\n            \n            \n              d\n              (\n              \n                T\n                \n                  \u2212\n                  1\n                \n              \n              )\n            \n          \n        \n        =\n        \u2212\n        \n          \n            \n          ", "               \n                  \u2212\n                  1\n                \n              \n              )\n            \n          \n        \n        =\n        \u2212\n        \n          \n            \n              \u0394\n              \n                H\n                \n                  \n                    m\n                  \n                \n                \n                  \u2296\n                \n              \n            \n            R\n          \n        \n      \n    \n    {\\displaystyle {\\frac {d\\ln K}{d(T^{-1})}}=-{\\frac {\\Delta H_{\\mathrm {m} }^{\\ominus }}{R}}}\n  \n\nAt first sight this appears to offer a means of obtaining the standard molar enthalpy of the reaction by studying the variation of K with temperature. In practice, however, the method is unreliable because error propagation almost always gives very large errors on the values calculated in this way.\n\n\n== Effect of electric and magnetic fields ==\nThe effect of electric field on equilibrium has been studied by Manfred Eigen among others.", "gives very large errors on the values calculated in this way.\n\n\n== Effect of electric and magnetic fields ==\nThe effect of electric field on equilibrium has been studied by Manfred Eigen among others.\n\n\n== Types of equilibrium ==\n\nEquilibrium can be broadly classified as heterogeneous and homogeneous equilibrium. Homogeneous equilibrium consists of reactants and products belonging in the same phase whereas heterogeneous equilibrium comes into play for reactants and products in different phases.  \n\nIn the gas phase: rocket engines\nThe industrial synthesis such as ammonia in the Haber\u2013Bosch process (depicted right) takes place through a succession of equilibrium steps including adsorption processes\nAtmospheric chemistry\nSeawater and other natural waters: chemical oceanography\nDistribution between two phases\nlog D distribution coefficient: important for pharmaceuticals where lipophilicity is a significant property of a drug\nLiquid\u2013liquid extraction, Ion exchange, Chromatography\nSolubility", "tween two phases\nlog D distribution coefficient: important for pharmaceuticals where lipophilicity is a significant property of a drug\nLiquid\u2013liquid extraction, Ion exchange, Chromatography\nSolubility product\nUptake and release of oxygen by hemoglobin in blood\nAcid\u2013base equilibria: acid dissociation constant, hydrolysis, buffer solutions, indicators, acid\u2013base homeostasis\nMetal\u2013ligand complexation: sequestering agents, chelation therapy, MRI contrast reagents, Schlenk equilibrium\nAdduct formation: host\u2013guest chemistry, supramolecular chemistry, molecular recognition, dinitrogen tetroxide\nIn certain oscillating reactions, the approach to equilibrium is not asymptotically but in the form of a damped oscillation .\nThe related Nernst equation in electrochemistry gives the difference in electrode potential as a function of redox concentrations.\nWhen molecules on each side of the equilibrium are able to further react irreversibly in secondary reactions, the final product ratio is determined ", "de potential as a function of redox concentrations.\nWhen molecules on each side of the equilibrium are able to further react irreversibly in secondary reactions, the final product ratio is determined according to the Curtin\u2013Hammett principle.\nIn these applications, terms such as stability constant, formation constant, binding constant, affinity constant, association constant and dissociation constant are used. In biochemistry, it is common to give units for binding constants, which serve to define the concentration units used when the constant's value was determined.\n\n\n== Composition of a mixture ==\nWhen the only equilibrium is that of the formation of a 1:1 adduct as the composition of a mixture, there are many ways that the composition of a mixture can be calculated. For example, see ICE table for a traditional method of calculating the pH of a solution of a weak acid.\nThere are three approaches to the general calculation of the composition of a mixture at equilibrium.\n\nThe most basi", " table for a traditional method of calculating the pH of a solution of a weak acid.\nThere are three approaches to the general calculation of the composition of a mixture at equilibrium.\n\nThe most basic approach is to manipulate the various equilibrium constants until the desired concentrations are expressed in terms of measured equilibrium constants (equivalent to measuring chemical potentials) and initial conditions.\nMinimize the Gibbs energy of the system.\nSatisfy the equation of mass balance. The equations of mass balance are simply statements that demonstrate that the total concentration of each reactant must be constant by the law of conservation of mass.\n\n\n=== Mass-balance equations ===\nIn general, the calculations are rather complicated or complex. For instance, in the case of a dibasic acid, H2A dissolved in water the two reactants can be specified as the conjugate base, A2\u2212, and the proton, H+. The following equations of mass-balance could apply equally well to a base such as ", "asic acid, H2A dissolved in water the two reactants can be specified as the conjugate base, A2\u2212, and the proton, H+. The following equations of mass-balance could apply equally well to a base such as 1,2-diaminoethane, in which case the base itself is designated as the reactant A:\n\n  \n    \n      \n        \n          T\n          \n            \n              A\n            \n          \n        \n        =\n        \n          [\n          A\n          ]\n          +\n          [\n          H\n          A\n          ]\n          +\n          [\n          \n            H\n            \n              2\n            \n          \n          A\n          ]\n        \n        \n      \n    \n    {\\displaystyle T_{\\mathrm {A} }=\\mathrm {[A]+[HA]+[H_{2}A]} \\,}\n  \n\n  \n    \n      \n        \n          T\n          \n            \n              H\n            \n          \n        \n        =\n        \n          [\n          H\n          ]\n          +\n          [\n          H\n          A\n          ]\n          +\n          2\n          [\n     ", "         H\n            \n          \n        \n        =\n        \n          [\n          H\n          ]\n          +\n          [\n          H\n          A\n          ]\n          +\n          2\n          [\n          \n            H\n            \n              2\n            \n          \n          A\n          ]\n          \u2212\n          [\n          O\n          H\n          ]\n        \n        \n      \n    \n    {\\displaystyle T_{\\mathrm {H} }=\\mathrm {[H]+[HA]+2[H_{2}A]-[OH]} \\,}\n  \n\nwith TA the total concentration of species A. Note that it is customary to omit the ionic charges when writing and using these equations.\nWhen the equilibrium constants are known and the total concentrations are specified there are two equations in two unknown \"free concentrations\" [A] and [H]. This follows from the fact that [HA] = \u03b21[A] [H], [H2A] = \u03b22[A] [H]2 and [OH] = Kw[H]\u22121\n\n  \n    \n      \n        \n          T\n          \n            \n              A\n            \n          \n        \n        =\n        \n          [\n          ", "\u03b21[A] [H], [H2A] = \u03b22[A] [H]2 and [OH] = Kw[H]\u22121\n\n  \n    \n      \n        \n          T\n          \n            \n              A\n            \n          \n        \n        =\n        \n          [\n          A\n          ]\n        \n        +\n        \n          \u03b2\n          \n            1\n          \n        \n        \n          [\n          A\n          ]\n          [\n          H\n          ]\n        \n        +\n        \n          \u03b2\n          \n            2\n          \n        \n        \n          \n            [\n            A\n            ]\n            [\n            H\n            ]\n          \n          \n            2\n          \n        \n        \n      \n    \n    {\\displaystyle T_{\\mathrm {A} }=\\mathrm {[A]} +\\beta _{1}\\mathrm {[A][H]} +\\beta _{2}\\mathrm {[A][H]} ^{2}\\,}\n  \n\n  \n    \n      \n        \n          T\n          \n            \n              H\n            \n          \n        \n        =\n        \n          [\n          H\n          ]\n        \n        +\n        \n          \u03b2\n          \n            1\n       ", "          \n            \n              H\n            \n          \n        \n        =\n        \n          [\n          H\n          ]\n        \n        +\n        \n          \u03b2\n          \n            1\n          \n        \n        \n          [\n          A\n          ]\n          [\n          H\n          ]\n        \n        +\n        2\n        \n          \u03b2\n          \n            2\n          \n        \n        \n          \n            [\n            A\n            ]\n            [\n            H\n            ]\n          \n          \n            2\n          \n        \n        \u2212\n        \n          K\n          \n            w\n          \n        \n        [\n        \n          H\n        \n        \n          ]\n          \n            \u2212\n            1\n          \n        \n        \n      \n    \n    {\\displaystyle T_{\\mathrm {H} }=\\mathrm {[H]} +\\beta _{1}\\mathrm {[A][H]} +2\\beta _{2}\\mathrm {[A][H]} ^{2}-K_{w}[\\mathrm {H} ]^{-1}\\,}\n  \n\nso the concentrations of the \"complexes\" are calculated from the free concentrations and t", "}=\\mathrm {[H]} +\\beta _{1}\\mathrm {[A][H]} +2\\beta _{2}\\mathrm {[A][H]} ^{2}-K_{w}[\\mathrm {H} ]^{-1}\\,}\n  \n\nso the concentrations of the \"complexes\" are calculated from the free concentrations and the equilibrium constants.\nGeneral expressions applicable to all systems with two reagents, A and B would be\n\n  \n    \n      \n        \n          T\n          \n            \n              A\n            \n          \n        \n        =\n        [\n        \n          A\n        \n        ]\n        +\n        \n          \u2211\n          \n            i\n          \n        \n        \n          p\n          \n            i\n          \n        \n        \n          \u03b2\n          \n            i\n          \n        \n        [\n        \n          A\n        \n        \n          ]\n          \n            \n              p\n              \n                i\n              \n            \n          \n        \n        [\n        \n          B\n        \n        \n          ]\n          \n            \n              q\n              \n                ", " \n                i\n              \n            \n          \n        \n        [\n        \n          B\n        \n        \n          ]\n          \n            \n              q\n              \n                i\n              \n            \n          \n        \n      \n    \n    {\\displaystyle T_{\\mathrm {A} }=[\\mathrm {A} ]+\\sum _{i}p_{i}\\beta _{i}[\\mathrm {A} ]^{p_{i}}[\\mathrm {B} ]^{q_{i}}}\n  \n\n  \n    \n      \n        \n          T\n          \n            \n              B\n            \n          \n        \n        =\n        [\n        \n          B\n        \n        ]\n        +\n        \n          \u2211\n          \n            i\n          \n        \n        \n          q\n          \n            i\n          \n        \n        \n          \u03b2\n          \n            i\n          \n        \n        [\n        \n          A\n        \n        \n          ]\n          \n            \n              p\n              \n                i\n              \n            \n          \n        \n        [\n        \n          B\n        \n        \n     ", "   \n        \n          ]\n          \n            \n              p\n              \n                i\n              \n            \n          \n        \n        [\n        \n          B\n        \n        \n          ]\n          \n            \n              q\n              \n                i\n              \n            \n          \n        \n      \n    \n    {\\displaystyle T_{\\mathrm {B} }=[\\mathrm {B} ]+\\sum _{i}q_{i}\\beta _{i}[\\mathrm {A} ]^{p_{i}}[\\mathrm {B} ]^{q_{i}}}\n  \n\nIt is easy to see how this can be extended to three or more reagents.\n\n\n==== Polybasic acids ====\n\nThe composition of solutions containing reactants A and H is easy to calculate as a function of p[H]. When [H] is known, the free concentration [A] is calculated from the mass-balance equation in A.\nThe diagram alongside, shows an example of the hydrolysis of the aluminium Lewis acid Al3+(aq) shows the species concentrations for a 5 \u00d7 10\u22126 M solution of an aluminium salt as a function of pH. Each concentration is shown as a percenta", "le of the hydrolysis of the aluminium Lewis acid Al3+(aq) shows the species concentrations for a 5 \u00d7 10\u22126 M solution of an aluminium salt as a function of pH. Each concentration is shown as a percentage of the total aluminium.\n\n\n==== Solution and precipitation ====\nThe diagram above illustrates the point that a precipitate that is not one of the main species in the solution equilibrium may be formed. At pH just below 5.5 the main species present in a 5 \u03bcM solution of Al3+ are aluminium hydroxides Al(OH)2+, AlOH+2 and Al13(OH)7+32, but on raising the pH Al(OH)3 precipitates from the solution. This occurs because Al(OH)3 has a very large lattice energy. As the pH rises more and more Al(OH)3 comes out of solution. This is an example of Le Ch\u00e2telier's principle in action: Increasing the concentration of the hydroxide ion causes more aluminium hydroxide to precipitate, which removes hydroxide from the solution. When the hydroxide concentration becomes sufficiently high the soluble aluminate", "tration of the hydroxide ion causes more aluminium hydroxide to precipitate, which removes hydroxide from the solution. When the hydroxide concentration becomes sufficiently high the soluble aluminate, Al(OH)\u22124, is formed.\nAnother common instance where precipitation occurs is when a metal cation interacts with an anionic ligand to form an electrically neutral complex. If the complex is hydrophobic, it will precipitate out of water. This occurs with the nickel ion Ni2+ and dimethylglyoxime, (dmgH2): in this case the lattice energy of the solid is not particularly large, but it greatly exceeds the energy of solvation of the molecule Ni(dmgH)2.\n\n\n=== Minimization of Gibbs energy ===\nAt equilibrium, at a specified temperature and pressure, and with no external forces, the Gibbs free energy G is at a minimum:\n\n  \n    \n      \n        d\n        G\n        =\n        \n          \u2211\n          \n            j\n            =\n            1\n          \n          \n            m\n          \n        \n        ", "s at a minimum:\n\n  \n    \n      \n        d\n        G\n        =\n        \n          \u2211\n          \n            j\n            =\n            1\n          \n          \n            m\n          \n        \n        \n          \u03bc\n          \n            j\n          \n        \n        \n        d\n        \n          N\n          \n            j\n          \n        \n        =\n        0\n      \n    \n    {\\displaystyle dG=\\sum _{j=1}^{m}\\mu _{j}\\,dN_{j}=0}\n  \n\nwhere \u03bcj is the chemical potential of molecular species j, and Nj is the amount of molecular species j. It may be expressed in terms of thermodynamic activity as:\n\n  \n    \n      \n        \n          \u03bc\n          \n            j\n          \n        \n        =\n        \n          \u03bc\n          \n            j\n          \n          \n            \u2296\n          \n        \n        +\n        R\n        T\n        ln\n        \u2061\n        \n          \n            A\n            \n              j\n            \n          \n        \n      \n    \n    {\\displaystyle \\mu _{j}=\\mu _{j}^{\\ominus }+", " +\n        R\n        T\n        ln\n        \u2061\n        \n          \n            A\n            \n              j\n            \n          \n        \n      \n    \n    {\\displaystyle \\mu _{j}=\\mu _{j}^{\\ominus }+RT\\ln {A_{j}}}\n  \n\nwhere \n  \n    \n      \n        \n          \u03bc\n          \n            j\n          \n          \n            \u2296\n          \n        \n      \n    \n    {\\displaystyle \\mu _{j}^{\\ominus }}\n  \n is the chemical potential in the standard state, R is the gas constant T is the absolute temperature, and Aj is the activity.\nFor a closed system, no particles may enter or leave, although they may combine in various ways. The total number of atoms of each element will remain constant. This means that the minimization above must be subjected to the constraints:\n\n  \n    \n      \n        \n          \u2211\n          \n            j\n            =\n            1\n          \n          \n            m\n          \n        \n        \n          a\n          \n            i\n            j\n          \n        \n        \n  ", "          \n            j\n            =\n            1\n          \n          \n            m\n          \n        \n        \n          a\n          \n            i\n            j\n          \n        \n        \n          N\n          \n            j\n          \n        \n        =\n        \n          b\n          \n            i\n          \n          \n            0\n          \n        \n      \n    \n    {\\displaystyle \\sum _{j=1}^{m}a_{ij}N_{j}=b_{i}^{0}}\n  \n\nwhere aij is the number of atoms of element i in molecule j and b0i is the total number of atoms of element i, which is a constant, since the system is closed. If there are a total of k types of atoms in the system, then there will be k such equations. If ions are involved, an additional row is added to the aij matrix specifying the respective charge on each molecule which will sum to zero.\nThis is a standard problem in optimisation, known as constrained minimisation. The most common method of solving it is using the method of Lagrange multipliers (altho", " molecule which will sum to zero.\nThis is a standard problem in optimisation, known as constrained minimisation. The most common method of solving it is using the method of Lagrange multipliers (although other methods may be used).\nDefine:\n\n  \n    \n      \n        \n          \n            G\n          \n        \n        =\n        G\n        +\n        \n          \u2211\n          \n            i\n            =\n            1\n          \n          \n            k\n          \n        \n        \n          \u03bb\n          \n            i\n          \n        \n        \n          (\n          \n            \n              \u2211\n              \n                j\n                =\n                1\n              \n              \n                m\n              \n            \n            \n              a\n              \n                i\n                j\n              \n            \n            \n              N\n              \n                j\n              \n            \n            \u2212\n            \n              b\n              \n  ", "  i\n                j\n              \n            \n            \n              N\n              \n                j\n              \n            \n            \u2212\n            \n              b\n              \n                i\n              \n              \n                0\n              \n            \n          \n          )\n        \n        =\n        0\n      \n    \n    {\\displaystyle {\\mathcal {G}}=G+\\sum _{i=1}^{k}\\lambda _{i}\\left(\\sum _{j=1}^{m}a_{ij}N_{j}-b_{i}^{0}\\right)=0}\n  \n\nwhere the \u03bbi are the Lagrange multipliers, one for each element. This allows each of the Nj and \u03bbj to be treated independently, and it can be shown using the tools of multivariate calculus that the equilibrium condition is given by\n\n  \n    \n      \n        0\n        =\n        \n          \n            \n              \u2202\n              \n                \n                  G\n                \n              \n            \n            \n              \u2202\n              \n                N\n                \n                  j\n           ", "       \n                \n                  G\n                \n              \n            \n            \n              \u2202\n              \n                N\n                \n                  j\n                \n              \n            \n          \n        \n        =\n        \n          \u03bc\n          \n            j\n          \n        \n        +\n        \n          \u2211\n          \n            i\n            =\n            1\n          \n          \n            k\n          \n        \n        \n          \u03bb\n          \n            i\n          \n        \n        \n          a\n          \n            i\n            j\n          \n        \n      \n    \n    {\\displaystyle 0={\\frac {\\partial {\\mathcal {G}}}{\\partial N_{j}}}=\\mu _{j}+\\sum _{i=1}^{k}\\lambda _{i}a_{ij}}\n  \n\n  \n    \n      \n        0\n        =\n        \n          \n            \n              \u2202\n              \n                \n                  G\n                \n              \n            \n            \n              \u2202\n              \n                \u03bb\n          ", " \n            \n              \u2202\n              \n                \n                  G\n                \n              \n            \n            \n              \u2202\n              \n                \u03bb\n                \n                  i\n                \n              \n            \n          \n        \n        =\n        \n          \u2211\n          \n            j\n            =\n            1\n          \n          \n            m\n          \n        \n        \n          a\n          \n            i\n            j\n          \n        \n        \n          N\n          \n            j\n          \n        \n        \u2212\n        \n          b\n          \n            i\n          \n          \n            0\n          \n        \n      \n    \n    {\\displaystyle 0={\\frac {\\partial {\\mathcal {G}}}{\\partial \\lambda _{i}}}=\\sum _{j=1}^{m}a_{ij}N_{j}-b_{i}^{0}}\n  \n\n(For proof see Lagrange multipliers.) This is a set of (m + k) equations in (m + k) unknowns (the Nj and the \u03bbi) and may, therefore, be solved for the equilibrium concentrations ", "}N_{j}-b_{i}^{0}}\n  \n\n(For proof see Lagrange multipliers.) This is a set of (m + k) equations in (m + k) unknowns (the Nj and the \u03bbi) and may, therefore, be solved for the equilibrium concentrations Nj as long as the chemical activities are known as functions of the concentrations at the given temperature and pressure. (In the ideal case, activities are proportional to concentrations.) (See Thermodynamic databases for pure substances.) Note that the second equation is just the initial constraints for minimization.\nThis method of calculating equilibrium chemical concentrations is useful for systems with a large number of different molecules. The use of k atomic element conservation equations for the mass constraint is straightforward, and replaces the use of the stoichiometric coefficient equations. The results are consistent with those specified by chemical equations. For example, if equilibrium is specified by a single chemical equation:,\n\n  \n    \n      \n        \n          \u2211\n        ", "equations. The results are consistent with those specified by chemical equations. For example, if equilibrium is specified by a single chemical equation:,\n\n  \n    \n      \n        \n          \u2211\n          \n            j\n            =\n            0\n          \n          \n            m\n          \n        \n        \n          \u03bd\n          \n            j\n          \n        \n        \n          R\n          \n            j\n          \n        \n        =\n        0\n      \n    \n    {\\displaystyle \\sum _{j=0}^{m}\\nu _{j}R_{j}=0}\n  \n\nwhere \u03bdj is the stoichiometric coefficient for the j th molecule (negative for reactants, positive for products) and Rj is the symbol for the j th molecule, a properly balanced equation will obey:\n\n  \n    \n      \n        \n          \u2211\n          \n            j\n            =\n            1\n          \n          \n            m\n          \n        \n        \n          a\n          \n            i\n            j\n          \n        \n        \n          \u03bd\n          \n            j\n          \n", "     1\n          \n          \n            m\n          \n        \n        \n          a\n          \n            i\n            j\n          \n        \n        \n          \u03bd\n          \n            j\n          \n        \n        =\n        0\n      \n    \n    {\\displaystyle \\sum _{j=1}^{m}a_{ij}\\nu _{j}=0}\n  \n\nMultiplying the first equilibrium condition by \u03bdj and using the above equation yields:\n\n  \n    \n      \n        0\n        =\n        \n          \u2211\n          \n            j\n            =\n            1\n          \n          \n            m\n          \n        \n        \n          \u03bd\n          \n            j\n          \n        \n        \n          \u03bc\n          \n            j\n          \n        \n        +\n        \n          \u2211\n          \n            j\n            =\n            1\n          \n          \n            m\n          \n        \n        \n          \u2211\n          \n            i\n            =\n            1\n          \n          \n            k\n          \n        \n        \n          \u03bd\n          \n            j\n  ", "m\n          \n        \n        \n          \u2211\n          \n            i\n            =\n            1\n          \n          \n            k\n          \n        \n        \n          \u03bd\n          \n            j\n          \n        \n        \n          \u03bb\n          \n            i\n          \n        \n        \n          a\n          \n            i\n            j\n          \n        \n        =\n        \n          \u2211\n          \n            j\n            =\n            1\n          \n          \n            m\n          \n        \n        \n          \u03bd\n          \n            j\n          \n        \n        \n          \u03bc\n          \n            j\n          \n        \n      \n    \n    {\\displaystyle 0=\\sum _{j=1}^{m}\\nu _{j}\\mu _{j}+\\sum _{j=1}^{m}\\sum _{i=1}^{k}\\nu _{j}\\lambda _{i}a_{ij}=\\sum _{j=1}^{m}\\nu _{j}\\mu _{j}}\n  \n\nAs above, defining \u0394G\n\n  \n    \n      \n        \u0394\n        G\n        =\n        \n          \u2211\n          \n            j\n            =\n            1\n          \n          \n            m\n          \n        \n       ", "ove, defining \u0394G\n\n  \n    \n      \n        \u0394\n        G\n        =\n        \n          \u2211\n          \n            j\n            =\n            1\n          \n          \n            m\n          \n        \n        \n          \u03bd\n          \n            j\n          \n        \n        \n          \u03bc\n          \n            j\n          \n        \n        =\n        \n          \u2211\n          \n            j\n            =\n            1\n          \n          \n            m\n          \n        \n        \n          \u03bd\n          \n            j\n          \n        \n        (\n        \n          \u03bc\n          \n            j\n          \n          \n            \u2296\n          \n        \n        +\n        R\n        T\n        ln\n        \u2061\n        (\n        {\n        \n          R\n          \n            j\n          \n        \n        }\n        )\n        )\n        =\n        \u0394\n        \n          G\n          \n            \u2296\n          \n        \n        +\n        R\n        T\n        ln\n        \u2061\n        \n          (\n          \n            \n        ", "        )\n        =\n        \u0394\n        \n          G\n          \n            \u2296\n          \n        \n        +\n        R\n        T\n        ln\n        \u2061\n        \n          (\n          \n            \n              \u220f\n              \n                j\n                =\n                1\n              \n              \n                m\n              \n            \n            {\n            \n              R\n              \n                j\n              \n            \n            \n              }\n              \n                \n                  \u03bd\n                  \n                    j\n                  \n                \n              \n            \n          \n          )\n        \n        =\n        \u0394\n        \n          G\n          \n            \u2296\n          \n        \n        +\n        R\n        T\n        ln\n        \u2061\n        (\n        \n          K\n          \n            c\n          \n        \n        )\n      \n    \n    {\\displaystyle \\Delta G=\\sum _{j=1}^{m}\\nu _{j}\\mu _{j}=\\sum _{j=1}^{m}\\nu _{j}(\\mu _{", "n\n        \u2061\n        (\n        \n          K\n          \n            c\n          \n        \n        )\n      \n    \n    {\\displaystyle \\Delta G=\\sum _{j=1}^{m}\\nu _{j}\\mu _{j}=\\sum _{j=1}^{m}\\nu _{j}(\\mu _{j}^{\\ominus }+RT\\ln(\\{R_{j}\\}))=\\Delta G^{\\ominus }+RT\\ln \\left(\\prod _{j=1}^{m}\\{R_{j}\\}^{\\nu _{j}}\\right)=\\Delta G^{\\ominus }+RT\\ln(K_{c})}\n  \n\nwhere Kc is the equilibrium constant, and \u0394G will be zero at equilibrium.\nAnalogous procedures exist for the minimization of other thermodynamic potentials.\n\n\n== See also ==\n\n\n== References ==\n\n\n== Further reading ==\n\nVan Zeggeren, F.; Storey, S. H. (1970). The Computation of Chemical Equilibria. Cambridge University Press. Mainly concerned with gas-phase equilibria.\nLeggett, D. J., ed. (1985). Computational Methods for the Determination of Formation Constants. Plenum Press.\nMartell, A. E.; Motekaitis, R. J. (1992). The Determination and Use of Stability Constants. Wiley-VCH.\n\n\n== External links ==\n Media related to Chemical equilibria at Wikimed", " Constants. Plenum Press.\nMartell, A. E.; Motekaitis, R. J. (1992). The Determination and Use of Stability Constants. Wiley-VCH.\n\n\n== External links ==\n Media related to Chemical equilibria at Wikimedia Commons"]