Page 1 : I PUC Chemical Equilibrium, Chemical Equilibrium, Introduction:, It is a common observation that most of the reactions when carried out in closed vessel do not go to, completion under given set of conditions of temperature and pressure. In fact all such cases in the initial stage,, only the reactants are present but as the reaction proceeds, the concentration of the reactant decreases and that, of products increases. Finally a stage is reached when no further change in concentration of reactants and, products and the reaction appears to have stopped. This state is called equilibrium state., Equilibrium can be established in both physical and chemical processes chemical equilibrium plays, very important role in numerous biological and environmental processes for example equilibrium involving, oxygen molecules and the protein haemoglobin plays a important role in the transport and delivery of oxygen, from lungs to muscles in our body., Chemical equilibrium:, A particular state of a reversible reaction in which the rate of forward reaction is equal to the rate of, backward reactions so that the concentrations of reactants and products remain constant is called chemical, equilibrium., Reversible reaction:, A reaction in which only the reactants react to form products but after the products react to form, reactants under the same conditions is called reversible reactions., , Irreversible reactions:, A reaction which cannot take place in the reverse direction is called an irreversible reaction., , Equilibrium:, A reaction (or a process) is said to be in equilibrium when the rate of forward reaction becomes equal, to the rate of backward reaction., Types of equilibrium:, There are two types of equilibrium, 1) Physical equilibrium, 2) Chemical equilibrium, 1) Physical equilibrium:, The equilibrium set up in physical process is called physical equilibrium OR Physical equilibrium in, which the opposing processes involves physical changes., Ex: Equilibrium involving melting of ice to water, dissociation of sugar in water., 2) Chemical equilibrium:, A particular state of a reversible reaction in which the rate of forward reaction is equal to the rate of, backward reactions so that the concentrations of reactants and products remain constant is called chemical, equilibrium., By Manik. R. W. (9008747357), , Personal Use Only, , Page 1

Page 2 : I PUC Chemical Equilibrium, Types of physical equilibrium:, There are five types of physical equilibrium, , The equilibrium in which the rate of melting of the solid is equal to the rate of freezing of the liquid is, called solid-liquid equilibrium., The equilibrium that exists between ice and water is an example of solid-liquid equilibrium., , The equilibrium in which the rate which the solute particles go into the solution phase is equal to, therate which solute particles the solution get into the solid phase is called solid – solution equilibrium., , The equilibrium in which the rate of vaporisation of a solid is equal to that of combustion of vapour, into solid is called solid – gas equilibrium., , The equilibrium in which rate of vaporisation is equal to the rate of condensation of vapour is called, liquid – vapour equilibrium., , The equilibrium in which the rate at which molecules of a gas go into a solution phase is equal to the, rate at which the molecule of the gas in the solution phase goes back to the gas phase is called gas – solution, equilibrium., Soda water in air tighten bottle., , General characteristics of physical and chemical equilibrium:, Dynamic nature of equilibrium:, Equilibrium is dynamic in nature not static that is even after equilibrium is attained, the forward, reaction as well as backward reaction takes place at equal speed., At equilibrium: The concentration of the reactants and products becomes constant., At chemical equilibrium: Can be established only in an closed system., Chemical equilibrium: Can be attained from either direction., , By Manik. R. W. (9008747357), , Personal Use Only, , Page 2

Page 3 : I PUC Chemical Equilibrium, Effect of catalyst on equilibrium:, A catalyst does not change the state of equilibrium. The only effect is that equilibrium is attained, quickly., Law of mass action:, The law of mass action was given by Norwegian chemists, Gildberg and Waage in 1864 to co-relate, the rate of a chemical reaction and the concentration of the reacting species., According to the law, “At a constant temperature, the rate of a chemical reaction is directly proportional to the product of the, molar concentration of the reactants each rise to a power equal to its corresponding stoichiometric coefficient, that appear in balanced chemical equation”., OR, “The law state that the rate at which a substance reacts is directly proportional to its active mass, hence, the rate of a chemical reaction is directly proportional to the product of active masses of all the reactants”., For example: Let us consider a reaction, According to rate law, rate of reaction  [ A]a [ B]b  k[ A]a [ B]b, Here a and b represent the stoichiometric coefficients of the reactants A and B, k is the rate constant., While square bracket indicates the molar concentration of the species involved in it., NOTE: Molar concentration means the number of moles of the substance divided by particles of the solution., It is also called active mass., Laws of chemical equilibrium and equilibrium constant:, Statement: For a reversible reaction in a state of equilibrium, the rate of the product of the molar, concentration of the products to that of the molar concentration term raised to a power equal to its, stoichiometric coefficient in the balanced chemical equation is a constant known as the equilibrium constant, provided the temperature us constant., Let us consider a general reversible reaction., By the law of mass action, the rate of forward reaction rf is given by, , rf  [ A]a [ B]b, rf  k f [ A]a [ B]b, ,     (1), , Where kf is the rate constant of the forward reaction. Similarly the rate of the backward reaction is., rb [C ]c [ D ]d, rb  kb [C ]c [ D]d, ,     (2), , Where kb is the rate constant of the backward reaction. But at equilibrium, The rate of forward reaction  The rate of backward reaction, rf  rb, ,     (3), , From equation (1) and (2), , k f [ A]a [ B]b  kb [C]c [ D]d, [C ]c [ D]d, , kb, [ A]a [ B]b, kf, , k , , But, , [C ]c [ D]d, [ A]a [ B]b, , By Manik. R. W. (9008747357), , kf, kb, , k, ,     (4), , k is called the equilibrium constant., Personal Use Only, , Page 3

Page 4 : I PUC Chemical Equilibrium, Definition of equilibrium constant (k):, It is defined as the ratio of the product of the equilibrium molar concentration of the products to that of, the reactants with each concentration term raised to a power equal to its stoichiometric coefficient in the, balanced chemical equation., OR, It is also defined as the ratio of the rate constant of the forward reaction to the rate constant of the, backward reaction., Reaction Quotient:, The concentration ratio that is the ratio of the product of concentration of the products to that of the, reactants is known as concentration quotient and denoted as Q., , [C ]c [ D]d, Q, [ A]a [ B]b, Equilibrium constant in terms of pressure:, For a reaction involving gases the partial pressure of the reactants and products are proportional to the, concentrations., Therefore the law of chemical equilibrium can be expressed in terms of P instead of concentrations., Thus for a general gaseous reversible reaction,, We can write the equilibrium constant as, , kp , , PCc PDd, PAa PBb, ,     (1), , Where PA, PB, PC and PD are the partial pressure of A, B,C and D respectively and kP is the equilibrium, constant in terms of P., Relationship between kp and kc:, From the ideal gas equation, nRT, P,  [ M ]RT, V, Now PA  [ A]RT , PB  [ B]RT ,, , kp , , PCc PDd, PAa PBb, , kp , , {[C ]RT }c {[ D]RT }d, {[ A]RT }a {[ B]RT }b, , Pc  [C ]RT ,, , PD  [ D]RT, , [C ][ D]RT c  d, [ A][ B]RT a b, [C ][ D], kp , RT ( c  d )( a b ), [ A][ B], , kp , , k p  kc RT, , ng, , Where ng = (Number of moles of gaseous products)-(Number of moles of gaseous reactants), i) If ng=0 k p  kc, ii) If ng=Positive k p  kc, iii) If ng=Negative k p  kc, By Manik. R. W. (9008747357), , Personal Use Only, , Page 4

Page 5 : I PUC Chemical Equilibrium, Units of kp and kc:, Unit of k p   units of pressure , , n, ,   atm , , n, , Unit of kc   units of concentration    molL1 , n, , n, , Types of equilibrium:, Homogenous equilibrium:, Equilibrium is said to be homogeneous if reactants and products are in the same phase., , Heterogeneous equilibrium:, Equilibrium is said to be heterogeneous if reactants and products are in the different phase., , Equilibrium constant for the reaction:, Equilibrium constant for the reversible reaction is the inverse of the equilibrium constant for the, reaction in the forward reaction., 1, k rev , kc, Numerical problems, Q 1) write the expression of equilibrium constant for the reaction, , Ans: solution, a), , kc , , b) kc , , [ HI ]2, [ H 2 ][ I 2 ], [ NO]4 [ H 2O]6, [ NH 3 ]4 [O2 ]5, , Q 2) the following concentration were obtained for the formation of NH3 from N2 and H2 at equilibrium, at 500K. [N2]=1.5x10-2M [H2]=3.0x10-2M and [NH3]= 1.2x10-2M calculate the equilibrium constant., Ans: Solution, For the reaction, , kc , , [ NH 3 ]2, [ N 2 ]1[ H 2 ]3, , kc , , [1.2 102 ]2, [1.5 102 ]1[3.0 102 ]3, , kc  0.106 104, , kc  1.06  103, , Q 3) At equilibrium the concentration of N2=3.0x10-3M, O2=4.2x10-3M and NO=2.8x10-3M in a sealed, vessel at 800K. What will be kc for the reaction., Ans: Solution:, , [ NO]2, kc , [ N 2 ]1[O2 ]1, , [2.8 103 ]2, kc , [3.0 103 ]1[4.2 103 ]1, , By Manik. R. W. (9008747357), , kc  0.622, , Personal Use Only, , Page 5

Page 6 : I PUC Chemical Equilibrium, Factors influencing Equilibrium constant:,  Equilibrium constant expression depends on the stoichiometry of the reaction not on its mechanism. Ex,, , SO2  O2 , kc , 2, SO3 , 2, , 1, , [SO ][O ] 2, kc  2 2, [SO3],  kc  ( kc ) 2, , kc  kc, ,  If a reaction is multiplied by 2, kc becomes square of the original value = (kc)2.,  If reaction is reversed, the value of kc is inversed = 1, kc,  If the reaction can be expressed in two steps or more, equilibrium constant is equal to the product of, equilibrium constants of the step reactions, e.g., , kc  k1  k2,  The numerical values of kc and kp for reactions change with change of temperature.,  Variation of kp values with temperature can be show by vant Hoff isochore as;, , k2 H  T2  T1 , , , , k1, R  TT, 1 2 ,  Where H is heat of reaction and T1, T2 are temperature in Kelvin, pressure is constant.,  Case I, if H=0, that is heat is neither evolved nor absorbed,, k, 2.303log 2  0, k1, 2.303log, , or, , k2, 1, k1, , or, , k2  k1, , That is equilibrium constant remains same at different temperatures.,  Case II, if H=+ve, that is heat is absorbed,, k, 2.303log 2  ve, k1, , or, , log k2  log k1, , or k2  k1, , That is equilibrium constant increases with increase in temperature.,  Case III, if H=-ve, that is heat is evolved., k, 2.303log 2  ve, k1, , or, , log k2  log k1, , or k2  k1, , That is equilibrium constant decreases with increases in temperature.,  Equilibrium constant values change with units of pressure and concentration. If n=0 then values of kp or, kc remain constant, irrespective of units., If n  0 then values of kp or kc change along with change in the unit used to express pressure or, concentration., By Manik. R. W. (9008747357), , Personal Use Only, , Page 6

Page 7 : I PUC Chemical Equilibrium, Applications of Equilibrium constant:, 1) Predicting the extent of a reaction:, The magnitude of equilibrium constant k indicates the extent to which a reaction can go. In other, words, it is a measure of the completion of a reversible reaction. Larger the value of k, greater will be the, equilibrium concentration of the components on the right hand side of reaction relative to those on the left, hand side, that is the reaction proceeds to a greater extent for example, consider the reaction:, , [CO2 ]2, k,  2.3 1022 molL1, 2, [CO] [O2 ], For this reaction, the value of k is very large which means that the reaction goes almost to completion, and it can be concluded that carbon dioxide is more stable than carbon monoxide and oxygen under these, conditions., 2) Predicting the direction of a reaction:, The equilibrium constant helps in predicting the direction in which a reaction can proceed at any stage., By substituting the concentration of substances that exist in a reaction mixture we can calculate the reaction, quotient, Q and comparing the value of Q with the equilibrium constant, k, we can predict whether the, reaction will proceed towards products or towards reactants., Case I. If Q < kc the reaction will proceeds in the forward direction., Case II. If Q > kc the reaction will proceeds in the backward direction, Case III. If Q = kc the reaction mixture is already at equilibrium., Relation of G and equilibrium constant:,  G0 is the free energy change when all the products and reactants are in their standard states.,  Although chemical equilibrium is defined kinetically (rate of forward and backward reaction is equal),, its properties can be studied thermodynamically, that is from the free energies (G) of the reactants and, products., G=Difference in Gibbs free energy between products and reactants. If G=negative, the reaction is, spontaneous that is reaction proceeds in the forward direction., If G=Positive, the reaction is non-spontaneous that is reaction proceeds in the backward direction., If G=0, reaction has achieved equilibrium, at this point, there is no longer any free energy left to drive the, reaction., A mathematical expression of this thermodynamic view of equilibrium can be described by the following, equation., , G  G 0  RT ln Q .......(1), Where G0=standard free energy difference between the products and reactants, Q is reaction quotient., T=Absolute temperature is Kelvin., R=Universal gas constant., At equilibrium, when G=0 and Q=k then equation (1) becomes., 0  G 0  RT ln k, or, , G  RT ln k, 0, , or, , G 0, ln k , RT, , Le Chaterlier’s principle:,  Le Chaterlier a French Chemist, studied the effect of concentration, temperature and pressure on several, equilibria and summed up his conclusion as:, By Manik. R. W. (9008747357), , Personal Use Only, , Page 7

Page 8 : I PUC Chemical Equilibrium, “When a system is in equilibrium, physical or chemical, is subjected to a change of concentration,, temperature or pressure, the equilibrium shifts in the direction that tends to undo the effect of the change”.,  According to this principle it can be concluded that, - An increase in pressure always favours the reaction where volume or moles decreases., - An increase in temperature always favours the endothermic process, - An increase in concentration of reactants always favours the forward reaction., - An increase in concentration of products always favours the backward reaction.,  Consider the equilibrium:, , , - An increase in pressure to this equilibrium will favour forward reaction., - As it is an exothermic process, increase in temperature will favour backward reaction., - Increase in concentration of reactants favours forward reaction, whereas increase in concentration of, products will favour backward reaction.,  On the basis of Le Chaterlier’s principle, to get a better yield of NH3 in above equilibrium concentration, of reactants and pressure should be high while temperature should be kept low and NH3 formed should be, continuously removed., Le Chaterlier’s principle on physical equilibria:,  Effect of pressure on boiling point: An increase in pressure on solvent, vapour equilibrium,, favour the backward reaction., That is boiling point of solvent will increase.,  Effect of pressure on melting points: An increase in pressure on a solid, liquid equilibrium,, favours the backward reaction because volume of liquid is more than solid hence melting point of solid, will increase., Exception: For ice, water equilibrium, melting point is forward on applying pressure because, volume of ice is greater than that of water.,  Effect of pressure on solubility of gases: an increase in pressure on gas+solvent, solution, equilibrium, always increase the solubility of gases as volume of forward reaction is less than backward, reaction.,  Henry’s law: explain quantitatively the solubility of gases in liquids it states. The amount of gas, dissolved per unit volume of solvent is directly proportional to its pressure., a( s ) P( s ) or a( s )  kP( s ), Where k is Henry’s constant, its value depends on the nature of gas solvent and temperature., Henry’s law is valid only at low concentration and low pressures for gases which do not react chemically, with the solvent., Volume of gas dissolved (V(s)) in a given mass of solvent at any temperature is directly proportional to, the pressure (P(s)) of the gas above the solvent, that is,, V( s ) P( s ),  Effect of temperature on solubility of solid: Solute+solvent, solution; H=+ve an increase in, temperature will increase the solubility, solute+solvent, solution; H=-ve, an increase in, temperature will show a decrease in solubility.,  Effect of adding an inert gas: in a reaction at equilibrium, np=nv, if an inert gas is added constant, volume or constant pressure, there is no change in equilibrium., In a reaction where np>nv addition of an inert gas has no effect at constant volume but, at constant, pressure, the equilibrium shifts in the forward direction e.g. dissociation of PCl5., , By Manik. R. W. (9008747357), , Personal Use Only, , Page 8

Page 9 : I PUC Chemical Equilibrium, Le Chaterllier’s principle on chemical equilibria:, Consider the reaction,, , ,  Effect of temperature: the dissociation of PCl5 is endothermic. Hence increase of temperature increases, in the rate of formation of PCl3 and Cl2.,  Effect of pressure: increase of pressure will favour the backward reaction, as it is accompanied by an, increase in the number of molecules.,  Effect of concentration: Addition of PCl3 or Cl2 or removal of PCl5 will shift the equilibrium in forward, direction., , ,  Effect of temperature: the dissociation of HI is an endothermic reaction. Hence increase of temperature, favours the formation of H2 and I2.,  Effect of pressure: As the number of reactant molecules is equal to the number of product molecules,, hence pressure has no effect on the equilibrium.,  Effect of concentration: Addition of HI or removal of H2 and I2 favour the forward reaction., Reaction, , Favourable condition for forward direction, High pressure low temperature and isolation of NH3 by, liquification., High pressure, low temperature and isolation of SO2., High temperature and isolation of NO., Low pressure, high temperature and isolation of PCl5 and Cl2., High temperature and isolation of H2 and I2., , Q) For a reaction in the equilibrium state the rate constant for the forward and backward reactions are, 2.38x10-4 and 8.15x10-5 respectively. What is the value of the equilibrium., Ans: Solution:, kf, 2.38 104, k, k, k  2.92, kb, 8.15 105, Ionic equilibrium:, The equilibrium involving ions in aqueous solution is called ionic equilibrium., Degree of ionisation or degree of dissociation:, Degree of ionisation is the fraction of total number of molecules of an electrolyte which ionises into, ions, it is denoted by ., Number of molecules of electrolyte which ionises, , Totatl number of molecules of the electrolyte, The degree of dissociation for weak electrolyte:, The reaction of acetic acid with water OR Hydrolysis of acetic acid., , Initial concentration, Final concentration, , C, C(1-), , By Manik. R. W. (9008747357), , 0, C, Personal Use Only, , 0, C, Page 9

Page 10 : I PUC Chemical Equilibrium, CH 3COO    H 3O  , ka , CH 3COOH , , ka , , C  C, C (1   ), , ka , , C 2, C (1   ), , ka , , 2, 1, , Since  for CH3COOH is very small as compared to 1.,  1- may be taken 1, k a  C 2, , 2 , , ka, C, , , , ka, C, , Ostwald’s dilution law:, “The degree of dissociation of a weak electrolyte is inversely proportional to the square root of its molar, concentration”., Concept of Acid and Bases:, There are three theories to explain about the concept of acid and bases., 1) Arrhenius theory, 2) Bronsted Lowry theory., 3) Lewis theory., 1) Arrhenius theory:, According to Arrhenius, Acid: An acid is a substance, which produce hydrogen ion in the aqueous solution., , Base: Base is a substance, which produce hydroxyl ions in the aqueous solution., , Bronsted - Lowry concept:, According to this theory, Acid: Acid is a substance which has a tendency to donate a proton to any other substance., Base: Base is a substance which can accept the proton., HCl is losses proton hence it is an acid NH3 accepts a proton and it is a base. OR Acid is a proton donor and, base is proton acceptor., , Conjugate base of the acid:, An acid after donating the proton becomes base, which is called conjugate bases of the acid., Cl- is the conjugate base of HCl., Conjugate acid of the base:, A base after accepting the proton behaves as an acid which is called conjugate acid of the base., NH4+ is the conjugate acid of ammonia., By Manik. R. W. (9008747357), , Personal Use Only, , Page 10

Page 11 : I PUC Chemical Equilibrium, Conjugate acid-base pair:, “A pair of an acid and a base that differs by a proton is called conjugated acid base pair”., , Lewis theory:, According to Lewis, Acid: An acid is a substance which can accept electron pair., Base: Base is a substance which can donate electron pair., Ex: For Lewis Acids:,  Natural molecules BF3, AlCl3, FeCl3,  Simple cation Fe3+, Ag+, H+, etc., Ex: Lewis bases: NH3, H2O, R–OH, F-, CN-, H- etc., Electrolytes:, Compounds that conduct electricity either in solution state or in fused state are called electrolytes., , Electrolytes are of two types, 1) Strong electrolyte, 2) Weak electrolyte, 1) Strong electrolyte:, Electrolyte in which dissociation proceeds to almost completion are called strong electrolyte., Ex: All salts like NaCl, KCl, MgSO4 etc, Strong acids like HCl, H2SO4, HNO3 etc, Strong bases like NaOH and KOH, 2) Weak electrolyte: Electrolytes in which dissociation is partial is called weak electrolyte., Ex: Weak acids like H2CO3, CH3COOH etc., Weak bases like Al(OH)3, NH4OH, Ca(OH)2 etc., Ionic products of water:, Consider a conjugate acid-base pair., , ka , ,  NH 3   H  ,  NH 4 , ,  NH 4  OH  , kb ,  NH 3 , , .......(1), , .......(2), , Adding equation (1) and (2), kw   H   OH  , ........(3), From equation (1), (2) and (3), we get, kw  ka  kb, , This shows that the product of dissociation constant for an acid and its conjugate base is the ionic, product of water., By Manik. R. W. (9008747357), , Personal Use Only, , Page 11

Page 12 : I PUC Chemical Equilibrium, Definition of ionic product of water:, It is defined as the product of molar concentration of hydrogen and hydroxyl ions., Concept of PH:,  A new term (PH) was used by ssorensen to express the hydrogen ion concentration., It is defined as the negative logarithm to base ten of the molar hydrogen ion concentration of an, aqueous solution., P H   log10  H  , , H   10 p, For [H+]=10-7 solution is neutral PH=7, [H+]>10-7 solution is acidic PH<7, [H+]<10-7 solution is alkaline or basic PH>7,  PH is a measure of acidic nature of a solution, but as the PH goes up the acidic nature decreases.,  When PH decreases by a single unit the acidic nature shows a ten fold increases and when P H goes, down by two unit acidic nature increases by a factor of 100., H, , PH Scale:, PH scale is a negative logarithmic scale used to express hydronium ion concentration in molarity., Prove that PH + POH = 14 = pkw at 298K, We know that,  H   OH    kw  1014, Taking log on both sides, log  H   OH    log kw  log1014, , Multiply both sides by negative sign,  log  H   OH     log kw   log1014,  log  H      log[OH  ]   log kw   log1014, , p H  pOH  14 at 298K, Dissociation constant of weak acids and weak bases:, Let us consider the dissociation of a weak acid HA as., At t=0, C, 0, 0, At t , C(1-α), 0, 0, Dissociation constant of acid, ,  H    A , ka ,  HA, C  C, C 2 2, C 2, ka , ka , C (1   ), C (1   ), 1, Similarlly for the dissociation of weak base BOH as, ka , ,  B   OH  , kb ,  BOH , , By Manik. R. W. (9008747357), , kb , , C 2, 1, , Personal Use Only, , Page 12

Page 13 : I PUC Chemical Equilibrium, Dissociation constant for polyprotic acids and bases:, These acids involve number of steps to ionize completely where number of steps is equal to the, number of replaceable hydrogen atoms. The value of ionisation constant of each step is definite and constant., , The overall dissociation constant (k) is given as k = k1 x k2 x k3 where k1> k2>k3., Relative strength of weak acids and weak bases:, Let us consider a weak acid HA, which undergoes hydrolysis as, At t=0, At t, , C, C(1-α), , 0, Cα, , 0, Cα, , ka,  kaC, C, If value of ka1, ka2 and C1, C2 for two different weak acids are known we can compare the relative strength of, two weak acids., ,   H 3O   C  C, , Relative strength of acids , , ka1, [ H 3O  ]1, C1, , , , [ H 3O ]2, ka2, C2, , Similarly for two different weak bases, kb1, [OH  ]1, C1, , , , [OH ]2, kb2, C2, , Relation between ka and kb:, We know that, ka x kb = 10-14, For conjugate acid base pair taking negative log10,  log ka  ( log kb )   log1014, , pka  pkb  14, pka  pkb  pkw, , Dissociation constant of acid ka:, It is defined as “the ratio of product of molar concentration of dissociated ions to the molar concentration of, undissociated acid molecules at a given temperature., Definition of dissociation constant of base (kb):, It is defined as the ratio of product of molar concentration of dissociated ions to the molar concentration of, undissociated base molecules., Definition of pka: Pka is defined as negative logarithm to the base 10 of dissociation constant of acid., pka   log10 ka, , Definition of pkb: Pkb is defined as negative logarithm to the base 10 of dissociation constant of base., pkb   log10 kb, , By Manik. R. W. (9008747357), , Personal Use Only, , Page 13

Page 14 : I PUC Chemical Equilibrium, Common ion effect:, The suppression of degree of dissociation of weak electrolyte by adding strong electrolyte having, common ion is called common ion effect., Ex: Consider a weak electrolyte say acetic acid. The dissociation equilibrium is, , Let a little of sodium acetate be added to the solution, sodium acetate being a strong electrolyte, dissociate completely as, So on adding of sodium acetate in the equilibrium increases. According to Le-Chaterlier principle, to, oppose this increase in concentration of CH3COO- ions the equilibrium favours backward reaction and shifts, to left. As a result [H+] decreases and [CH3COOH] increases. Thus the degree of dissociation of CH3COOH is, suppressed by the addition of CH3COONa., Hydrolysis of salts:, It was observed that different salts give different types of solution in water. Ex NaCl gives a neutral, solution, CuSO4 gives an acidic solution where as Na2CO3 gives an alkaline solution., This difference in behaviour was explained in terms of hydrolysis by Arrhenius., Definition: hydrolysis can be defined as the process in which a salt react with water to give back the acid and, base., , Types of salts undergo (hydrolysis):,  Salts of strong acid and strong bases.,  Salt of weak acids and strong bases.,  Salt of strong acids and weak bases.,  Salt of weak acid and weak bases., Salt of strong acid and strong bases:, In the salt of strong acid and strong base no hydrolysis takes place, only ionisation takes place., ,  The resulting solution is neutral, [H+] = [OH-], Salt of weak acids and strong bases:, , Since OH- ions remains in solution the resulting solution will be alkaline., Salt of strong acids and weak bases:, Since H+ ions remains in solution the resulting solution will be acidic in nature., By Manik. R. W. (9008747357), , Personal Use Only, , Page 14

Page 15 : I PUC Chemical Equilibrium, Salt of weak acids and weak bases:, , This involves both cationic and anionic hydrolysis and as both acid and base produced are weak, the, resulting solution is almost neutral., Buffer solutions:, The pH of a solution changes when acid or base is added to it. However in certain chemical reactions, or biological processes, the pH of the medium has to be kept constant. This is achieved with the help of buffer, solution. A Buffer solution may be defined as;, “A solution which resists a change in the pH value on the addition of a small amount of acid or base to it”., Types of buffer:, Simple buffer:, Simple buffers are solutions of weak acid and weak base in water. Ex CH3COO-, NH4CN etc., proteins, and amino acids a mixture of an acid salt and normal salt. OR In simply these are the salts of weak acids and, weak bases., Ex: Na2HPO4 + Na3PO4., Mixed buffers: The mixed buffers as the name suggests are the mixture of two components., Basic buffers: Mixture contains equimolar amounts of a weak base and its salt with strong acid., Ex: NH4OH and NH4Cl, Acidic buffers: Mixture contains equimolar amount of a weak acid and its salt with a strong base., Ex: NaHCO3 + H2CO3., Handerson – Hassel batch equation:, It is used to calculate the pH of a buffer solution. Ex a buffer structure of HA and NaA (acidic buffer), can be calculated as fallows., , By law of mass action, , [ H  ][ A ], [ HA], Taking log on both sides, ka , , [H  ] , , log[ H  ]  log ka  log, , [ HA], [ A ], , pH  pka  log, , ka [ HA], [ A ],  log[ H  ]   log ka  log, , [ HA], [ A ], , [ Salt ], [ Acid ], , Buffer capacity:, Buffer capacity is the property of a buffer to resist a change in pH. It is defined as the number of moles of the, acid or base which, when added to 1 litre of the solution changes its pH by unity. That is, Marks of acid or base added / litre, Buffer Capacity , Changein pH, , Buffer Capacity , , dn, dpH, , By Manik. R. W. (9008747357), , Personal Use Only, , Page 15

Page 16 : I PUC Chemical Equilibrium, Table blow gives some common buffers and their nature:, Buffer Pairs, pH, Nature, CH3COOH + CH3COO 4.74, Acidic, 2H2C2O4 + C2O4, 1.42, Strongly acidic, 2HCO3 + CO3, 7.38, Basic, 2H3BO3 + BO3, 7.0 – 9.1 Basic, +, NH4 + NH3, 9.25, Basic, Solubility product:,  Certain compounds like AgCl, BaSO4, PbSO4 etc. Are sparingly soluble and thus solution form a, saturated solution in water.,  Since their solubilities are low, even their saturated solution is very dilute and the salt get ionised to a, small extent. An equilibrium exist between the ions and undissolved salt., Or a general formula:,  Applying law of mass action, [ A y ] x [ B  x ] y, k, [ Ax By ], , k[ Ax By ]  [ A y ]x [ B x ]y, Or, Since concentration of solid is taken to be constant, therefore:, [ A y ]x [ B x ]y  ksp, Ksp is called the solubility product and is constant for a given solute.,  Solubility product is defined as the product of molar concentrations of ions (formed at a given, temperature) raised to the power equal to the number of times each occurs in the equilibrium constant.,  Solubility product value varies with temperature as solubilities of salts vary with change in temperature.,  Solubility products of a few compounds can be expressed as;, , BaSO4 : ksp  [ Ba2 ][SO42 ], Fe(OH )3 : ksp  [ Fe3 ][OH  ]3, PbI3 : ksp  [ Pb2 ][ I  ]2, Relation between solubility and solubility product:,  Solubility is a term applicable to all solutes and it can be decreased by addition of a common ion.,  Solubility product is significant only for sparingly soluble electrolytes and is constant at a given, temperature.,  If solubility of a sparingly soluble salt, Ax, By is S moles/litre, then, , Thus, [ A y ]  xS and [ B  x ]  yS, , ksp  [ A y ][ B x ], ksp  ( xS ) y ( yS ) x, y, x, ( x y ), or x  y  S, , By Manik. R. W. (9008747357), , Personal Use Only, , Page 16

Page 17 : I PUC Chemical Equilibrium, Types of salts, , Example, , 1:1 (AB), , AgCl, BaSO4 etc, , 1:2 (AB2), , PbCl2, Ca(OH)2 etc, , 2:1 (A2B), 1:3 (AB3), , Ag2CrO4, Ag2CO3 etc, Fe(OH)3, Al(OH)3 etc, , Relationship between solubility, (S) and solubility product (ksp), S  k sp or k sp  S 2, S, , ksp, , S, , ksp, , S, , ksp, , 4, , or ksp  4S 2, , 4, , or ksp  4S 2, , 27, , or ksp  27S 2, , Applications of solubility product:,  Solubility of sparingly soluble salts can be calculated from ksp value.,  Useful in predicting precipitation in ionic reactions. If the ionic product of electrolyte exceeds its k sp, the precipitation will occur.,  Qualitative analysis of mixture is based on the concept of solubility product and ion effect. Ex, precipitation of group–I reactants as chlorides, group–II as sulphides etc.,  In manufacture of NaHCO3 by solvay’s ammonia soda process where NH4HCO3 is added to brine to, facilitate separation of NaHCO3.,  Salting out soaps from their saturated solution is facilitated by adding sodium chloride., Ionic product and solubility product:, For a solution of a salt at a specified concentration, the product of the concentration of the ions, related, to the proper power is called the ionic product kip., For a saturated solution in aqueous solution in equilibrium with excess solid, the ionic product is equal, to its solubility product., Case I: kip=ksp saturated solution, that means no more solute can be dissolved., Case II: kip<ksp unsaturated solution, that means more salt can be dissolved in it., Case III: kip>ksp supersaturated solution, that means the solution is holding more salt than that can be, dissolved in it and precipitation takes place, which continues till the ionic product becomes equal to the, solubility product., Effect of common ion on solubility:, Addition of a common ion lowers the solubility of a sparingly soluble salt. Let S be the solubility of a, salt AgCl in pure water and suppose x gm equivalent of NaCl is added to a litre of the saturated solution. Let, now the solubility of AgCl becomes, , ksp  [ Ag  ][Cl  ]  S ' (S '  x), or, , ksp  S 2, , or, , S ' ( S '  x)  S 2, , When the addition of a common ion or an inert salt leads to a considerable increase in the solubility of, a salt, then it can be considered that a complex ion has been formed., Applications of common ion effect and solubility product:, 1) Precipitation of group I radicals:, The group reagents of this group is dilute HCl, the precipitation of radicals occurs because the ionic, product of the chlorides of these ions exceed their corresponding solubility products., 2) Precipitation of the sulphides of group II:, H2S is a weak electrolyte and used for the precipitation of various sulphides of group II and IV., , By Manik. R. W. (9008747357), , Personal Use Only, , Page 17

Page 18 : I PUC Chemical Equilibrium, H2S is padded in presence of HCl, HCl is a strong electrolyte. Thus in presence of HCl, ionization of, H2S (weak electrolyte) suppresses due to the presence of common ion (H+). As a result only the ionic product, of the sulphides of group II radicals exceed their corresponding solubility product and hence only group II, radicals are precipitated out leaving the sulphides of group III and IV etc., 3) Precipitation of the hydroxides of group III:, , The group reagent of group III is NH4OH, it is a weak electrolyte. Thus addition of NH4Cl suppresses, the ionization of NH4OH. As a result only ionic product of group III radicals exceed their corresponding, solubility product and hence only group III radicals are precipitated out leaving the hydroxides of other groups, in the solution., Problems on Chemical Equilibrium, Q 1) For the equilibrium, , the value of the equilibrium constant, Kc is, 3.75 x 10 at 1069K. Calculate the Kp for the reaction at this temperature?, Ans: Solution:, -6, , We know that, K p  Kc ( RT ), , ng, , For the above reaction, ng  npg  nRg, , n  (2  1)  2  1, , K p  3.75 106 (0.08311069), K p  0.000333, Q 2) Find the value of Kc for each of the following equilibria from the value of Kp:, , Ans: Solution:, , K p =1.8 x 102, , ng  npg  nRg, , n  3  2  1, K p  Kc ( RT ), , KC , , ng, , KP, 1.8 x 102, ,  4.33 104, ( RT )1 (0.0831 500)1, , Q 3) At 450K, KP=2.0 x 1010 bar for the given reaction at equilibrium, , ., , What is KC at this temperature?, , Ans: Solution:, For the given reaction,, , ng  npg  nRg, , n  2  3  1, K p  Kc ( RT ), , ng, , Or KC  K P ( RT ), , ng, , By Manik. R. W. (9008747357), , Personal Use Only, , Page 18

Page 19 : I PUC Chemical Equilibrium, KC  (2.0 100 bar )(0.0831L bar K 1 mol 1 )  ( 1) (450 K ), KC  7.48 1011 Lmol 1, Q 4) For the following equilibrium, KC =6.3 x1014 at 1000K., ., Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc for, the reverse reaction?, , Ans: Solution:, For the reverse reaction, 1, 1, KC' , ,  1.59 1015, 14, KC 6.3 10, ,  NH 3  O2   Write the balanced chemical, , 4, 6,  NO   H 2O , 4, , Q 5) the equilibrium constant expression for a gas reaction is K, , 5, , equation corresponding to this expression., , Ans: Solution:, Q 6) One mole of H2O and one mole of CO are taken in 10L vessel and heated to 725K. At equilibrium, 40% of, water (by mass) reacts with CO according to the equation., Calculate the equilibrium constant for the reaction., , ., , Ans: Solution:, At equilibrium, Initial moles, Equilibrium moles, ,  H 2O  , , 1, 1–x, , 1, 1 –x, , 1  0.40, molL1  0.06molL1, 10, , 0, x, , , , , , 0, x, , Active mass , , n, , V, , CO  0.06molL1, H2  , , 0.40, molL1  0.04molL1, 10, , CO2   0.04molL1,  H 2 CO2   0.04  0.04  0.44, K,  H 2O CO  0.06  0.06, Q 7) A sample of pure PCl5 was introduced into an evaculated vessel at 473K. After equilibrium was attained, concentration of PCl5 was found to be 0.5 x 10-1mol L-1. If value of KC is 8.3 x 10-3, what are the concentrations of, PCl5 and Cl2 at equilibrium?, , Ans: Solution:, Equilibrium concentration, ,  PCl3 Cl2 ,  PCl5 ,  x  x , 8.3 10 4 ,  0.05, , 0.05M, , X, , X, , KC , , x 2  8.3 104  0.05, x  2 102, , By Manik. R. W. (9008747357), , Personal Use Only, , Page 19

Page 20 : I PUC Chemical Equilibrium, ,  PCl3   2 102 M, , and Cl2   2 102 M, , Q 8) At 700K, equilibrium constant for the reaction:, is 54.8. If 0.5molL-1 of, HI(g) is present at equilibrium at 700K, what are the concentrations of H2(g) and I2(g) assuming that we initially, started with HI(g) and allowed it to reach equilibrium at 700K?, , Ans: Solution:, Equilibrium conc., , x, , x, , 0.5, ,  HI ,  H 2  I 2 , 2, 0.5, , KC ,  54.8,  x  x , 2, , KC , , x  0.068,  H2   0.068molL1, ,  I2   0.068molL1, Q 9) KP=0.04 atm at 899K for the equilibrium shown below. What is the equilibrium concentration of C2H6 when, it is placed in a flask at 4.0atm pressure and allowed to come to equilibrium?, , Ans: Solution:, Initial pressure, At equilibrium, , KP , , 4.0atm, 4-p, , 0, p, , 0, p, , PC2 H4  PH2, ,  0.04 , , PC2 H6, p2, 4 p, , p 2  0.16  0.04 p, , 0.04  0.0016  4(0.16) 0.04  0.89, , 2, 2, 0.80,  0.40, Taking positive value, p , 2, C2 H 6 eq  4  0.04atm  3.60atm, , p, , Q 10) Deduce the units of equilibrium constant KC and KP for the following reactions., , Ans: Solution:, ,  HI , KC ,  H 2  I 2 , 2, ,  HI  , KC ,  H 2  I 2 , 2, , (mol L1 )2, (mol L1 ) (mol L1 ), ,  nounit, , In above reaction KP = KC = no unit, , By Manik. R. W. (9008747357), , Personal Use Only, , Page 20

Page 21 : I PUC Chemical Equilibrium, NO2 , , (mol L1 ) 2, KC , ,  N 2O4  (mol L1 ), 2, , KC , , 2, PNO, 2, , PN 2O4, , , ,  mol L1, , (bar ) 2,  bar, bar, , Q 11) The value of KP for the reaction,, , is 3.0at 1000K. If initially, PCO2=0.48 bar and PCO=0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and, CO2., , Ans: Solution:, For the reraction, let ‘x’ be the decrease in pressure of CO2, then, Initial pressure, At equilibrium, , KP , 3, , 0.48bar, (0.48-x)bar, , 0, 2x bar, , 2, PCO, PCO2, , (2 x)2, (0.48  x), , 4 x 2  3(0.48  x), 4 x 2  1.44  3 x, 4 x 2  3x  1.44  0, a  4, b  3, c  1.44, ,  b , x, , b2  4ac, , , , 2a,  3  5.66 , x, 8, 2.66, x,  0.33, 8, , (as value of x cannot be negative hence we neglect that value), , The equilibrium partial pressure are,, , PCO  2(0.33)  0.66bar, PCO2  0.48  x  0.48  0.33  0.15bar, is 2 x10-3. At a given time, the composition of, reaction mixture is [A]=[B]=[C]=3x10 M. in which direction will the reaction proceed?, , Q 12) The value of KC for the reaction, -4-, , Ans: Solution:, For the given reaction, the reaction quotient QC is given be,,  BC , QC , 2,  A, As, ,  A   B  C   3104 M, , 3 10 3 10   1, ,  3 10 , 4, , QC, , 4, , 4 2, , By Manik. R. W. (9008747357), , Personal Use Only, , Page 21

Page 22 : I PUC Chemical Equilibrium, QC  KC so the reaction will proceed in the reverse direction., , As, , Q 13) A mixture of 1.57mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20L reaction, vessel at 500K. At this temperature, the equilibrium constant, KC for the reaction, is 1.7 x 102. Is the reaction mixture at equilibrium? If not, what is the, direction of the net reaction?, , Ans: Solution:, 2, ,  NH 3 , QC , 3,  N 2  H 2 , 2, ,  8.31, , molL1 , ,  20, , ,  2.38  103, 3,  1.57,  1.92, , molL1 , molL1 , ,  20,  20, , , Since, QC  K C , the reaction mixture is not at equilibrium., Since, QC  KC , the net reaction will be in the backward direction., , Q 14) Equilibrium constant, Kc for the reaction, , . At 500K is 0.061. at a, particular time, the analysis shows that composition of the reaction mixture of 3.0 mol L N2, 2.0 mol L-1 H2 and, 0.5mol L-1 NH3. Is the reaction at equilibrium? If not in which direction would the reaction tend to proceed to, reach equilibrium?, -1, , Ans: Solution:, Conc. (mol L-1), , 3.00, , QC , , 2.00, , 0.500, ,  NH 3    0.500   0.0104, 3, 3,  N 2  H 2   3.00  2.00 , 2, , 2, , Since, as QC  KC , the reaction is not at equilibrium., Since, QC  KC , the reaction would tend to proceed in forward direction to attain equilibrium., , Q 15) The value of G0 for the phosphorylation of glucose in glycolysis is 13.8kJmol-1. Find the vlue of, Kc at 298K., Ans: Solution:, 1, G 0  2.303RT log KC, or, G 0  2.303RT log, KC, , 13.8 1000  2.303  8.314  298log KC, log, , 1, 13.8 1000, , K C 2.303  8.314  298, , log, , 1,  2.418, KC, , 1,  Antilog (2.418), KC, 1,  2.623 102, KC, ,  KC , , 1,  3.814 103, 2, 2.623 10, , By Manik. R. W. (9008747357), , Personal Use Only, , Page 22

Page 23 : I PUC Chemical Equilibrium, Q 16) Hydrolysis of sucrose gives,, for the reaction is 2 x 103 at 300K. Calculate G0 at 300K., , . Equilibrium constant KC, , Ans: Solution:, G 0  2.303RT log KC, , G 0  2.303  8.314  300log 2 1013, G 0  2.303  8.314  300 13  0.3010, G 0  7.64 104 Jmol 1, Q 17) Calculate the pH of 0.001M HCl, Ans: Solution:, 0.001M, , 0.001M, ,  H    0.001M  103 M, , pH   log  H  , , pH   log103, pH  (3log10), pH  3, Q 18) Calculate the pH of 0.01M H2SO4 by assuming complete ionisation., Ans: Solution:, ,  H    0.02M  2  102 M, , pH   log  H  , pH   log 2 102   log 2  log10 2 , pH   log 2  (2 log10)   log 2  2, pH  0.3010  2, pH  1.6990, , Q 19) Calculate the pH of 0.02M Ba(OH)2 solution by assuming complete ionisation., Ans: Solution:, ,  OH    0.04M  4  102 M, pOH   log OH  , pOH   log  4 102 , pOH   log 4  log102 , pOH   log 4  (2 log10), pOH  0.6021  2, pOH  1.3979, We know that, , By Manik. R. W. (9008747357), , Personal Use Only, , Page 23

Page 24 : I PUC Chemical Equilibrium, pH  pOH  14, pH  14  pOH, pH  14  1.3979, pH  12.6021, , Q 20) Calculate the pH of 0.1M CH3COOH. The dissociation constant of acetic acid is 1.8 x 10-5., Ans: Solution:, Acetic acid is weak acid dissociated partially as, ,  H    Ka  C  1.8 105  0.1,  H    1.314 103 M, , pH   log  H  , , pH   log1.314 103  3  0.1587, pH  2.8431, Q 21) Calculate the pH of 0.002M acetic acid if it is 2.3% ionised at this solution., Ans: Solution:,  H    c, % dissociation 2.3, , ,  0.023 moles, 100, 100,  H    c,   H    0.002  0.023  4.6 105 M, pH   log  H  , , pH   log 4.6 105, , pH  (log 4.6  log105 )  0.6628  (5log10), pH  0.6628  5  4.3372, Q 22) Calculate the pH of 0.1M NH3 solution. The ionisation constant Kb for NH3 is 1.8 x 10-5 at 298K., Ans: Solution:, , OH    Kb  C  Kb   NH3 , , OH    1.8 105  0.1  1.8 106  1.341103 M, pOH   log OH  , pOH   log 1.341 103 , pOH   log1.341  (3log10), pOH  0.1587  3, , pOH  2.8413, pH  pOH  14, pH  14  pOH, pH  14  2.8413, , pH  11.1587, , By Manik. R. W. (9008747357), , Personal Use Only, , Page 24

Page 25 : I PUC Chemical Equilibrium, Q 23) The concentration of hydrogen ion in a sample of soft drink is 3.8 x 10-3M. What is its pH?, Ans: Solution:, pH   log  H  , , pH   log(3.8 103 )   log 3.8  3, pH  0.5798  3  2.4202  2.42, Q 24) The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen of hydrogen ion in, it., Ans: Solution:, pH   log  H  , Or, , log  H    3.76  4.24, ,  H     Antilog 4.24  1.738 104 M, Q 25) It has been found that the pH of a 0.01M solution of monobasic orgainic acid is 4.15. Calculate, the concentration of the anion, the ionisation constant of the acid and its pKa., Ans: Solution:, pH   log  H  , , 4.15   log  H  , log  H    4.15,  H     Antilog (5.85)  7.08 105 M,  A    H    7.08  105 M, ,  H    A   7.08 105  7.08 105 , Ka , ,  5.01107, 103,  HA, , pK a   log K a, pK a   log(5.01107 )  7  0.6998  6.302, , Q 26) The pH of 0.005M codeine (C15H21NO3) solution is 9.95. Calculate the ionisation constant and, pKb., Ans: Solution:, pH  9.95,  pOH  14  9.95  4.05, pOH   log OH  , ,  log OH    4.05, , log OH    4.05  5.95, OH    8.91105 M, ,  BH   OH   OH  , Kb , ,  B,  B, , By Manik. R. W. (9008747357), , 2, ,  BH    OH  , , Personal Use Only, , Page 25

Page 26 : I PUC Chemical Equilibrium, Kb, , 8.9110 , , ,  1.588 106, , Kb, , 8.9110 , , ,  1.588 106, , 5 2, , 5 105, , 5 2, , 5, , 5 10, pKb   log Kb, , pKb   log(1.588 106 )  6  0.1987  5.7991, , Q 27) The dissociation constant of formic acid 1.77 x 10-4 at 298K. Calculate its pKa and pKb of its, conjugate base., Ans: Solution:, pK a   log K a, , pK a   log1.77 104, pK a  (log1.77  log104 ), pK a   log1.77  4, pK a  0.2480  4, pK a  3.752, pK a  pK b  14, , We knowthat, pKb  14  pK a, , pKb  14  3.752, pKb  10.248, Q 28) The pKa of acetic acid is 4.7447. Calculate its Ka and Kb of its conjugate base., Ans: Solution:, pK a   log K a, ,  log K a  4.7447, log K a  4.7447, log K a  5.2553, K a  Antilog (5.2553), K a  1.800 105, We knowthat, , K a  K b  1014, , 11014,  0.555  109, 5, 1.8 10, K b  5.55 1010, , Kb , , Q 29) Determine the degree of ionisation and pH of a 0.05M of ammonia solution. The ionisation, constant of ammonia is 1.77 x 10-5. Also, calculate the ionisation constant of the conjugate acid of, ammonia., Ans: Solution:, , , , Kb, c, , 1.77 105, 0.05,   0.01881 moles, , , ,  OH    0.05  0.018, , By Manik. R. W. (9008747357), , Personal Use Only, , Page 26

Page 27 : I PUC Chemical Equilibrium, OH    9.4 104 M, Kw, 1014,  H   , ,  1.06 1011, OH   9.4 104, pH   log  H  , pH   log(1.06 1011 )  10.97, Now, using the relation for conjugate acid-base pair K a  Kb  K w, Using the value of Kb of NH3 is 1.77 x 10-5., We can determine the concentration of conjugate acid NH4+., , Ka, , Kw, 1014, ,  5.64 1010, 5, Kb 1.77 10, , Q 30) The pH of 0.004M hydrazine solution is 9.7. Calculate its ionisation constant Kb and pKb., Ans: Solution:, , pH   log  H  ,  H    Antilog ( pH ),  H    Antilog (9.7),  H    Antilog (10.3)  1.995 1010 M, , Kw, 11014, OH  , ,  5.01105 M, 10, ,  H  1.995 10, , , The concentration of the corresponding hydrazinium ion is also the same as that of hydroxyl ion. The concentration of, both these ions is very small so the concentration of the undissociated base can be taken equal to 0.004M., Thus,, ,  NH 2 NH 3  OH   (5.01105 ) 2, Kb , ,  6.275  107, 0.004,  NH 2 NH 2 , pK b   log K b   log(6.275  107 ), pK b   log 6.275  7, pK b  0.7976  7, pK b  6.2024, , Q 31) Calculate pH of a 1.0 x 10-8M solution of HCl., Ans: Solution:, K w  OH    H 3O    1014, The H3O+ concentration is generated (i) from the ionisation of HCl dissolved and (ii) from ionisation of H2O. In these, very dilute solutions, both sources of H3O+ must be considered., (i), ,  H 3O    108 M, , (ii), ,  H 3O    107 M, ,   H 3O    108  107 M,  H 3O    1.1  107 M, , By Manik. R. W. (9008747357), , Personal Use Only, , Page 27

Page 28 : I PUC Chemical Equilibrium, pH   log  H 3O  , , pH   log1.1  107, , pH  [log1.1 log107 ], pH  0.0414  7  6.958, Q 32) The pH of 0.1M monobasic acid is 4.50. Calculate the concentration of species H+, A- and HA at, equilibrium. Also, determine the value of Ka and pKa of the monobasic acid., Ans: Solution:, , pH   log  H  ,  H    Antilog ( pH ),  H    Antilog (4.5),  H    Antilog 5.5,  H    3.162 105 M   A ,  H    A , Ka ,  HA, Thus,, ,  HAequi  0.1  (3.16 105 ), , 0.1M, , (3.16 105 )(3.16 105 ), 0.1, 8, K a  110, , Ka , , pK a   log K a, pK a   log(108 )  8, Q 33) The degree od ionisation of a 0.1M bromoacetic acid solution is 0.132. calculate the pH of the, solution and the pKa of bromoacetic acid., Ans: Solution:, ,  H    c,  H    0.1 0.132  0.0132  1.32 102 M, pH   log  H  , , pH   log1.32 102  1.88, K a  c 2, K a  0.1 (0.132)2, K a  1.74 103, Q 34) The pKa of acetic acid and pKb of ammonia hydroxide are 4.76 and 4.75 respectively. Calculate, the pH of ammonia acetate solution., Ans: Solution:, 1, pH  7   pK a  pKb , 2, 1, pH  7   4.76  4.75, 2, 1, pH  7   0.01  7  0.005  7.005, 2, By Manik. R. W. (9008747357), , Personal Use Only, , Page 28

Page 29 : I PUC Chemical Equilibrium, Q 35) Calculate the solubility of A2X3 in pure water assuming that neither kind of ion reacts with water., The solubility product of A2X3 Ksp)=1.1 x 10-23., Ans: Solution:, If S is the solubility of salt, then in solution, ,  A3   2 S, ,  X 2   3S, 2, ,  K sp   A3   X 2 , K sp  (2 S ) 2 (3S )3, , 3, , K sp  108S 5, 1, , 1, , 1,  K sp  5  1.11023  5,  25 5, 5, S ,  1 105 molL1,  ,   1 10,  108 ,  108 , Q 36) The values of Ksp of two sparingly soluble salts Ni(OH)2 and AgCN are 2.0 x 10-15 and 6 x 10-17, respectively . which salt is more soluble? Explain., Ans: Solution:, , , , , , K sp   Ag   CN    6 1017, 2, , K sp   Ni 2  OH    2 1015, Let  Ag    S1 , then CN    S1, Let  Ni 2   S2 , then OH    2S2, , S12  6 1017 S1  7.8 109, ( S2 )(2S2 )2  2 1015 , S2  0.58 104, Ni(OH)2 is more soluble than AgCN., , Q 37) A saturated solution of silver chloride contains 1.46 x 10-3gdm-3 at 298K. What is the solubility, product of AgCl at this temperature?, Ans: Solution:, Gram molecular mass of AgCl = 107.8 + 35.5 = 143.3g, The solubility of AgCl = 1.46 x 10-3gdm-3, , , 1.46 103,  102 105 moldm3, 143.3, , At equilibrium,, , K sp  S 2, K sp  (1.02 105 ) 2  1.04 1010 mol 2 dm6, , Q 38) Solubility product constant of PbI2 in water at a given temperature is 3.4 x 10-11. Calculate its, solubility in grams per dm3., Ans: Solution:, Let S be the solubility of CaF2 in water. It is a AB2 type ternary salt., , By Manik. R. W. (9008747357), , Personal Use Only, , Page 29

Page 30 : I PUC Chemical Equilibrium,  Solubility, S , , 3, , Solubility product 3 3.4 1011, , 4, 4, , S  3 0.85 1011, S  3 8.5 1010, S  (8.5 1010 ), , 1, , 3, , Take log on both sides, 1, , log S  log(8.5 1010 ) 3, 1, 1, 1, log S  (8.5  10)  (0.7079  10)  (9.2921), 3, 3, 3, log S  3.0973, S  Antilog (3.0973), S  Antilog 4.9027, S  7.993 104, , Solublity in gram per dm3  7.993 104  461.19,  0.3686 g dm3, Q 39) The solubility product of AgCl is 2.8x10-10 at 298K. Calculate the solubility of AgCl in (i) pur, water (ii) 0.1M AgNO3 solution and (iii) 0.1M HCl solution., Ans: Solution:, i., , K sp of AgCl   Ag   Cl  , , S, ,  Solubility, S  K sp  2.8 10, , S, 10, ,  1.67 105 moldm3, , ii., ,  Solubility of AgCl inthe presence of 0.1M  AgNO3 , , K sp, c, , , , 2.8 1010,  2.8 109 moldm3, 0.1, , iii., , 2.8 1010,  Solubility of AgCl inthe presence of 0.1M HCl , ,  2.8 109 moldm3, c, 0.1, Q 40) The solubility product of AgCl at a particular temperature is 1.08 x 10-10 mol-1 dm-6. Calculate its, solubility in 0.01M HCl., Ans: Solution:, NOTE: In the presence of a common ion with the concentration of ‘c’ moldm-3, the solubility ‘S’ of the, K, sparingly soluble salt is generally given by sp ., c, K sp, , Let ‘S’ be the solubility of AgCl in water., , By Manik. R. W. (9008747357), , Personal Use Only, , Page 30

Page 31 : I PUC Chemical Equilibrium, , Solubility product of AgCl, K sp   Ag   Cl  , , 1.08 x 1010   S  S  0.01, Assuming S to be small when compared to 0.01, S  0.01  0.01, 1.08 1010,  1.08 108 moldm 3, 0.01, Q 41) the solubility product of BaSO4 at 298K is 2 x 10-5mol2dm-6. Predict whether BaSO4 will, precipitate when (i) equal volumes of 10-3M solution of Ba2+ ions and 10-3M solution of SO42- ions are, mixed (ii) equal volumes of 2 x 10-2M solution [Ba2+] ions and SO42- ions are mixed., Ans: Solution:, i., After mixing the two solutions in equal volumes, ionic concentrations become half of their original, values., Solubility ( S ) , , 103,  Ba 2    SO42  ,  5 104 moldm3, 2, Ionic prodct of BaSO2   Ba 2   SO42 , Ionic prodct of BaSO2  (5 104 )(5 104 )  2.5 107 mol 2 dm3  K sp of BaSO4, Since the ionic product is less than the solubility product value, BaSO4 will not precipitate., , ii., , 2 102,  1102 moldm3, 2, 2, Ionic prodct of BaSO2   Ba   SO42 , After mixing  Ba 2    SO42  , , Ionic prodct of BaSO2  (1102 )(1102 )  1104 mol 2 dm3  K sp of BaSO4, Since, ionic product value exceeded the solubility product value, BaSO4 will precipitate., , By Manik. R. W. (9008747357), , Personal Use Only, , Page 31