CHEMICAL AND IONIC EQUILIBRIUM - Lecture Notes

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 CHAPTER - 00 CHEMICAL & IONIC EQUILIBRIUM Teaching Notes This chapter comprises the following main topics: 1. Equilibrium constants KC and KP 2. Le Chatlier’s principle 3. Solubilities and solubility product constants (Ksp) of sparingly soluble salts 4. Common ion effect 5. Theories of acids and bases : a) Arrhenius concept b) Bronsted-Lowry concept c) Lewis concept 6. Ionic product of water (Kw) 7. The pH scale and pH calculations 8. Buffer solutions 9. Salt hydrolysis Topics (1) and (2) come under chemical equilibrium, and topics (3) to (9) under ionic equilibrium. Thus, the major part of this chapter is on ionic equilibrium Law of mass action (Guldberg & Waage) The law states that the rate of a chemical reaction is directly proportional to the product of the molar concentrations (active masses) of the reactants Consider the reversible reaction : A + B  C + D Let r1 and r2 be the rates of the forward and backward reactions respectively. Let [A], [B], [C] and [D] be the molar concentrations (moles per litre) of A,B,C and D respectively By the law of mass action, r1 α [A][B] r1 = K1 [A ][B] . The proportionality constant k1 is called rate constant of the forward reaction Similarly, r2α [C][D] r2 = K2 [C][D]. The constant k2 is called rate constant of the backward reaction 1

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 ∴K1 [A][B] = k2 [C][D] K1 = [C][D] = KC K2 [A][B] kC is the equilibrium constant of the reaction expressed in terms of the molar concentrations of the reactants and products In general, for a reversible reaction of the type : m1A + m2B  n1C + n2D Kc = [C]n1 [D]n2 [A]m1 [ ]B m2 Eg. N2(g) + 3H2(g)  2NH3(g) mol L−1 2 mol L−1 4 = L2mol−2 (( ))[ [ ][ ] ]KC = NH3 2 N2 H2 3 As all the species in the above equilibrium are gaseous, the equilibrium constant of the reaction can be also expressed in terms of their partial pressure. This is called KP . KP = P2 atm2 = atm−2 NH3 atm4 PN2 × PH32 For the reaction : A(g) + B(g)  C(g) + D(g) KC = [C][D] ( mol )L−1 2 = No units Kp = PC × PD atm2 = No.units [A][B] ( mol )L−1 2 PA × PB atm2 The KC and KP of a given reaction are related as ( )KP = KC RT ∆ng ∆ng = change in the no. of gaseous moles = No. of moles of gaseous products - No. of moles of gaseous reactants If ∆ng = 0 , KP = KC Eg. N2(g) + O2(g)  2NO(g) ∆ng = 2 − 2 = 0 ∴KP = KC If ∆ng > 0, KP > KC Eg. PCl5(g)  PCl3(g) + Cl2(g) ∆ng = 2 −1 = 1 ∴KP > KC If ∆ng < 0, KP < KC Eg. N2(g) + 3H2(g)  2NH3(g) ∆ng = 2 − 4 = −2 ∴KP < KC 2

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 Reaction quotient (QC ) and Equilibrium constant (KC ) or ( QP & KP ) Consider the reversible reaction P + Q  R + S (all gases), at some stage (say at time ‘t’) before reaching equilibrium At time ‘t’, reaction quotient QC = [R]t [S]t [P]t [Q]t [ ] [ ]R S eqm. eqm PQ [ ] [ ]At equilibrium, equilibrium constant, KC = eqm eqm Evidently, QC ≠ KC ∴ The system is not in equilibrium at time ‘t’. If QC < KC , the forward reaction occurs to a greater extent so that QC increases and eventually becomes equal to KC and now the system reaches equilibrium If QC > KC at time ‘t’, the backward reaction occurs to a greater extent inorder to decrease QC and bring it equal to KC and the system now reaches equilibrium (In terms of partial pressures of gaseous reactants and products, we have QP instead of QC and KP instead of KC ) Eg. At a certain temperature, the KC of the reaction SO2 + NO2  SO3 + NO (all gases) is 16. If 1 mol each of all the 4 gases is taken in a 1 litre vessel at this temperature, which of the following statements is true ? 1) the system is now in equilibrium 2) the forward reaction occurs 3) the backward reaction occurs 4) more data is needed to predict what happens [ ] [ ]SO3 eqm KC = NO = 16(given) eqm [ ] [ ]SO2 eqm NO2 eqm Initially (i.e. at t = 0), [SO2 ]0 = [NO2 ]0 = [SO3 ]0 = [NO]0 = 1 mol L−1 ∴ At t = 0, QC = [SO3 ]0 [NO]0 = 1×1 = 1 [SO2 ]0 [NO2 ]0 1×1 QC ≠ KC ∴ The system is not in equilibrium now. QC < KC . Therefore, the forward reaction occurs leading to an increase in the concentrations of both SO3 and NO (products) and a decrease in the concentrations of both SO2 and NO2 (reactants) until QC becomes equal to KC and then only equilibrium would be attained. Answer: option(2) 3

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 Note : In heterogenous equilibria, the molar concentrations (active masses) of pure solids and pure liquids are taken as unity [ ] ( )Eg. CaCO3(s)  CaO(s) + CO2(g). KC = CO2 eqm KP = PCO2 eqm [ ] [ ] ( ) ( )NH4SH(s)  NH3(g) + H2S(g). KC = = NH3 × H2S eqm .KP PNH3 × PH2S eqm eqm eqm Degree of dissociation (α) The degree of dissociation of a compound at a given temperature is the fraction that undergoes dissociation out of 1 mol of that compound, at that temperature Eg. 2 moles of PCl5(g) enclosed in a sealed container were heated to constant temperature. The degree of dissociation of PCl5 at this temperature is 0.4. This statement means that out of 1 mol of PCl5 , 0.4 mol dissociates at the said temperature (i.e. 40% dissociation). Therefore out of the 2 moles of PCl5 heated, 2× 0.4 = 0.8mol dissociates, leaving 2 – 0.8 = 1.2 moles of PCl5 undissociated % dissociation = α × 100 The maximum possible value of α is 1 which corresponds to 100% dissociation  Now α = 100 = 1  100 Degree of dissociation (α) of a gaseous compound from vapour density data is α= D−d (n −1)d D is the normal vapour density of the substance (before dissociation) i.e. Mol.mass 2 d is the observed V.D. (after dissociation) n is the no. of moles of the gaseous products obtained from the dissociation of 1 mol of the gaseous reactant Le Chatlier’s principle states that when a system in equilibrium is subjected to a constraint (a change in T or P or concentration), the system will readjust inorder to nullify the effect of the applied constraint A+B  C + D. KC = [ ]C [ ]D eqm eqm Consider the reversible reaction [ ]A [ ]B . Adding more of the reactant A or eqm eqm B to the above equilibrium, will shift the equilibrium to the right and produce more C and D inorder to maintain the constancy of KC . Similarly, adding more of the product C or D to the above equilibrium will shift the equilibrium to the left and produce more A and B inorder to maintain the constancy of KC . Therefore, the KC or KP of a reaction cannot be changed by adding to it more of reactant or product. 4

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 Adding a catalyst to a reversible reaction will speed up the forward and backward reactions equally with the result that the final position of the equilibrium contains unchanged. ∴ The KC or KP of a reaction does not change in the presence of a catalyst. (The only advantage of adding a catalyst here is that equilibrium would be reached sooner) The only factor on which the magnitude of KC or KP of a reaction depends is temperature. ∴ for the same reaction KC or KP will have different values at different temperatures. (The magnitude of KC or KP of a reaction also depends on the stoichiometry of the chemical equation of the reaction. Eg. N2 + 3H2  2NH3 (all gases). KC = [ [NH3 ]2 ]3 = 49 at 750 K . Therefore 1 N 2 + 3 H2  NH3 (all N2] [H2 2 2 gases) K / = [ [NH3 ] ]3 = KC = 49 = 7 at 150 K C ] [H1/2 2 N2 2 The van’t Hoff equation gives the relation between the equilibrium constant of a reaction and temperature. The equation is  K2  = ∆H  1 − 1  where T2 > T1 log  K1  2.303R  T1 T2      K1 is the equilibrium constant of the reaction at T1 and K2 that at T2 . ∆H is the enthalpy change in the reaction. This may be +ve or –ve depending on whether the reaction is endothermic or exothermic. When temperature is raised from T1 to T2 , the equilibrium constant of a reaction may increase from K1 to K2 or decrease from K1 to K2 . This depends on the sign of ∆H If the ∆H of a reaction is +ve (∆H > 0) (endothermic reaction), then, when the temperature is increased from T1 to T2 , the KC of the reaction increases from K1 to K2 . This means that increasing the temperature favours an endothermic reaction. Conversely, if the ∆H of a reaction is −ve (∆H < 0) (exothermic reaction), then, when the temperature is raised from T1 to T2 , the KC of the reaction decreases from K1 to K2 . This means that increasing the temperature is unfavourable for an exothermic reaction. Exothermic reactions are favoured by keeping the reaction temperature low 5

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 T2 > T1 & nKC2 > nKC1 or KC2 > KC1 ∴Endothermic (∆H > 0) T2 > T1 & nKC2 < nKC2 or KC2 < KC1 ∴Exothermic (∆H < 0) Summary : Increasing the temperature favours an endothermic reaction. Decreasing the temperature favours an exothermic reaction Increasing the pressure will shift an equilibrium towards lesser number of gaseous moles (or towards lesser volume). Decreasing the pressure will shift an equilibrium towards larger number of gaseous moles (or towards larger volume) Changing the pressure will have no effect on the equilibrium if the number of gaseous moles on the two sides of the chemical equation of the reaction are equal (i.e. when ∆ng = 0 ) (Decreasing the volume of the system means increasing the pressure. This will shift the equilibrium towards smaller number of gaseous moles. Increasing the volume of the system means decreasing the pressure. This will shift the equilibrium towards larger number of gaseous moles). Effect of adding an inert gas a) At constant volume & temperature Eg. N2(g) + 3H2(g) He 2NH3(g) at constant V & T Adding an inert gas (say, He) to the above gaseous equilibrium maintained at constant V & T increases the pressure. But since V is kept constant, the molar concentrations of N2 , H2 & NH3 all remain unchanged. Hence the equilibrium is not disturbed. Therefore, it need not shift either to the right or left. (On the other hand, if pressure is increased by compressing the gaseous mixture of N2 , H2 & NH3 , since V decreases, the molar concentrations of all the species change. This disturbs the equilibrium It shifts towards lesser number of gaseous moles (in left to right) and produces more NH3 . This is one of the ways by which the yield of NH3 is increased in the Haber process) b) At constant pressure & temperature 6

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 Eg. N2(g) + 3H2(g) He 2NH3(g) at constant P & T Adding an inert gas (say, He) to the above gaseous equilibrium kept at constant P & T increases the volume. So P tends to decrease. Inorder to keep P constant in the increasing volume, the number of gaseous molecules in the system increases. This means that the equilibrium shifts from right to left. i.e more & more NH3 dissociates and changes into N2 & H2 . This leads to a decrease in the concentration of NH3 Some applications of Le Chatlier’s principle 1. Formation of diamond from graphite C(graphite)  C(diamond) ∆H > 0(endothermic) ↓↓ Lower density Higher density ∴L arg er volume ∴Smaller volume As the formation of diamond from graphite is endothermic and accompanied by a decrease in volume, its is favoured by a high temperature and high pressure 2. Melting of ice H2O (s)  H2O () ∆H > 0 ↓↓ Lower density Higher density ∴L arg er volume ∴Smaller volume ∴ Melting of ice is favoured at high temperature and high pressure. 3. Manufacture of NH3 by the Haber process N2(g) + 3N2(g)  2NH3(g)∆H < 0(exothermic) As the formation of NH3 from N2 and H2 (forward reaction) is exothermic, the equilibrium will shift to the right and produce more NH3 when the temperature is kept low. But at low temperature, the reaction becomes too slow which we cannot tolerate in an industrial process. Hence, to compensate this defect, a catalyst (iron oxide) is used in the presence of promoters K2O and Al2O3 and an optimum temperature of 700K at which the catalyst functions best is employed. (Formerly used catalyst: Fe powder , Promoter: Molybdenum) As the forward reaction (formation of NH3 ) from N2 and H2 takes place with a decrease in the number of gaseous moles, a high pressure (200 atm) also is maintained. 4. Contact process for manufacture of SO3 (and from that, H2SO4 ) 7

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 2SO2(g) + O2(g)  2SO3(g) ∆H < 0 As the formation of SO3 from SO2α O2 is exothermic, keeping the temperature low favours it theoretically. But now the rate of the forward reaction by which SO3 is being formed becomes too slow. Inorder to compensate for this, a catalyst (V2O5 ) is used at an optimum temperature of 720K. Although the forward reaction is accompanied by a decrease in the number of gaseous moles and is therefore favoured by high pressure, employing a high P is avoided because at high P, the acidic SO3 &SO2 gases cause rapid corrosion of the metallic parts of the contact plant 5. Manufacture of nitric oxide gas (NO) from N2 & O2 by Birkeland-Eyde process (and from that, nitric acid) N2(g) + O2(g)  2NO(g) ∆H > 0 ( )As the number of gaseous moles on both sides are equal i.e. ∆ng = 0 , change of pressure has no effect on this equilibrium. As the forwards reaction (formation of NO) is endothermic, a high temperature (1500K to 2500K) favours it. Using excess of the reactants N2 & O2 also shifts the equilibrium to the right & produce more of NO (Nitric oxide) (Note : Please be careful not to confuse the vant Hoff equation viz. log  K2  = ∆H  1 − 1  with the Arrhenius equation viz  K1  2.303R  T1 T2      log  K2  = Ea  1 −1  which gives the relation between the rate constant of a reaction and  K1  2.303R  T1 T2      temperature. Ref : Chemical kinetics) Thermodynamics of Equilibrium (for details see : chemical energetics and thermodynamics) ∆G = ∆Go + 2.303RT log QC At equilibrium, ∆G = 0 and QC = KC Now ∆Go = −2.303RT log KC ∆G = Gibbs Free Energy change ∆Go = Standard Gibbs Free Energy change QC = Reaction quotient at any instant before reaching equilibrium KC = Equilibrium constant IONIC EQUILIBRIUM 8

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 These are equilibria involving electrolytes (acids, bases or salts) in aqueous solution where all the important species are ions. Weak electrolytes are only very little ionised in their aq. solutions (α << 1) . Eg. Weak acids like HCOOH, CH3COOH, C6H5COOH, HCN, H2CO3, H2C2O4 , H3PO4 etc. & weak bases like NH3(aq) or NH4OH, CH3NH2 , C6H5NH2 , C5H5N (pyridine), N2H4 (hydrazine), Ca (OH)2 etc. All sparingly soluble salts like AgCl, AgBr, AgI, CaSO4 , BaSO4 , CaCO3 etc. Eg. α = 0.01 means only 0.01×100 = 1% ionisation Strong electrolytes are completely or almost completely ionised in their aq. solutions and so far them α = 1 or very close to 1 ( α = 1 means 1×100 = 100% ionisation). Strong acids like HClO4 , HI, HBr, HCl, HNO3, H2SO4 etc. Strong bases like NaOH, KOH, Ba(OH)2 etc and all water soluble salts of strong acids (SA) & strong bases (SB), SA & WB, WA & SB or WA & WB Eg. NaCl, KCl, KNO3 etc. α= Ka  c for weak acids  Ostwald 's dilution law, applicable to weak electrolytes only α= Kb for weak bases c (strong electrolytes are completely ionised in their aq. solutions and so for them, α = 100 = 1 . This is 100 why Ostwald’s dilution law is not valid for strong electrolytes) α is the degree of ionisation or dissociation i.e. % ionisation 100 c is the concentration of the solution in mol L−1 (molarity) Ka is the ionisation constant or dissociation constant of the weak acid Kb is the ionisation constant or dissociation constant of the weak base Eg. CH3COOH(aq)  CH3COO(−aq) + H + aq (a weak acid) Ka = CH3COO−  H+  = 1.6 ×10−5 . A very small value of Ka for CH3COOH shows that it is a very [CH3COOH] weak acid pKa = − log Ka 9

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 Ka of HCOOH at 298K = 2 ×10−4 . Here Ka = HCOO−  H+  = 2 ×10−4 at 298K [HCOOH] Ka of HCOOH > Ka of CH3COOH . This means that in their aq. solutions of the same concentration, HCOOH undergoes ionisation to a greater extent than CH3COOH does. Therefore HCOOH is a stronger acid than CH3COOH although both have small values of Ka and both are weak acids. Ka values of different weak acids are different. Larger the value of Ka , smaller is pKa and stronger is the acid. Eg. For the weak acid HA, Ka = 1×10−5 at 298K. Its pKa = − log1×10−5 = 5 For the weak acid HB, Ka = 1×10−3 at 298K. Its pKa = − log1×10−3 = 3 ∴ HB is a stronger acid than HA NH4OH(aq)  NH + (aq) + OH− ( aq ) Kb =  NH +  OH−  = 1.8×10−5 4 4 (a weak base) [ NH 4OH ] at 298K pKb = − log Kb . Kb values of different weak bases are different Larger the value of Kb , smaller is pKb and stronger is the base. Other eg. of weak bases are : CH3NH2 + H2O  CH3NH3+ (aq) + OH− (aq) C5H5N + H2O  C5H5NH+ (aq) + OH− (aq) Derivation of Ostwald’s dilution law Consider a weak acid like CH3COOH in its aq. solution. Being a weak acid, it is only very little ionised. The equilibrium that exists in the solution is CH3COOH (aq)  CH3COO− (aq) + H+ (aq) Initial concentration: C mol/L Nil Nil Eqm. concentrations: c − cα or c (1− α) mol / L cαmol / L cα mol / L (Let α be the degree of dissociation of the acid at this concentration and temp. This means that out of 1 mol of the acid, α mole dissociates. Therefore, out of c moles of the acid, cα mol dissociates producing cα mol each of CH3COO− and H+ and leaving c − cα or c (1− α) mol of the undissociated acid) 10

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 Ka = CH3COO−  H+  = cα × cα = cα2 1− α [CH3COOH] c (1− α) As α is very small compared to unity (because CH3COOH is a very weak acid), 1− α ≅ 1 Now Ka = cα2 , α2 = Ka and α = Ka c c ( )Calculation of solubility product constants Ksp of sparingly soluble salts from their solubilities in water and vice versa 1) AB type of salts (binary salts) eg. AgCl, AgBr, AgI, BaSO4 , CaCO3 etc. Let the solubility of the salt in water at a given temperature be x molL−1 . (This means that the maximum amount of the salt that can dissolve in 1 Litre of water at this temperature is xmol) AgCl(s)  Ag+ (aq) + Cl− (aq) x molL−1 x molL−1 The Ksp of a salt at a given temperature is defined as the product of the molar concentrations of its ions in a saturated solution of the salt at that temperature. Thus, Ksp of AgCl = Ag+  Cl−  is a saturated solution ( )= x molL−1 × xmolL−1 = x2mol2L−2 M2 ( )1 ∴ Solubility in water, x = Ksp 2 molL−1 2. AB2 or A2B type of salts (ternary salts) eg. PbCl2 , CaF2 , Ag2SO4 etc. Let the solubility of the salt in water at a given temperature be x molL−1 PbCl2 (s)  Pb2+ (aq) + 2Cl−1 (aq) x molL1 2x molL−1 Ksp of PbCl2 = Pb2+  Cl− 2 in a saturated solution ( )= x × (2x)2 = 4x3mol3L−3 M3  Ksp  1 3  4  ∴ Solubility in water, x =   molL−1 11

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 3. AB3 or A3B type of salts. eg. FeCl3, Ag3PO4 etc. Let the solubility of the salt in water at a given temperature be x molL−1 Ag3PO4 (s)  3Ag+ (aq) + PO34− (aq) 3x molL−1 x molL−1 Ksp of Ag3PO4 = Ag+ 3 PO34−  in a saturated solution ( )= (3x)3 × x = 27x4mol4L−4 M4  Ksp  1 4  27  Solubility in water, x =  molL−1  4. A2B3 or A3B2 type of salts Eg. Bi2S3, Ca3 (PO4 )2 Let the solubility of the salt in water at a given temperature be x molL−1 Ca3 (PO4 )2 (s)  3Ca2+ (aq) + 2PO34− (aq) 3x molL−1 2x molL−1 Ksp of Ca3 (PO4 )2 = Ca2+ 3 PO34− 2 in a saturated solution ( )= (3x)3 x (2x)2 = 108x5mol5L−5 M5  Ksp  1 5  108  ∴ Solubility in water, x =   molL−1 Suppose the salt in AB and its ions in its aq. solutions are A+ and B− If A+  B−  = Ksp of AB, the solution is saturated If A+  B−  < Ksp of AB, the solution is unsaturated If A+  B−  > Ksp of AB, the solution is supersaturated and precipitation of some of the salt from the solution will immediately occur inorder to avoid supersaturation Condition of precipitation. A salt will precipitate out from a solution only when the product of the molar concentrations of its ions (ionic product) exceeds its solubility product Eg. Ksp of AgCl at room temperature = 1×10−10 M2 12

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 Case - 1 Question : When equal volumes of these 2 solutions are mixed, will AgCl precipitate out or not ? When equal volumes of two solutions are mixed, the volume doubles and therefore their concentrations are halved Thus, Ag+  after mixing becomes 2 ×10−5 M = 1×10−5 M 2 Similarly Cl−  after mixing becomes 2 ×10−7 M = 1×10−7 M 2 Ionic product Ag+  Cl−  = 1×10−5 M ×1×10−7 M = 1×10−12 M2 < Ksp of AgCl ∴ AgCl will not precipitate out from the solution Case 2 Question : Same On mixing the 2 solutions in equal volumes, the ionic product, Ag+  Cl−  = 1×10−5 M ×1×10−3 M = 1×10−8 M2 > Ksp of AgCl ∴ Some AgCl will precipitate out from the solution in order to avoid supersaturation Common ion effect Adding a strong electrolyte to a weak electrolyte having a common ion suppresses the dissociation of the weak electrolyte. This is called common ion effect Eg. 1) The strong electrolyte CH3COONa suppresses the dissociation of the weak electrolyte (weak acid) CH3COOH by common ion effect and makes it a weaker acid 13

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 CH3COOH(aq) CH3COO- (aq) + H+ CH3COONa(aq) CH3COO- (aq) + Na+(aq) common ion 2) The strong acid HCl suppresses the dissociation of the weak acid H2S through common ion effect and keeps s2−  at a very low level. H2S 2H+ + S2- HCl H+ + Cl- common ion 3) The solubility of AgI in a KI solution becomes certainly less than that in water because of common ion effect (the common ion is I− ) Theories of Acids & Bases Arrhenius Concept. An Arrhenius acid is a substance which when dissolved in water dissociates giving H+ ions from its own molecules. Eg. HCl, H2SO4 , CH3COOH etc. An Arrhenius base is a substance which when dissolved in water dissociates giving OH− ions from its own molecules. Eg. NaOH, NH4OH, KOH, Ba (OH)2 etc. Bronsted - Lowry concept This concept defines an acid as a proton donor ( H+ donor), and a base as a proton acceptor. Therefore, acids and bases can be classified as conjugate pairs Conjugate pairs Acid Base HCl Cl− CH3COOH CH3COO− NH + NH3 4 14

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 H2O OH− H3O+ H2O H2CO3 HCO3− HCO3− CO32− i) a conjugate acid-base pair differs by one proton ii) if an acid is strong, its conjugate base is weak and vice versa. eg. Cl− is weak base because its conjugate acid HCl is a strong acid. The weakest conjugate base is ClO − because its conjugate acid HClO4 (perchloric acid) is the strongest Bronsted 4 acid iii) Species like H2O, HCO3−,HSO4− etc. which can be proton donors as well as proton acceptors are said to be amphiprotic Eg. Acid1 – Base1 is a conjugate pair. Acid2 – Base2 is another conjugate pair The position of the above equilibrium (i.e. whether it lies more to the right or more to the left) depends on the relative strengths of the two Bronsted acids and the two Bronsted bases [For a conjugate acid-base pair, Ka of the acid x Kb of the base = Kw OR pKa of the acid + pKb of the base = pKw Lewis concept of acids & bases A Lewis acid is an electron pair acceptor. Therefore it is either an electron deficient species or a molecule or ion whose central atom has vacant d-orbitals into which one or more electron pairs can be accepted. Eg. BF3, BeF4 , AlCl3,SiF4 , H+ , Ag+ , Cu2+ etc. A Lewis base is an electron pair donor. Therefore, it must have at least one lone pair of electrons. Eg. NH3, RNH2, R2NH, R3N, C5H5N, H2O, ROH, ROR, OH-, F-, Cl-, Br-, I-, CN - etc. 15

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 NH3 + BF3 (H3N : BF3) (L.base) e deficient (L.acid) BeF2 + 2F- [BeF4]2- e deficient (L.base) (L.acid) AlCl3 + Cl- [AlCl4]- e deficient (L.base) (L.acid) SiF4 + 2F- [SiF6]2-. L. Acid L.Base ( SiF4 is not e deficient, but the valence shell of Si (z = 14; 2,8,4) being the 3rd shell has vacant d- orbitals for accepting e– pairs from Lewis bases like F− ) Compare : CF4 + F− → No change (CF4 is neither e– deficient nor has the valence shell of carbon (z = 6; 2,4) being the 2nd shell d-orbital. This is why CF4 is not a Lewis acid although SiF4 is) SnCl4 + 2Cl− → [SnCl6 ]2− (L.acid) (L.base) Compare : CCl4 + Cl− → No change (Reason : same as CF4) Cu2+ + 4NH3 → Cu ( NH3 )4 2+ L.acid L.base IONIC PRODUCT OF WATER (Kw) Water undergoes dissociation to a very small extent as shown below. H2O  H+ + OH− . The ionic product of water at a given temperature is defined as the product of the molar concentrations of H+ and OH− ions in water at that temperature Kw = H+  OH−  and pKw = − log Kw ( )At 298K (25oC), Kw = H+  OH−  = 1×10−14 mol2L−2 M2 and pKw = − log1×10−14 = 14 16

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 When temperature is raised, some more water molecules dissociate into H+ and OH− ions. ∴ Kw increases with temperature, but pKw decreases because it is the negative logarithm of Kw. Thus at temperatures above 298K, K w > 1×10−14 M2 and pKw < 14 (eg. At 323K (50oC), Kw = 5×10−14 M2 . Now pKw = − log 5×10−14 = −0.7 +14 = 13.3 ) (log5 = 0.6990 ≅ 0.7) The pH scale (SORENSEN) This is a convenient way of expressing the acidity or alkalinity of aq. solutions 1 − log H3O+  log pH = –log [H+] or or H+  pOH = –log[OH–] (both the concentrations H+  & OH−  must be in molL–1 At 298K, Kw = H+  OH−  = 1×10−14 M2 and pKw = 14 In water & all neutral aq. solutions at 298K, H+  = OH−  = 1×10−14 M2 = 10−7 M ∴ pH of water at 298K = − log10−7 = 7.0 pOH of water at 298K = − log10−7 = 7.0 As pH = pOH, the medium is neutral For water at 298K, pH + pOH = 7 + 7 = 14 (pKw at 298K) The result above shows that the pH and pOH of water at a given temperature is half of pKw at that temperature eg. At 298K, pKw = 14 ∴ pH of water at 298K = 14 = 7.0 pOH of water at 298K = 14 = 7.0 OR Neutral 2 2 pH of 298 K = 7.0. Neutral pOH at 298K = 7.0 At323K, pKw = 13.3 ∴ pH of water at 323K = 13.3 = 6.65 . pOH of water at 323K = 13.3 = 6.65 . As 2 2 pH = pOH, the medium is still neutral OR neutral pH at 323K = 6.65. Neutral pOH at 323K = 6.65 The results above show that the pH and pOH of water vary with temperature because Kw and pKw vary with temperature For water and all neutral aq. solutions, H+  = OH−  & pH = pOH At 298K, for water & all neutral aq. solutions H+  = OH−  = 10−7 M , and pH = pOH = 7 and pH + pOH = 7 + 7 = 14 (pKw at 298K) 17

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 In acidic solutions at 298K, H+  > 10−7 M and ∴ OH−  < 10−7 M inorder to maintain the constancy of Kw. eg. if H+  = 10−2 M, OH−  = 10−12 M Now, pH = 2 and pOH = 12 pH + pOH = 2 + 12 = 14 (pKw at 298 K) In alkaline solutions at 298 K, OH−  > 10−7 M and ∴ ∴ H+  < 10−7 M inorder to maintain the constancy of Kw. Eg: if OH−  = 10−3 M, H+  = 10−11M Now, pOH = 3 and pH = 11. pH + pOH = 11 + 3 = 14 (pKw at 298K) Summary : For neutral aq. solutions at 298K, pH = 7 & pOH = 7 For acidic aq. solutions at 298K, pH < 7 & pOH > 7 For basic aq. solutions at 298K, pH > 7 & pOH < 7 pH Scale at 298K pH calculations pH = − log H+  pOH = − log OH−  H+  in a monobasic acid = cα mol / L  (monoprotic acid) Acids eg.HCl, HNO3, CH3COOH  H+  in a dibasic acid = 2cα mol / L   (diprotic acid) eg.H2SO4 , H2C2O4 18

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021  OH−  in a monoacidic base = cα mol / L  Bases eg.NaOH, KOH, NH4OH  OH−  in a diacidic base = 2cα mol / L  eg.Ba (OH)2 , Ca (OH)2 ‘c’ is the concentration of the solution in mol/L (molarity). α is the degree of dissociation For strong acids & strong bases, α = 1 (100% ionisation), For weak acids and weak bases α << 1 because of very little ionisation. In such cases, α can be obtained from Ostwald’s dilution law. α = Ka for weak acids α = Kb for weak bases c c Note : a) If H+  coming from an acid in its aq. solution is less than 10−6 M , H+  coming from the medium (i.e water with H+  =10−7 M ) also should be included for calculating the pH of the solution eg. pH of a 10−7 M aq. solution of HCl at 298K = ? H+  from the strong monoprotic acid, HCl = cα mol / L = 10−7 ×1 = 10−7 M H+  from water at 298K = 10−7 M Total H+  = 1×10−7 M +1×10−7 M = 2 ×10−7 M pH of the solution = − log 2×10−7 = −0.3 + 7 = 6.7 (log 2 = 0.3010 = 0.3) b) If H+  coming from an acid in tis aq. solution is 10−6 M or > 10−6 M , then H+  coming from water can be ignored eg. pH of a 10−5 M aq. solution of HCl = ? H+  from the strong monoprotic acid, HCl = cα mol / L = 10−5 ×1 = 10−5 M ( )Compared to this, H+  coming from water 10−7 M being smaller can be ignored Now pH of the solution = − log10−5 = 5 c) If OH−  coming from a base in its aq. solution is less than 10−6 M , OH−  coming from the 19

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 medium (i.e water with OH−  = 10−7 M ) also should be considered for calculating the pH of the solution eg. pH of a 10−8 M of aq. solution of NaOH at 298 K = ? (log1.1 = 0.0414 = 0.04) OH−  from the strong monoacidic base, NaOH = cα mol / L = 10−8 ×1 = 10−8 M OH−  from water at 298K = 10−7 M Total OH−  in the solution = 10−8 M +10−7 M = 0.1×10−7 M +1×10−7 M = 1.1×10−7 M pOH of the solution = − log1.1×10−7 = −0.04 + 7 = 6.96 pH = 14 - 6.96 = 7.04 d) When OH−  coming from the base in its aq. solution is 10−6 M or > 10−6 M , then OH−  coming from water (10−7 M ) can be ignored eg. pH of a 10−6 M aq. solution of NaOH at 298K = ? OH−  from the strong monoacidic base NaOH = cα = 10−6 ×1 = 10−6 M Compared to this OH−  coming from water (10−7 M ) being smaller can be ignored Now pOH of the above solution = -log 10–6 = 6 pH of the solution at 298K = 14-6 = 8 BUFFER SOLUTIONS A buffer solution is one which has the ability to maintain its pH reasonably constant even when a small amount of an acid or base is added to it. In other words, a buffer solution is one which is able to resist any large change in its pH even when a little acid or base in added to it. Acidic buffers (pH < 7) are prepared by mixing a weak acid and its salt of a strong base eg. CH3COOH + CH3COONa, HCOOH + HCOOK, HNO2 + NaNO2 , citric acid + sodium citrate, H2CO3 + NaHCO3 etc. Basic buffers (pH > 7) are made by mixing a weak base and its salt of a strong acid eg. NH4OH + NH4Cl, NH4OH + ( NH4 )2 SO4 , C6H5NH2 + C6H5NH3+Cl− (aniline + anilinium chloride), C5H5N + C5H5NH+Cl− (pyridine + pyridinium chloride) etc. An example of a neutral buffer (pH = 7) is an aq. solution of ammonium acetate (CH3COONH4 ) . This is the salt of the weak acid CH3COOH and the weak base NH4OH whose Ka and Kb are nearly equal. 20

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 Henderson’s equations for pH of buffers A) For acidic buffers : pH = pK a + log [salt] OR [acid] pH = pK a + log [anion of the salt ] [Acid] pKa = –log Ka where Ka is the dissociation constant of the weak acid used in the preparation of the buffer. eg. pH = pK a + log [CH3COONa ] OR pH = pK a + log CH3COO−  [CH3COOH] [CH3COOH] [salt] By suitably adjusting the concentration ratio [acid] , we can prepare acidic buffers of different pH values. The most efficient acidic buffer in one for which pH = pKa. To prepare such a buffer, the ratio [salt] [acid] must be 1. i.e. [salt] = [Acid] B) For basic buffers : pOH = pK + [salt] OR pOH = pK + log [cation of salt ] [Base] [Base] b b pKb = –logKb where Kb is the dissociation constant of the weak base used in the preparation of the buffer Here pH of the buffer = 14 – pOH eg. pOH = pK b + log [ NH [ NH 4Cl] ( aq ) OR pOH = pKb + log  NH+4  4OH]or NH3 NH3 (aq) [Salt] By suitably adjusting the concentration ratio [Base] , we can prepare basic buffers of different pOH [Salt] and pH. The best basic buffer is one for which pOH = pKb. To prepare such a buffer, the ratio [Base] must be 1. i.e. [Salt] = [Base] Buffer Capacity The buffer capacity of a buffer solution is the number of moles of a strong acid or strong base that is required inorder to change the pH of 1 litre of the buffer by one unit 21

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 Buffer capacity = No.of moles of acid or base required change in pH Larger the buffer capacity, the more efficient the buffer is. For an acidic buffer, the buffer capacity is maximum which its pH = pKa (the best acidic buffer) For a basic buffer, the buffer capacity is maximum when its pOH = pKb (the best basic buffer) (Refer : Henderson’s equations) Range of an acidic buffer is pH = pKa ± 1 [pH = pKa when [salt] = pKa + 1 when [S] = 10 . pH = pKa – 1 when [S] =1 [acid] = 1. pH [A] 1 [A] 10 Range of a basic buffer is pOH = pKb ± 1 pOH = pKb when [Salt] = 1. pOH = pKb + 1 when [S] = 10 [Base] [B] 1 pH = pKb – 1 when [S] = 1 [B] 10 SALT HYDROLYSIS Some salts, when dissolved in water, undergo hydrolysis and produce acidic or basic solutions. Some others do not hydrolyse and therefore their aq. solutions are neutral Type 1 : Salts of strong acids and strong bases when dissolved in water do not undergo hydrolysis. Therefore, their aq. solutions are neutral (pH = 7). This is because neither the cation nor the anion of these salts react with water eg. NaCl NaOH (S.B) ∴ No hydrolysis. Hence aq. NaCl is neutral HCl (S.A) KNO3 KOH (S.B) HNO3 (S.A) ∴ No hydrolysis. Hence aq. KNO3 is neutral Na2SO4 NaOH (S.B) ∴ No hydrolysis. Hence aq. Na2SO4 is neutral H2SO4 (S.A) Type 2 : Salts of S.A. and W.B. when dissolved in water undergo hydrolysis and produce acidic solutions (pH <7) 22

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 Hydrolysis constant, Kh = Kw . Degree of hydrolysis ; αh = Kh = Kw Kb c Kb ×c pH of the salt solution = 1 ( pK w − pK b − log c) where c is the concentration (molarity) of the salt 2 solution eg. NH4Cl NH4OH (W.B) ∴ aq. NH4Cl is acidic due to hydrolysis H2SO4 (S.A) Here, the cation of the salt reacts with water & produces acidity. NH4Cl (aq )  NH + (aq ) + Cl− (aq ) 4 NH + + HOH  NH4OH + H+ (cation hydrolysis) 4 (W.B.) (S.A.) Cu(OH)2 (W.B) CuSO4 ∴ aq. CuSO4 is acidic due to hydrolysis. (cation hydrolysis) H2SO4 (S.A) Type 3 : Salts of W.A. and S.B. also when dissolved in water undergo hydrolysis and produce basic solutions (pH>7) Kh = Kw αh = Kh = Kw pH = 1 ( pK w + pK a + log c) Ka c Ka ×c 2 CH3COOH (W.A) eg. CH3COONa ∴ aq. CH3COONa is basic due to hydrolysis. NaOH (S.B) CH3COONa(aq)  CH3COO− (aq) + Na+ (aq) CH3COO− + HOH  CH3COOH + OH− (Anion hydrolysis) (W.A) (S.B) KCN KOH (S.B) ∴ aq. KCN is basic due to hydrolysis. (anion hydrolysis) HCN (W.A) 23

BBrilliant STUDY CENTRE REPEATERS CHEMISTRY (ONLINE) -2021 Na2CO3 NaOH (S.B) H2CO3 (W.A) ∴ aq. Na2CO3 is basic due to hydrolysis. (anion hydrolysis) Type 4 : Salt of W.A. and W.B. when dissolved in water give : 1) Neutral solutions if Ka of the acid = Kb of the base 2) Acidic solutions if Ka of the acid > Kb of the base 3) Basic solutions if Kb of the base >Ka of the acid Here, Kh = Kw αh = Kh = Kw pH = 1 ( pK w + pK a − pK b ) Ka ×Kb Ka ×Kb 2 This equation contains no concentration term ‘C’. Therefore the pH of the solution is independent of its concentration. eg. CH3COONH4 CH3COOH (W.A) Ka = 1.6 × 10-5 at 298K NH4OH (W.B) Kb = 1.8 × 10-5 at 298K As Ka of the acid (CH3COOH) ≅ Kb of the base ( NH4OH) , aq. solution of CH3COONH4 is neutral CH3COONH 4 ( aq )  CH3COO− ( aq ) + NH + ( aq ) 4 NH + + HOH  NH4OH + H+ 4 (cation hydrolysis) (W.B) CH3COO− + HOH  CH3COOH + OH− (anion hydrolysis) (W.A) 24