Chemical Equilibrium Notes

Equilibrium CHEMICAL EQUILIBRIUM INTRODUCTION CHEMICAL REACTION : Symbolic representation of any chemical change in terms of reactants and products is called chemical reaction. Types of chemical reaction : (a) On the basis of physical state : Homogeneous reaction Heterogeneous reaction All reactants and products are in same phase. Reactants and products are in two or more phases.   N2(g) + 3H2(g)   2NH3(g)   Zn(s) + CO2(g)   ZnO(s) + CO(g) (b) On the basis of direction : Reversible reaction (i) Chemical reaction in which products can be Irreversible reaction Chemical reaction in which products cannot be converted back into reactants. converted back into reactants.   N2 + 3H2   2NH3 AgNO3 + NaCl  AgCl + NaNO3   3Fe + 4H2O   Fe3O4 + 4H2 NaCl + H2SO4  NaHSO4 + HCl   H2 + I2   2HI Zn + H2SO4  ZnSO4 + H2 (ii) Proceed in forward as well as Proceed only in one direction (forward direction) backward direction. 77 Equilibrium (iii) Possible in closed container Generally possible in open container (iv) These attain equilibrium These do not attain equilibrium. (v) Reactants are never completely Reactants are nearly completely converted into converted into products products. (vi) Generally thermal dissociations are held Generally thermal decompositions are held in closed vessel. in open vessel.   (1) PCl5(g)   PCl3(g)+Cl2(g) 2KClO3(s) 2KCl(s) + 3O2(g)   (2) 2HI(g)   H2(g) + I2(g) (vii) Neutralisation reaction except strong Neutralisation reaction of strong acid and acid and strong base. strong base.   HCl + NH4OH   NH4Cl + H2O HCl + NaOH  NaCl + H2O (c) On the basis of speed: Fast reactions Slow reactions (i) Generally these reactions are ionic reactions. HCl ( Acid) + NaOH (Base)  NaCl (Salt) Generally these reactions are molecular reactions. H2 + I2  2HI + H2O (Water) (ii) Rate determination is impossible. Rate determination is possible. (d) On the basis of heat : Exothermic reaction (i) Heat is evolved in these type of chemical reactions Endothermic reaction Heat is absorbed in these type of chemical reactions. R  P + x kcal R  P – x kcal (ii) Change in heat energy (with product) Q = (+) ve Q = (–) ve (iii) Change in enthalpy or internal energy H = (–) ve H = (+) ve    Note (i) Whenever questioner doesn't ask about direction then we take forward direction only. (ii) In a reversible reaction if forward reaction is exothermic then the backward reaction will be endothermic and vice-versa. 78 Equilibrium Characteristics of equilibrium : CHEMICAL EQUILIBRIUM AND (a) Chemical equilibrium is dynamic in nature i.e. the ACTIVE MASS reaction although appears to be stopped but actually A. CHEMICAL EQUILIBRIUM : takes place in both the directions with the same speed. The most important characteristic property of a (b) Chemical equilibrium can be approached from both reversible reaction is that it always attains a state of sides chemical equilibrium.      H2 + I2 or H2 + I2   2HI 2HI  Consider a general reversible reaction in a closed At equilibrium, each reactant and product has a fixed vessel concentration and this is independent of the fact       whether the reaction starts from forward direction or backward direction with the reactant or with the where Rf = forward rate of reaction product. Rb = backward rate of reaction (c) Equilibrium is not affected by the presence of catalyst. Initially rate of forward reaction is maximum and rate The catalyst only helps in attaining equilibrium rapidly. of backward reaction is zero, but as the concentration (d) The measurable properties of the system like of products increases the rate of backward reaction temperature, concentration, colour, density etc. don't also starts increasing and rate of forward reaction undergo any change with time at the chemical decreases due to decrease in concentration of equilibrium conditions. reactant. (e) Homogeneous equilibrium is the equilibrium in which At a certain stage, rate of forward reaction becomes the reactants and products are in the same phase. equal to the rate of backward reaction called    equilibrium state.         At equilibrium state : (f) Heterogeneous equilibrium is the equilibrium in which Rate of forward reaction (Rf) = Rate of backward the reactants and products are in two or more phases. reaction (Rb)    ZnO(s) + CO(g) Zn(s) + CO2(g)  B. ACTIVE MASS : The term active mass means the concentration of the reactants expressed in moles per litre (molar concentration) or the pressure of the reacting gas in atmosphere. In case of gases and solutions, the molar concentration means the number of gram molecules present per litre. Active mass is usually expressed by enclosing the symbol of the reactant in square bracket [ ]. Active mass The concentrations of the reactants and products do not change with time. At this stage, number of moles of substances produced per second in the forward reaction is equal to the number of moles of substances which disappear per second in the backward reaction. = =    =   79 Equilibrium The active mass of solids and pure liquids is a Ex.2 8.5 g ammonia is present in a vessel of 0.5 litre constant quantity (unity) because it is an intensive capacity then find out the active mass of ammonia? property i.e. number of molecules present per unit volume do not change because density and molecular Sol.      weight of solids and pure liquids are constant. But it does not apply for gaseous substances because for Ex.3 Assertion : The active mass of pure solids and them number of molecules present per unit volume pure liquids is taken unity.. change with change in volume of vessel. Reason: The active mass of pure solids and pure liquids depends on density and molecular mass. The Molar concentration =  density and molecular mass of pure solids and pure liquids are constant.  = (where = density in gram/litre) Sol. (A) =   (where  = density in g/mL) LAW OF MASS ACTION OR LAW OF CHEMICAL EQUILIBRIUM Active mass = Following other names of active mass can also be used : (i) mole/litre (ii) gram mole/litre (iii) gram molecules/litre (iv) molarity (v) Concentration (vi) Effective concentration (vii) active quantity (viii) n/v (ix) C (x) M (xi) [ ] Ex.1 In any chemical reaction, equilibrium is supposed to be establish when : (1) Mutual opposite reaction undergo. (2) concentration of reactants and resulting products are equal. (3) Velocity of mutual reactions become equal. (4) The temperature of mutual opposite reactions becomes equal. Sol. (3) The law of mass action was given by Guldberg and Waage (1864). It states that the rate of a chemical reaction is directly proportional to the product of active masses of the reacting substances. A. Derivation of equilibrium constant : Consider a reversible homogeneous chemical reaction which has attained equilibrium state at a    C+D particular temperature : A + B  Let the active masses of A, B, C and D be [A] [B] [C] and [D] respectively at equilibrium. According to law of mass action : Rate of forward reaction (Rf)  [A] [B] Rate of backward reaction (Rb)  [C] [D] Rf = kf [A] [B] and Rb= kb[C] [D] Where kf and kb are forward and backward rate or velocity constants respectively. At equilibrium state : Rf = Rb kf [A] [B] = kb [C] [D]                  ;    Kc is known as equilibrium constant. Kc has a definite value for every chemical reaction at particular temperature. 80 Equilibrium For a general reaction Put all these values in KP   m1A + m2B + m3C   n1M + n2N + n3O                       P = atmosphere B. Rel ati on b et w een K P and KC : Consider a reversible homogeneous chemical equilibrium reaction   m1A + m2B   n1C + n2D According to law of mass action (L.M.A.) For an ideal gas Where – and  PV = nRT P = Pressure in atm V = Volume in litres n = Number of gaseous moles R = Gas constant = 0.082 litre atm/mole/K T = Temperature in kelvin  = active mass  RT T = molar concentration or active mass PA = [A]RT, PB = [B] RT, PC = [C] RT, PD = [D] RT   n = (n1 + n2)  (m1 + m2) = total number of moles of gaseous products – total number of moles of gaseous reactants   For reaction A + B   C+D                 [ ] = Moles/litre   KP                                 The equilibrium constant at a given temperature is the ratio of the rate constants of forward and backward reactions.     So K P The KC is expressed by the units (mole litre1)n and KP by (atm)n. Three cases may arise : KP = KC (RT)0 = KC (a) When n = 0   For example : (i) N2 + O2   2NO   (ii) H2 + I2   2HI KC and KP are unit less in this case. (b) When n = +ve, KP > KC For example : (i) PCl5      PCl3 + Cl2   (ii) 2NH3     N + 3H  2 2              (c) When n = ve, KP < KC for example :    (i) N2 + 3H2   2NH3  (d) Special point :         If              (ii) PCl3 + Cl2   PCl5         then  or KP = KC (1)n or KP = KC For any value of n 81 C. Possible values of K :    K=   K = 1 when [Product] = [Reactant] K > 1 when [Product] > [Reactant] K < 1 when [Product] < [Reactant] D. Application of K : (a) Stability of reactants and products : Stability of reactants increases when K decreases Stability of products increases when K increases Ex : In the following reaction which one oxide is more stable. 24   2XO (g)   X2(g) + O2(g) ; K1 = 1 × 10   2XO2 (g)   X2(g) + 2O2(g) ; K2 = 2.5 × 1010  K1 > K2 So the stability of XO2 > XO (b) Time taken to attain equilibrium increases when K decreases E. Factors affecting the equilibrium constant : (a) The mode of representation of the reaction Consider the reversible chemical equilibrium reaction Equilibrium (b) Stoichiometry of the reaction : When a reversible reaction can be written with the help of two or more stoichiometric equation then the value of equilibrium constant will be numerically different in these cases. For example the dissociation of NO2 can be represented as :   2NO2   N2 + 2O2      ..........(1)   (i) If reaction (1) is divided by 2 : Then the reaction becomes :   NO2           Thus, the two equilibrium constants are related as  So if reaction is divided by n then   A + B   C+D The equilibrium constant for the reaction            (ii) If reaction (1) is multiplied by 2 : Then the reaction becomes – If the reaction is reversed   4NO2     C + D   A+ B  The equilibrium constant for the reaction is           The equilibrium constant    is actually the reciprocal of KC   Thus, the two equilibrium constants are related as Thus, the two equilibrium constants are related as  So if reaction is multiplied by n then  82 (c) If a reaction can be expressed as the sum of F. two or more reactions then overall KC will be equal to the product of the individual equilibrium constants of the reactions.   O2(g)   SO3(g) K1 Ex. SO2(g) +   NO2(g)   NO(g) + O2(g)  K2 Equilibrium None affecting factors of the equilibrium constant : The value of equilibrium constant is independent of the following factors : (a) Concentration of reactants and products. (b) Pressure (c) Volume (d) The presence of a catalyst. (e) Presence of inert materials. then   SO2(g) + NO2(g)   SO3(g) + NO(g)  K So   (d) Temperature : The value of equilibrium constant changes with the change of temperature. If K1 and K2 be the equilibrium constants of a reaction at absolute temperatures T1 and T2K and H is the change in enthalpy then :          So log K2 – log K1 =    XeO4 + XeF6   XeOF4 + XeO3F2 constant= K2. Then equilibrium constant for the following   reaction will be– XeO4 + 2HF   XeO3F2 + H 2O (1) (2) K1 + K2 (3) (4)    ( According to vant hoff equation) If the temperature T2 is higher than T1 then   Ex.4 XeF6 + H2O   XeOF4 + 2HF constant = K1, Ans. (3)   Ex.5 If the value of Kc for P + Q   R + C is 10–2 and that of kf is 10–1, the rate constant for the backward reaction will be : (1) 100 times of Kc (2) equal to KC (i) When H = +ve (endothermic reaction) (3) 1000 times of Kc (4) None of the above log K2  log K1 > 0 or log K2 > log K1  Sol. (3) K2 > K1  The value of equilibrium constant increases when   Therefore,   temperature increases in case of endothermic Therefore, kb = 10 and Kc = 10–2 reactions. Therefore, kb = 1000 × Kc (ii) When  = ve (exothermic reaction) log K2  log K1 < 0 Ex.6 Assertion:- In the presence of catalyst, the value log K2 < log K1 of equilibrium constant K increases. K2 < K1 Reason :- Catalysts increases the rate of forward The value of equilibrium constant decreases when and backward reaction to same extent. temperature increases in the case of exothermic Ans. (D) reactions. 83 DISSOCIATION  Where  = Degree of dissociation in percentage x= Number of dissociates moles a= Given initial number of moles Equilibrium Let us start with 'a' moles of H2 and 'b' moles of I2 in a closed bulb of V volume. If at equilibrium x moles of each of H2 and I2 have reacted, then 2x moles of HI will be formed so active masses.     ;     ;    Applying law of mass action Ex.7 40% of PCl5 is not dissociated at 300°C. The reaction is carried out in a flask of 1 litre capacity. The value of KC would be : (1) 3.2 (2) 1.6 –1 (3) (3.2) (4) 0.9 Ans. (4) Ex.8 In the beginning of the reaction, A     B + C, 2  or                            When a = b = 1, x becomes degree of formation of HI or degree of dissociation of H2(or I2). moles of A are taken, out of which 0.5 moles gets dissociated. What is the amount of dissociation of A?   (1) 0.5 (2) 1 Let if x <<< 1 then 1– x ~ 1 (3) 0.25 (4) 4.2 Sol. (3)   So KC = 4x2 i.e.   A  + C  B Initial mole 2 0 0  Initially x = 0.5 At equilibrium the degree of dissociation is At equilibrium 2–x x x independent to the volume. Moles at equilibrium 2 – 0.5 0.5 0.5 Since, two moles dissociated into 0.5 (ii) The equilibrium constant KP can also be calculated Therefore, one mole will dissociated into 0.25 considering partial pressures of reactants and products at equilibrium. APPLICATIONS OF LAW OF MASS ACTION Total number of moles at equilibrium = (a  x) + (b [RELATION OF DISSOCIATION (X) WITH  x) + 2x = (a + b) VOLUME (V) AND PRESSURE (P)] If total pressure of the system at equilibrium be P A. HOMOGENEOUS GASEOUS REACTIONS then OF TYPE-I (n=0) :  Synthesis of HI : Partial pressure of H2 = , Partial pressure  (i) The formation of HI from H2 and I2 is represented  by following reaction of I2 =  , Partial pressure of HI =  1 1 2   H2 (g) + I2 (g)   2HI (g) Initial moles a b 0 Moles at equilibrium (a – x) (b – x) 2x                      84 Equilibrium or    Thus KP = KC Let if x <<< 1 then  So KP = 4x2 i.e. (ii). Expression for KP : Let the total pressure at equilibrium be P. Total number of moles at equilibrium = (a  x) + x + x = a + x 1– x ~ 1                 , ,         At equilibrium the degree of dissociation is independent to the pressure also. B. HOMOGENEOUS GASEOUS REACTIONS OF TYPE-II (n > 0) Apply law of mass action     When a = 1, x becomes degree of dissociation Dissociation of PCl5 :  (i) The dissociation of PCl5 takes place according to the equation 1 1 1   PCl5   PCl3 + Initial moles a If   x<<<1 Cl2 0    then ,    0 The degree of dissociation of PCl5 is inversely Moles at equilibrium (ax) x x proportional to the square root of the total pressure Let a moles of PCl5 be taken in a closed vessel of at equilibrium. volume V. At equilibrium x moles of PCl5 are dissociated into x moles of each PCl3 and Cl2. C. HOMOGENEOUS GASEOUS REACTIONS OF TYPE-III (n < 0)  ; ;    Synthesis of Ammonia :       (i) The formation of ammonia from nitrogen and Apply law of mass action hydrogen is represented by the equation : 1 3 2                          When a = 1, x becomes degree of dissociation   If x < < < 1   then 1  x 1  The degree of dissociation of PCl5 at equilibrium is directly proportional to the square root of the volume. N2 + 3H2  2NH3    Initial moles a b 0 Moles at equilibrium (ax) (b3x) 2x Let us start with 'a' moles of N2 and 'b' moles of H2 in a closed vessel of Volume V. At equilibrium x moles of N2 has combined with 3x moles of H2 and produced 2x moles of NH3. At equilibrium      ;     ;   85 Equilibrium                         According to Law of mass action                      If a = 1, b = 3 then   If x < < < 1 then                   If a = 1 , b = 3 then i.e.   At equilibrium, the degree of dissociation is reciprocal to the volume of vessel. (ii) Expression for KP : Total number of moles at equilibrium =   If x < < < 1     ,     then and  =   If total pressure is P at equilibrium then      i.e. ,    At equilibrium, the degree of dissociation is directly proportional to the pressure. Summary n=0 Effect   H2+I2   2HI x (v)0 (P)0 n>0 or +ve n<0 or –ve   PCl5   PCl3 + Cl2   N2 +3H2   2NH3     x (v)1/2          (i) Pressure increases x unchanged x decreases x increases (ii) Volume increases x unchanged x increases x decreases (a) constant pressure x unchanged x increases x decreases (b) constant volume x unchanged x unchanged (iii) Mixing of inert gas at x unchanged Special point :– If inert gas mixed at constant temperature D. Law of Mass Action as Applied to and constant volume in an equilibrium Heterogeneous Equilibrium : chemical reaction then total number of moles of gases In such cases the active mass of pure solids and are present in a container, increases i.e. total pressure pure liquids is taken as unity and the value of of gases increases but concentration in terms of equilibrium constant is determined by the gaseous moles/ litre and partial pressure of reacting substances only. substances are unchanged so dissociation (x) unchanged. 86 Equilibrium (i) The dissociation of CaCO3 in closed vessel. LE-CHATELIER'S PRINCIPLE A. PRINCIPLE :   CaCO3(s)   CaO(s) + CO2(g) ; KP = According to this principle, if a system at equilibrium   is subjected to a change of concentration, pressure (ii) PCl5(s)   PCl3(  ) + Cl2(g) ; KP = or temperature then the equilibrium is shifted in such   (iii) 2H2O(  )   2H2(g) + O2(g) ; a way as to nullify the effect of change. (a) Change in concentration: In an equilibrium  increasing the concentrations of reactants results in shifting the equilibrium in favour of products while   (iv) 3Fe(s) + 4H2O(g)   Fe3O4(s) + 4H2(g) increasing concentrations of the products results in shifting the equilibrium in favour of the reactants.  (b) Change in pressure : When the pressure on the system is increased, the volume decreases proportionately i.e. the total number of moles present per unit volume increases. According to LeChatelier's principle, the equilibrium shifts in that Ex.9 Two semple of HI each of 5 g were taken seperately direction in which there is decrease in number of into vessels of volume 5 litres and 10 litres moles. respectively at 27°C. The extent of dissociation of If there is no change in number of moles of gases in HI will be :a reaction then a pressure change does not affect (1) More in 5 litre vessel the equilibrium. (2) More in 10 litre vessel (c) Change in temperature : If the temperature of the (3) Equal in both vessel system at equilibrium is increased then reaction will (4) Nill in both proceed in that direction in which heat can be used. Thus increase in temperature will favour the forward Sol. (3) reaction for endothermic reaction. Ex.10 What will be the amount of dissociation, if the Similarly, increase in temperature will favour the volume is increased 16 times of initial volume in the backward reaction for exothermic reactions. Special point :- Le-Chatelier's principle is applicable   reaction PCl5   PCl3 + Cl2 ? for both chemical and physical equilibrium. (1) 4 times (2) times B. PHYSICAL EQUILIBRIUM : Physical reaction : Those reaction in which change (3) 2 times (4) times in only and only state of substance takes place without any chemical change is called physical Sol. (1)  =  Thus, 4 times. reaction. Ex : (a) Ice-water system (melting of ice) :   Ex.11 Assertion:– For the reaction, N2 + O2   Melting of ice is accompained by absorption of 2NO, increase in pressure at equilibrium has no effect heat (endothermic) and decrease in volume on the reaction.   Ice (s)   water () Reason :–The reaction is not accompanied by any (H2O) (H2O) change in number of moles of gaseous species. 1gram (1.09 mL) 1gram (1.01 mL) Ans. (A) Hence both increase of temperature and pressure will favour the melting of ice into water. 87 Equilibrium Ex.13 Which of the following conditions should be more favourable for increasing the rate of forward reaction (b) Water-water vapour system (Vapourisation of water) : Vapourisation of water is an endothermic and condensation of vapour into water is an exothermic reaction : Water ()     Vapour (g) (H2O) (H2O)   in the equilibrium H2   H + H (H = +ve) ? (1) 2000 C temperature and 760 mm pressure. (2) 3500 C temperature and 100 cm pressure. (3) 3500 C temperature and 1 mm pressure. (4) All are wrong. The equilibrium shifts towards right side when the Sol. In H     H + H, heat has to be provided to 2 temperature is increased so rise in temperature will dissociate H2 into H. Therefore, the reaction is increase the vapour. endothermic (H will positive). So, temperature The equilibrium shifts towards left side when the should be high. Since, one mole of H2 forms two pressure is increased (i.e. volume is decreased) so atoms of H, so volume is increasing (n is positive) increase in pressure will favour the rate of so pressure should be low for increasing the rate of condensation of vapour into water. forward reaction. Thus favourable conditions for conversion of water       into vapour are high tempeprature and low pressure. Ex.14 Assertion: (c) Solubility of gases :   ()   Aqueous solution () (Solute) (Solvent) (Solution) Effect of pressure  Solubility of gases which dissolve in a solvents increases with a decrease in volume, and with increasing pressure.   Ex.12 On applying pressure to the equilibrium ice   water, which phenomenon will happen : Forward reaction is favoured at low temperature and low pressure. Reason :– Reaction is endothermic. Ans. (C) FEW IMPORTANT POINTS A. Calculation of Degree of dissociation from vapour density : Ex. PCl5     PCl3 + Cl2 = (1) More ice will be formed  or  (2) More water will be formed Where : (3) Equilibrium will not be disturbed DT D = Principle or theoretical vapour density D0 d = Observed or practical vapour density (4) Water will evaporate Ans. (2) = Degree of dissociation 88 Equilibrium   PCl5   PCl3 + Cl2 Total moles Reversible reaction Let initial moles 1 0 0 Moles at equilibrium (1)   Volume at NTP 1 Vapour density DT  VT =22.4  1+ D0  V0 = 22.4 (1+) If is the degree of dissociation       or   or   So for a general reversible reactionn1A     n2 B + n3C       (1) 87 (2) 60 (3) 40 (4) 70 Sol. (1)    Ex.16 If PCl5 is 80% dissociated at 250° C then its vapour density at room temperature will be When B Ex.15 The vapour density of undecomposed N2O4 is 46. When heated, vapour density decreases to 24.5 due to its dissociation to NO2. The precentage dissociation of N2O4 at the final temperature is - (i) n > 0 then DT > D0 (ii) n = 0 then DT = D0 (iii) n < 0 then DT < D0 Reaction Quotient (Q) : Consider a general homogeneous reversible reaction    C+D A + B      Reaction Quotient (Q) =     ,    (Applied in any condition)     Equilibrium constant K =     ,    (1) 56.5 (2) 104.25 (3) 101.2 (4) 52.7 Sol. (2)  = = 104.25 Vapour density at room temperature (DT) is 104.25, which is fixed. Ex.17 Assertion: For a reaction concentration quotient Q is equal to K when the reaction is in equilibrium. Reason: If a catalyst is added to the reaction at equilibrium, the value of Q remains no longer equal to K. Ans. (C) (Applied only in equilibrium state) (i) When Q = K then reaction is in equilibrium state. (ii) When Q < K then rate of forward reaction increases. (iii) When Q > K then rate of backward reaction increases. 89 Equilibrium IONIC EQUILIBRIUM INTRODUCTION Formulae : (i) ln x = logex = 2.303 log10x = 2.303 log x (ii) log (x × y) = log x + log y   (iii) log   = log x – log y   (iv) log xy = y log x Ex. (i) log 6 = log (2 × 3) = log 2 + log 3 = 0.3010 + 0.4771 = 0.7781 (ii) log30 = log (3 × 10) = log 3 + log 10 = log 0.4471 + 1 = 1.4471 (iii) log1000 = log 103 = 3 log 10 =3×1=3 A. Some values of Log : log 1 = 0 log 2 = 0.3010 log 3 = 0.4771 log 4 = 0.6020 log 5 = 0.699 log 6 = 0.7781 log 7 = 0.8451 log 8 = 0.9030 log 9 = 0.9542 log 10 = 1 log 11 = 1.04 log 100 = 2 log 1000 = 3 Some terms which are used in ionic equilibrium : Antilog : Antilog (x) = 10x Ex. Antilog (2) = 102 = 100 Antilog (0.3010) = 100.3010 = 2 Antilog [log (2)] = Antilog (0.3010) = 2 pH - Scale : Given by – Sorenson pH Scale is called Sorenson scale. pH scale is a measuring scale used to measure strength of acid and base and its value is equal to – log[H+] i.e. pH = – log[H+] = log Ex. [H+] = 10–3 pH = – log 10–3 = + 3 log 10 = 3    Conclusion : If pH = x then [H+] = 10–x or Vice versa i.e. If [H+] = 10–x then pH = x pOH  It is equal to – log [OH–] i.e. pOH – log [OH–] pH scale at 25°C :     90 (2) Ex.18 If X=   , then find pX ? Sol. pX = – log X = – log               pX = –       (3) = – [ log a – log b – c ] (4) According to strength, ionic conductors are of two types – (1) Strong electrolytes – Those ionic conductors which are completely ionized in aqueous solution are called as strong electrolytes. Ex. Na+Cl–, K+Cl–, etc. For strong electrolyte the value of degree of ionisation is 100% i.e. = 1 Ex. (a) Strong acid  H2SO4, HCl, HNO3, HClO4, HBr, HI (b) Strong base  KOH, NaOH, Ba(OH)2, CsOH, RbOH (c) All Salts  NaCl, KCl, CuSO4.......... (2) Weak electrolytes – Those electrolytes which are partially ionized in aqueous solution are called as weak electrolytes. For weak electrolytes the value of is less than one. Ex. (a) Weak acids  HCN, CH 3COOH, HCOOH, H2CO3, H3PO3, H3PO2, B(OH)3, etc.  H3BO3 (boric acid) (b) Weak bases  NH 4 OH, Cu(OH) 2, Zn(OH)2, Fe(OH)3, Al(OH)3 etc. B. ARRHENIUS CONCEPT : (1) According to Arrhenius when an electrolyte dissolves in aqueous solution it spilts up into two oppositely charged particles i.e. cation and anion. (5) Equilibrium In an electrolytic solution (Aqueous solution of electrolyte), total +ve charge = total –ve charge i.e. solution is electrically neutral. In an electrolytic solution number of +ve ions may or may not be equal to the number of negative ions. Properties of an electrolytic solution are defind by its ions. Ex. Blue colour of CuSO4 aqueous solution is due to Cu+2 ion (dark blue colour) When electric current is pass in aqueous solution of electrolyte then cation shows migration towards cathode where as anion shows migration towards anode. Ionization is the reversible reaction. In it an equilibrium is set up between unionized moles and ionized moles. This condition of the reversible ionic reaction is known as ionic equilibrium.   AB (Aq.)   A+ (Aq.) + B– (Aq.) According to L.M.A.          Ex.19 Number of total ion's and resultant total charge in A2B3 electrolyte's :(1) Five and + 1 (2) Five and - 1 (3) 0 and 0 (4) None of them Sol. (4) Ex.20 If [H+] =   then find pH ?  Sol. pH = – log [H+] = – log    =–             = – [ log 5 – log 3 – 4] = – [ 0.699 – 0.4771 – 4] = – [ – 3.778] = 3.778 Ex.21 Asserssion : H2SO4 is a strong acid. Reason : H2SO4 undergoes almost completely ionised in aqueous solution. Sol. (A) 91 Equilibrium OSTWALD'S DILUTION LAW  Ostwald was the first to apply law of mass action to ionic equilibrium.  Ostwald dilution law is applicable only for weak electrolytes. Statement : According to Ostwald when solution of weak electrolyte is diluted then degree of ionisation of solution is increased, is called dilution law. Let   AB (Aq.)   A+ (Aq.) + B– (Aq.) Initial C 0 0 concentration Degree (C – C (C (C of ionisation () According to L.M.A.   Ionisation constant               <<<< 1 then K = C or 1 –  ~ 1  (K = constant, At constant temperature)  (C )  dilution  by dilution law dilution  At infinite dilution,  Thus for weak electrolytes the degree of ionisation is directly proportional to square root of dilution or inversely proportional to square root of concentration. This law is known as Ostwald Dilution Law. Note : (1) At infinite dilution the value of becomes equal to one. (2) The value of concentration should be in terms of normality. Application of Ostwald's Dilution Law :   (a) For mono basic weak acid (HA) (i). Ionisation constant (Ka)   HA   H+ + A– Initial concentration C 0 0 At equilibrium C – C C C If degree of ionisation is  Ka = Ka =         =   =   <<<<   Ka = C 2 For mono acidic weak base (BOH) Ionisation constant (Kb)   BOH   B+ + OH– Initial concentration C 0 0 At equilibrium C – C C C If degree of ionisation is  (b)      Kb =   Kb =    <<<<  Kb = C 2  Where (1 –          =   =       (1 –  Ka Ionisation constant of weak acid Kb Ionisation constant of weak base 92 Equilibrium + + – (ii). [H ] (Concentration of H ) [H+] = C  Ka = C 2 or  – [OH ] (Concentration of OH ) [OH–] = C  .......(2) Kb =C 2 or from Eq. (1) and (2) from Eq. (1) and (2) [H+] = [OH–] =  [H+] = (iii). pH = –log [H+] put the value of [H+]  pH =   ) = – log (Kb × C)1/2 pOH = pH = –  pH =  pOH =  In summary : 1. K a = C 2 or  pOH = – log [OH–] put the value of [OH–] ) = 2. 3. .........(2)  [OH–] =     Kb = C 2 1. [H+] = C=  pH = – log [H+] 2. [OH–] = C=  3. pOH = – log [OH–] or  Limitation of Ostwald Dilution Law : (1) It is not applicable for strong electrolytes. (2) It is not applicable for saturated solutions. Factors affecting the Value of Degree of ionisation : (1) Temperature  On increasing temperature, ionization increases so increases.    H2O = 81 D2O = 79 C6H6 = 2.5 CCl4 = 0 Ex.22 For which, dilution law is applicable : (1) NaCl (SASB) (2) HCl (SA) so on dilution,  (3) CH3COONa(WASB) (4) None Sol. (4) (2) Dilution   increases. (3) Nature of electrolytes Ex.23 Find out the value of of 10–2 M HCN solution (i) Strong electrolytes (ii) Weak electrolytes if [H+] = 10–3. = 100% < 100% Sol. [H+] = C (4) Nature of solvent  If Dielectric constant of solvent increases then the –3 –2 10 = 10  or = = 10–1 or = 0.1  value of increases. %= 10% 93 Ex.24 For 10–3 M HCN solution if is 10% then find out Ka and pH of the solution ? Sol. (i) Ka = C2 =   c. = 10–5  d. (ii) e. –5 ( Ka = 10 = 5)  f. pH =    = =  =4 Ex.25 For 10 M CH3COOH solution if Ka = 10–5 then find out : (i)  ii[H+] (iii) pH g. Equilibrium Number of H+ and OH– ions in 1 litre water Number of H+ ions = 10–7NA and number of OH– ions = 10–7 NA Number of H2O moles in 1 litre water 1000/18 = 55.5 moles Number of H2O molecules in 1 litre water = 55.5 NA (NA= avogadro's number) Molar concentration of H2O molecules in water = 55.5 mol L–1 In water (Number of H2O molecules : Number of H+ ions) = 55.5 NA : 10–7 NA Sol. (i) (Degree of ionisation) :– = 55.5 × 107 : 1 Ka = C 10–5 = 102 i.e. one H+ ion is obtained from 55.5 × 107 H2O molecules or    So degree of ionisation of water = 10–6 or = 10–3 (ii) [H+] = C= 10 × 10–3 = 10–2 = = = – log [H+] = – log 10–2 = 2 (iii) pH  Ex.26 Find out the pOH of 10–3 M NH4OH solution If Kb = 10–5 – Sol. [OH ] = =   =    K (Ionisation constant of water) H+ + OH– H2O  K= pOH = – log [OH–] = – log 10–4 = 4 log10 = 4 Ex.27 For 10–3 M H2CO3 if = 10% then find out the value of pH ? Sol. Normality = M × V.F. = 10–3 × 2 = 2 × 10–3 2 × 10–4 pH = – log [H+] = – log (2 × 10–4) = – log 2 – log 10–4 = – 0.3010 + 4 = 3.7 EXPLANATION OF WATER a. Nature of water is neutral. i.e. at 25°C pH = 7 and pOH = 7  pH = pOH b. Concentration of H+ and OH– ions in 1 litre water [H+] = 10–7 mol L–1 and [OH–] = 10–7 mol L–1  Hence, water is a very weak electrolyte. = 10–4 [H+] = C = 2 × 10–3 × 0.1 =  = 18 × 10–10 = 1.8 × 10–9       K= h.        or Ionic product of water Represented by [ Kw ] K × [H2O] = [H+] [OH–] At 25°C – KW = 10–7 × 10–7 = 10–14 K [H2O] = KW  (always) KW = [H+] [OH–] taking –log on both sides –log KW = – log [H+] – log [OH–] 94 Equilibrium Mixing of ions : Nature of water is neutral so [pH = pOH] pKW = pH + pH pKW = pOH + pOH 2pH = pKW 2 pOH = pKW pH = pOH = At 25°C, KW = 10–14 or pKW = 14 (a) Common ion effect When a weak electrolyte having a common ion, is mixed with strong electrolyte then the degree of ionization of weak electrolyte is decreased. This effect is called common ion effect. (b) Odd ion effect When a weak electrolyte having a odd ion, is mixed with electrolyte then the degree of ionization of weak electrolyte is increased. This effect is effect is called common ion effect. pH + pOH = 14 or pH = pOH = 7 i. Main Points :pH = pOH = At 25°C, pKW = pH + pOH =14 j. Effect of temperature :  H2O   H + OH + – On increasing temperature rate of the forward reaction increases i.e. increases or [H+] and [OH– ] increases or [H+][OH–] increases i.e. Kw increases means pH decreases or pOH decreases. Common ion : On mixing CH3COONa with CH3COOH solution Initially   CH3COOH   CH3COO– + H+ On mixing   CH3 COONa   CH3 COO– + Na+   CH3 COOH   CH3 COO– + H+ Odd ion : On mixing, NaOH with CH3 Initially –   CH3 COOH  CH3COO + H+   On mixing, NaOH   Na+ + OH– H2O   CH3 COOH   CH3 COO– + H+ T = [H+] [OH–] = KWpKW  At 25° C, KW = 10–14 At 90° C, KW = (100 × 10–14 ) = 10–12 KW  = pOH [H+] = [OH–] At 25°C At 90°C 10–14 10–12 14 12 7 6 10–7 10–6 Ex.28 Dissociation constant of water at 25° C is (1) 10–14 × (55.5)–1 (2) 10–7 × (18)–1 (3) 10–14 × (18)–1 (4) 10–7 × (55.4)–1 Sol. (1) Ex.29 What should be the number of H+ ions in 1 mL of distilled water, if the number of H+ ions in 1 L is 6.023 × 1016? Sol. Number of H+ ions in 1 mL distilled water =    Ex.30 Assertion : NaCl + HCl does not experience common ion effect. Reason : Both NaCl and HCl are strong Important point : On increasing temperature, both electrolytes. [H+] and [OH–] increases equally so water remains Sol. (A) neutral but neutral pH changes from 7 to 6 at 90°C. pH + pOH 14 12 95 SALTS, TYPES OF SALT AND CONJUGATE THEORY A. SALT : Salts are regarded as compounds made up of positive and negative ions. The positive part comes from a base while negative part from an acid. Salts are ionic compounds. i.e. A compound formed by the combination of acid and base is known as salt. Acid + Base Salt + Water HCl + NaOH NaCl (Salt) + H2O H = –ve B. TYPES OF SALT (a) Normal/general salts :– The salts formed by the loss of all possible protons. (replaceable hydrogen ions as H+) are called normal salts. Such a salt does not contain either a replaceable hydrogen ions or a hydroxyl (OH–) group. Equilibrium (e) Complex salts :– These are formed by combination of simple salts or molecular compounds or complex ions are present in salt molecules. These are stable in solid state as well as in solutions.       K4[Fe(CN)6]  4K+ + [Fe(CN)6]–4   [Co(NH3)6]SO4  [Co(NH3)6]+2 + SO4–2 (f) Mixed salts :– The salts which furnishes more than one cation or more than one anion when dissolved in water is called a mixed salt. Ex. : (i)    Ex. : NaCl, NaNO 3 , K 2 SO 4 , Ca 3 (PO 4 ) 2 , Na2HPO3, NaH2PO2 etc. (b) Acid salts :– Salt formed by incomplete neutralisation of poly basic acids are called acid salts. Such salts still contain one or more replaceable hydrogen ions (H+). Ex. : NaHCO3, NaHSO4, NaH2PO4, Na2HPO4 etc. (ii) (iii) (iv) (a) Types of general salts : (i) SASB (ii) SAWB (iii) WASB (iv) WAWB (c) Basic salts :– Salts are formed by incomplete C. Neutralization : neutralisation of poly acidic base are called basic 1. In this process heat is evolved by the system, means salts. Such salts still contain one or more hydroxyl the process is exothermic. ion group. 2. In this process acid and base neutralize each other Ex. : Zn(OH)Cl, Mg(OH)Cl, Fe(OH) 2 Cl, therefore the evolved heat is known as heat of Bi(OH)2Cl etc. neutralization. (d) Double salts :– The addition compounds 3. Heat of neutralization : formed by the combination of two simple salts are (a) When one equivalent of an acid reacts with one termed double salts. Such salts are stable in solid equivalent of a base then the evolved heat is known as heat of neutralization. state only. Unit of heat of neutralization Kilo calorie / Ex. : FeSO4.(NH4)2SO4.6H2O, equivalent or kJ/equivalent (Ferrous ammonium sulphate) (b) For strong acid and Strong base the heat of K2SO4 . Al2(SO4)3 . 24H2O etc. neutralization is – 13.7 Kilo calories / equivalent or (Potash alum) – 57.3 kilo joules / equivalent 96 Equilibrium Ex. HCl + NaOH  NaCl (Salt) + H2O;  E. Relation between conjugate acid-base pair :– H of neutralization = – 13.7 Kilo calories/ equivalent Example – H+ + OH–  H 2O + 13.7 Kilo calories/   CH3COOH   CH3COO– + H+ equivalent 1 eq. 1 eq. 1 eq. 13.7 Kilo calories Acid conjugate base n eq. n eq. n eq.  13.7 × n Kilo calories   NH4OH  NH4+ + OH–   Total amount of heat evolved during the Base conjugate acid neutralization. CH3COOH CH3COO– H = (– 13.7 × number of limiting equivalents) acid conjugate base Kilo calorie – +   (c) For Strong acid + weak base CH3COOH + H2O   CH3COO +H3O or Weak acid + strong base or Weak acid + weak base Heat of neutralization is always less than 13.7 Kilo calorie because some heat is required to ionize weak electrolytes. HCl + NH4OH NH4Cl + H2O H = – 12.9 Kilo calories/equivalent SA WB HCN + NH4OH NH4CN + H2O H = – 11.9 Kilo calories/equivalent WA   ...... (i)  ..... (ii) In both the reactions H2O in excess quantity so active mass of H2O is one. Now multiply the equation (i) and (ii) Ka × Kb = [H+] [OH–] we know [H+] × [OH–] = KW (Ionic product of water) WB D. Conjugate acid-base pair : Conjugate acid-base pairs means difference between two species of only one H+ or OH– ion.   CH3 COOH   CH3 COO– + H+ Weak acid Conjugate strong base (Ka = 1.85 × 10–5) or acceptor ion   NH4OH   Weak base (Kb = 1.85 × 10–5) HCl –   CH3COO–+H2O   CH3COOH + OH + OH– Conjugate strong acid or acceptor ion    Taking –log on both sides   we know that for water at 25°C, KW = 10–14 or pKW = 14 So or      Note : These relation are applicable for only conjugate acid-base pairs.     H+ + (Ka ~  Cl– Conjugate weak base (spectator ion) Accepting tendency of H+ ~ 0   NaOH   Na+ + OH– Strong base Conjugate weak acid (Kb ~  (spectator ion) Ex.31 Which salt does not an acidic salt :(1) HCOONa (3) NaHS (2) NaH2PO2 (4) (1) and (2) both Sol. (4) 97 Equilibrium Ex.32 Calculate the no. of ions in Mohr's salt [FeSO4.(NH4)2 SO4.6H2O] or Ferrous Ammonium Sulphate Sol. FeSO4. (NH4)2SO4 . 6H2O Fe+2 + SO4–2 + 2NH4+ + SO4–2 + 6 H2O Total number of ions = 5 Ex.33 Assertion :– Heat of neutralisation of perchloric acid HClO4 with NaOH is same as that of HCl with NaOH. Reason :–Both HCl and HClO4 are strong acids. Sol. (A) (i) In this type of salt hydrolysis, cation reacts with H2O therefore called as cationic hydrolysis. (ii) Solution is acidic in nature as [H+] is increased. (iii) pH of the solution is less than 7. (iv) Solution turns blue litmus paper red. Kh = Hydrolysis constant KW = Ionic product of water Ka = Ionisation constant of acid Kb = Ionisation constant of base h = Degree of hydrolysis C = Concentration of salt (concentration of ions) HYDROLYSIS OF SALTS Salt hydrolysis is defined as the process in which water reacts with salt to form acid and base. i.e. change in concentration of H+ and OH– ions of water.   Water + Salt   Acid + Base (a) Relation between Kh , KW and Kb   B+ + H2O   BOH + H+   NH4+ + H2O   NH4OH + H+ Hydrolysis constant [Kh] A. Hydrolysis of strong acid and strong base   [SA – SB] types of salt – Ex. NaCl, BaCl2, Na2SO4, KClO4, BaSO4, NaNO3, KBr, KCl etc.   NaCl + H2O   NaOH + HCl For weak Base   NH4OH   NH4+ + OH–      Na+ + Cl– + H2O   Na+ + OH– + H+ + Cl– H2O + .......(1)  – H + OH (It is not salt hydrolysis)      For water (ii) Aqueous solution of these type of salt is neutral in nature. (pH = pOH = 7) KW = [H+] [OH–]  B. Hydrolysis of strong acid and weak base H+ + OH– H2O ........(3) Now mutiplying Eq. (1) and (2) = Eq. (3)  (iv) No effect on litmus paper. ........(2)  (i) Hydrolysis of salt of [SA – SB] is not possible. (iii) pH of the solution is 7.  i.e.       ×         = [H+] [OH–] Kh × Kb = KW [SA - WB] types of salt – Ex. CaSO 4 , NH 4 Cl, (NH 4 ) 2SO 4, Ca(NO 3 ) 2, ZnCl2, CuCl2, CaCl2 , AgCl, AgI, AgNO3 etc   NH4Cl + H2O   NH4OH + HCl WB SA   NH4++Cl– + H2O   NH4OH + H+ + Cl–   NH4+ + H2O   NH4OH + H+  . .......(4) (b) Degree of hydrolysis – Represented by h   NH4+ + H2O   NH4OH + H+ Initial concentration of salt C C – Ch 0 Ch 0 Ch 98   Since     h <<<< 1           Ex. CH3 COONa, HCOONa, KCN, NaCN, K2CO3, BaCO3, K3PO4 etc.   CH3 COONa + H2 O    CH3 COOH + NaOH then (1 – h) 1   CH3 COO– +Na+ + H2O   ........(5)   Equilibrium C. Hydrolysis of weak acid and strong base [WA – SB] types of salt –   CH3 COOH +Na+ + OH–     CH3 COO– + H2O   CH3 COOH + OH– .......(6) (i) In this type of salt hydrolysis, anion reacts with water therefore called as anionic hydrolysis.       ........(7) (iii) pH of the solution is greater than 7. (c) pH of the solution : pH = – log [H+] [H+] = Ch = C  [H+] = (iv) Solution turns red litmus paper blue. (a) Relation between Kh , KW and Ka   CH3 COO– + H2O   CH3COOH + OH–   (ii) Solution is basic in nature as [OH–] increases. ........(8)  taking – log on both sides  – log [H+] = – log   pH =  [ log KW + logC – log Kb ] pH =  logKW  (–log Kb)          ........(9)       ........(2)  For water H2O H+ + OH– KW = [H+] [OH–] ........(3) Now multiply eq. (1) × eq. (2) = eq. (3)      pH = ........(1)        For weak acid   CH3COOH   CH3COO– + H+ pH = – log  logC         ×         = [H+][OH–] Kh × Ka = KW  ........(4) 99 Equilibrium (b) Degree of hydrolysis (h) :   CH3COO– + H2O   CH3 COOH + OH– Initial concentration of salt C 0 0 C – Ch Ch Ch             h <<<< 1 then   (1 – h) 1   ........(5)   or   ........(6) ........(7)  (c) pH of the solution [OH–] = Ch [OH–] = C ×     or            pH =   ........(9)          ........(1)       ........(2)  For weak acid CH3COOH CH3 COO– +H+ ........(8)  – log [H+] = – log  logC   CH3COO– + NH4+ + H2O   CH3COOH + NH4OH Solution is almost neutral but it may be acidic or basic depending upon the nature of acid and base and pH of the solution is near to 7. (a) Relation between Kh, KW, Ka and Kb   taking – log on both sides pH = – log  pKa + logC For weak base NH4OH NH4+ + OH–    pKa + D. Hydrolysis of weak acid and weak base (WA - WB) type of salt : Ex. CH 3 COONH 4 , AgCN, NH4CN, CaCO 3, [NH4]2CO3, ZnHPO3 etc.   CH3COONH4 + H2O   CH3COOH +NH4OH      pKW+ logKa  logC   CH3 COO– + NH4+ + H2O   CH3COOH + NH4OH KW = [OH–] [H+]  pH = pH = 7 +    Since pH =  logKW              ........(3) For water H2O H++ OH– KW = [H+] [OH–] ........(4) Multipy Eq. (1) × Eq. (2) × Eq. (3) = Eq. (4) [ log KW + logKa – logC ] 100 Equilibrium             ×              ×  + ........(5) (b) Degree of hydrolysis (h) – CH3 COO + NH Initial concentration of salt C C C – Ch C – Ch Kh =       =  Since  or 0 Ch   =   then (1 – h) 1 .......(6)    or  (c) pH of the solution from equation (3)                  .......(8) taking – log on both sides – log [H+] = – log       pH = – [ log (Kw × Ka) – log Kb ] pH = – [ log KW + log Ka – log Kb ]   h <<<< 1   0 Ch  Since h <<<< 1 then (1 – h) 1 [H+] = Ka × h [Now put the value of h from eq. (5)]        CH3COOH + NH4OH      + H2O   + 4      = [H ] [OH ]  –  – Kh × Kb × Ka = KW   .......(7) pH = – [ log KW] – pH = +   pKW +  [log Ka] – pKa – [ – log Kb ] pKb .......(9) Note : Degree of hydrolysis of weak acid and weak base [WA – WB] type salt does not depend on the concentration of salt.  101 Equilibrium For WA - WB types of salt : Ka > Kb Kb > Ka Ka = Kb 1. Hydrolysis Cationic-anionic Anionic-cationic Neutral hydrolysis 2. Nature Acidic Basic Neutral 3. pH pH < 7 pH > 7 pH = 7 Summary : Type of salts  SA SB - WA SB   SA WB   WA WB   [H+]  -  Ex.34 What is the pH of 1 M CH3COONa solution? Ka of acetic acid = 1.8 × 10–5, Kw=10–14 mol2 L–2 (1) 2.4 (2) 3.6 (3) 4.8 (4) 9.4 Sol. (4) Ex.35 Calculate the degree of hydrolysis of a mixture containing 0.1N NH4OH and 0.1N HCN If Ka = 10–5 and Kb = 10–5 Sol. Salt is [WA – WB]  h=  =  =    =    = 10–2 Ex.36 Assertion : An aqueous solution of NH4NO3 is acidic in characters. Reason : NH 4NO 3 in an aqueous solution undergoes anionic hydrolysis. Ans. (C) pH - 7       7+ pKa + log C 7– pKb – 7+ pKa – log C SOLUBILITY AND SOLUBILITY PRODUCT (K sp) A. SOLUBILITY (a) Definition (i). At constant temperature the maximum number of moles of solute which can be dissolved in a solvent to obtain 1 litre of solution (i.e. saturated solution) is called solubility.    mol L–1 (ii). The maximum weight of solute which can be dissolved in a given amount of solvent is known as solubility. (b) Important points :Solubility does not depend on amount of substances and volume of solution where as depends on the following – (i). Temperature (ii). Presence of common ion (iii). Nature of solvent (Molecular wt. of AgCl = 143.5) (Molecular wt. of BaSO4 = 233) Solubility can be expressed in terms of molarity. 102 Equilibrium B. SOLUBILITY PRODUCT(Ksp) : (b) General form   (a) At constant temperature product of AxBy   xA+y + yB–x concentrations of ions in a saturated solution of a 0 0 substance is called solubility product of that a–S xS yS substance. (Saturated solution is that solution in which K sp = [A+y ]x [B–x]y further dissolution of even a small amount of salt is = [xs]x × [ys]y = xx.sx. yy.sy not possible).  Ex. (i) solubility of AgCl    A gCl   Ag+ + Cl– Ex. (i) Al (SO ) 2 4 3 is S mol L–1 a 0 0 K sp = 22 × 33 × (S)2 + 3 = 4 × 27 × S5 After saturation a – S S S = 108 S5 According to L.M.A. (ii) Na2 KPO4 Solubility product at saturation in terms of Ksp = 22 × 11 × 11 (S)2+1+1 = 4S4 concentration of ions Ksp = [Ag+ ] [Cl–] (iii) NaKRbPO4 Solubility product in terms of solubility Ksp = 11 × 11 × 11 × 11 × (S)1 + 1 + 1 + 1 = S4 Ksp = (S) (S)  Ex. (ii) Ksp for CaCl2   CaCl2   Ca+2 + 2Cl– Initially a 0 0 Let solubility of CaCl2 a – S S 2S is S mol L–1 Solubility product in terms of concentration of ions Ksp = [Ca+2 ] [Cl–]2 In terms of solubility Ksp = (S) (2S)2 Ex.37 If solubility product of the base M(OH) 3 is 2.7 × 10 –11 , the concentration of OH–1 will be (1) 3 × 10–3 (3) 10–3 (2) 3 × 10–4 (4) 10–11 Sol. (1) Ex.38 Compound AB gets 80% ionised. What should be the solubility of AB, calculate if its solubility product is 6.4 × 10–9 ? Sol. AB     A+ s  0.8s B– 0.8s Ksp = [A+] [B–] Ex. (iii) Ksp for AlCl3 6.4 × 10–9   AlCl3   Al+3 + 3Cl– = 0.8s × 0.8s 0.64 s2 = 6.4 × 10–9 or Initially a 0 0    Let solubility of AlCl3 a – S S 3S = 10–8 is S mol L–1  s = 10–4 Solubility product in terms of concentration of Ex.39 Assertion : For a sparingly soluble salt, Ksp is the ions Ksp = [Al+3 ] [Cl–]3 maximum dissolved value of solute in a solution. In terms of solubility Reason : Ksp corresponds to the ionic product of Ksp = (S) (3S)3 the salt in a saturated solution.  Ans. (A) 103 (iv) Ksp of A2B3 type salt – PPLICATION OF SOLUBILITY PRODUCT (Ksp) Equilibrium A3B2 Di-tri or tri-di valency) Ex. Al2(SO4)3, Ba3 (PO4 )2 etc. A. To find out the solubility (S) : (i) K sp of AB Mono-mono, di-di, tri-tri valency) type salt – Ex. NaCl, BaSO4, CH3COONa, CaCO3, NaCN, KCN, NH4CN, NH4Cl etc.   AB(s)   A+ + B– 1 0 0 1 s s     2A+3 A2B3(s) 0 0 1 2s 3s K sp =       = 2s × 2s × 3s × 3s × 3s = 108s5       B. Common ion effect on solubility : Ksp = s2 or s = (ii) Ksp of AB2 or A2B Mono-di or di-mono valency) type salt – Ex. CaCl2, CaBr2, K2S, (NH4)2SO4, K2SO4, K2CO3 etc.   AB2(s)   A+2 + 2B– 1 0 0 1 s 2s Important point :- Solubility of substances always decreases in the presence of common ion. Ex. (i) Find out the solubility of AgCl in the presence of C NaCl solution ?   AgCl   Ag+ + Cl– S S S (Let solubility of AgCl is S mol L–1) Ksp = [Ag+] [Cl–] Ksp = [A+2] [B–]2 Ksp = S2 Ksp = s × (2s)2 = s × 4s2 = 4s3 In NaCl solution   NaCl   Na+   3B–2 1 Ksp = [A+] [B–] s=  +    C + C Cl– C [Cl–] = C (iii) Ksp of AB3 or A3B Mono-tri or tri-mono valency) type salt – Let solubility of AgCl in the presence of NaCl solution is S' mol L–1. Ex. FeCl3, AlCl3, K3PO4 etc.   AgCl   Ag+ S' S'   AB3(s)   A+3 1 0 0 1 s 3s Ksp =       =  + = 27s4 3B– + Cl– S'+C According to L.M.A. Ksp = [Ag+]' [Cl–]' Ksp = S' (S' + C) = S' 2 + S'C (Neglecting the higher power terms of S') Ksp = S' C  s=       104 Equilibrium Ex. (ii) Find out the solubility of CaCl2 solution in the Ex. (iii) Find out the solubility of NaCl in the presence presence of C NaCl solution ? of C CaCl2 solution ?   CaCl2   Ca+2+   CaCl2   2Cl– S S 2S –1 [Let solubility of CaCl2 is S mol L ] K sp +2 C = 4S3 For NaCl solution C C C   CaCl2   Ca+2 + S' S' Ksp = S' (2S' + C)2 = S' (4S' 2 + 4S'C + C2) Ksp = 4S' 3 + 4S' 2 C + S'C2 (Neglecting the higher power terms of S') Zero First Second  Ag+,Hg+1, C. Group precipitation : (i) Ksp > [ ] [ ] precipitation (ii) Ksp = [ ] (iii) Ksp < [ ] CH3 COOH  , Pb+2 Pb+2, Hg+2 ,Cu+2,Cd+2 Sn+2,Sn+4, As+3, Sb+3 Bi+3 Third Al+3, Fe+3, Cr+3 Fourth Mn+2,Co+2,Ni+2 Zn+2 Fifth Ba+2, Sr+2 , Ca+2 Sixth Mg+2 S'  Condition for precipitation (Group reagent) ,K+ ,Na+ Cl– S'+2C Ksp = S' (S' + 2C) = S' 2 + 2S'C (Neglecting the higher power terms of S') 2Cl– 2S'+C Ksp = [Ca+2]' [Cl–]' 2 Radicals     Na+ + Ksp = [Na+]' [Cl–]' [Cl–] = C According to L.M.A.  2C According to L.M.A. Cl– Let solubility of CaCl2 in the presence of NaCl solution is S' mol L–1. Group C NaCl S' Na+ + 2Cl– Let solubility of NaCl in the presence of CaCl2 solution is S' mol L–1. – 2 = [Ca ] [ Cl ]   NaCl   Ca+2+ By mixing of dilute HCl H2S gas passed in the presence of acidic medium NH4OH mixed in the presence of NH4Cl H2S gas passed in presence of basic medium (NH4)2 CO3 mixed in the presence of NH4Cl By mixing of Na2HPO4 [ ] [ ] Unsaturated No Saturated Super saturated  precipitation Forms of precipitation – Chloride AgCl,Hg2Cl2, PbCl2 Sulphide PbS,HgS, CuS,CdS, SnS, SnS2 ,As2S3, Sb2S3 , Bi2S3 Hydroxide Al(OH)3,Fe(OH)3 ,Cr(OH)3 Sulphide MnS,CoS, NiS, ZnS Corbonate BaCO3,SrCO3, CaCO3 Hydrogen phosphate (MgHPO4) 105 (iv) Important points : (a). Hg+1 (us) Unstable in aqueous solution. [Stable in dimer form (Hg2+2) Hg2Cl2]   Hg2Cl2   Hg2+2 + 2Cl– S S 2S +2 – 2 K sp = [Hg2 ] [Cl ] = (S) (2S)2 Equilibrium (e). In IV group, H 2S gas is passed in basic medium to increase S–2 ion concentration by odd ion effect, so that the ionic product of cations of group IV and S– 2 ions exceed the solubility product of their corresponding metal sulphide and hence gets precipitated.  (b). Hg+2 (ic) HgCl2  stable in aqueous solution   HgCl2   Hg+2 + 2Cl– S S 2S  So [S–2] KspIV < [Radicals of group IV] [S–2] (f). In V group, order of Ksp and precipitation is – BaCO3 < SrCO3 < CaCO3 (c). For precipitation of II group, H2S gas is passed in acidic medium to decrease S–2 concentration by D. Precipitation of soap and salt (NaCl) common ion effect. So that ionic product of cations (i) Precipitation of NaCl (common salt) : of group II and S–2 ions exceed the solubility product For the precipitation of NaCl, HCl gas passed in of their corresponding metal sulphides. Therefore saturated solution of NaCl to increase the only the cations of group II gets precipitated. But concentration of Cl– ion by common ion, so ionic the sulphides of the cation of the other groupes (III, product of concentration of Na+ and Cl– ion is more IV & V) are not precipitated under these conditions because their solubility products are quite high. than Ksp of NaCl and NaCl easily precipitated. When H2S gas is directly passed through solution Saturated solution of NaCl Ksp = [Na+] [Cl–] th nd then IV group is also precipitated with II group.   By passing of HCl gas HCl   H+ + Cl–   H2S   2H+ + S–2 due to common ion Ksp < [Na+] [Cl–]  nd KspII < KspIV < [radicals of II and IV group] (ii) Precipitation of soap : [S–2] When H2S gas pass in acidic medium [H+] then only For precipitation of soap, NaCl mixed in saturated IInd group radicals are precipitated. solution of soap to increase the concentration of Na+   H2S   2H+ + S–2 ion by common ion. So ionic product of concentration nd –2 of Na+ and strearate ions is more than Ksp of soap KspII < [radicals of II and IV group] [S ] < KspIV and soap is easily precipitated. (d). For precipitation of group III, NH4OH is added in the presence of NH4Cl to decrease the concentration Ex. C17H35COONa (Sodium stearate) of OH– by common ion effect so ionic product of Saturated solution of C17H35 cations of group III and hydroxides ions exceed the Ksp = [C17H35COO–] [Na+] solubility product of their corresponding metal hydroxides and only group III cations will By mixing of NaCl solution precipitated. On the other hand cations of group IV,   NaCl   Na+ + Cl– V and VI, which require large concentration of OH– due to common ion due to their high solubility product, will not be precipitated. Ksp < [C17H35COO–] [Na+]  106 Equilibrium Ex.40 Solubility products of M(OH)3 and M(OH)2 are 10 –23 and 10 –14 respectively. Which will be precipitated first on adding NH4OH, if M+2 and M+3 both the ions are in solution ? (1) M+2 (2) M+3 (3) Both M+2 and M+3 together (4) Precipitation will not take place. Sol. (2) Ex.41 Solubility of CaCl2 is 4 × 10–8, then find out its Ksp and its new solubility in the presence of 10–2 M Ca(OH)2.   Sol. CaCl2   Ca+2 + 2Cl– 1 s 2s (Initial solubility = s) 3 –8 3 Ksp = 4s = 4(4 × 10 ) = 256 × 10–24   Ca(OH)2   Ca+2 + 2OH– 0 0 0 C 2C     Ca+2 1 + 2Cl s' +C  2s' (New solubility = s') = [s' + C] [2s']2 = 4s'3 + 4s'2C (s'3 = negligible) 4s'2C or    – K sp = [Ca+2]' [Cl–]2 ' K sp = Then [H+]1 = and [H+]2 = According to given condition for isohydric solution pH of different solution is same. Hence [H+]1 = [H+]2 = C CaCl2 ......(i) FEW IMPORTANT POINTS A. Various forms of KW (a) KW = [H+] [OH–] for water [H+] = [OH–] (b) KW = [H+]2 (c) KW = [OH–]2 (d) KW = [H3O+] [OH–] (e) KW = [H3O+]2 {[H3O+] = [H+]} Hydronium Proton ion B. Isohydric solution – If different–different solution has same pH are called isohydric solution. Sp. point :– Isohydric conditions for two given weak acids HA1 and HA2 which has concentration C1 and C2 and ionisation constants are Ka1 and Ka2 respectively. .......(ii) C. Strength of Acids and Bases :– We known   If there are two weak acids HA1 and HA2 which has concentration C1 and C2, degree of ionisation 1 and 2 and ionisation constants Ka1 and Ka2 respectively then ratio of their strength of acids. We know [H+] = C = From equation (i) and (ii) s' 2 = =     = 64 × 10–22 s' = 8 × 10–11 mol L–1 Ex.42 Assertion : In acidic medium, Zn2+ is not precipitated by S–2 ions. Reason : Common ion effect reduced the concentration of S–2 ions to the minimum value. Sol. (A) if C1 = C2  Thus the relative strength of two acids of equimolar concentration can be compared by taking square root of the ratio of their ionisation constants. 107 Equilibrium D. Resultant concentration :– When two non reactive electrolytes AB and CD which has concentration C1 and C2 and volume V1 and V2 respectively are mixed in one container then total volume of container is (V1 + V2) and resultant concentration of each electrolyte is– Resultant concentration of AB   Resultant concentration of CD       Ionic product = [Ca+2] × [SO4–2] = (0.002) × (0.001) = 2 × 10–6 Hence ionic product of CaSO4 is 2 × 10–6, which is less than its solubility product. Therefore, precipitation will not take place. pH  INTRODUCTION If volume of both electrolytes are equal pH of some important substance :- i.e. V1 = V2 = V (i) Blood  7.4 then resultant concentration of each electrolyte is becomes half of initial concentration. (ii) Tear  7.4     ,       and (iii) Gastric Juice  1 to 3 (iv) Soft drink  2 to 4 (v) Acidic rain  6 (vi) Milk  6.3 to 6.6 (vii) Water  7 Sp. Point : ‘n’ different solutions of non reactive different electrolytes are mixed in equal volume then TYPES OF pH + – resultant concentration of each electrolyte is times. (i) pH of SA and SB. [H ] = C, [OH ] = C (ii) pH of WA and WB. [H+] = C, [OH–] = C (iii) pH of very dilute solution. Ex.43 pH values of two acids A and B are 4 and 5. The strengths of these two acids are related as :(1) The strengths of the two acids cannot be compared. (Consider H+ & OH– of H2O also) (iv) pH of salt solution. (a) SA SB Type Salt (always 7) (2) Acid B is 10 times stronger than acid A. (b) SA WB Type Salt (< 7) (3) Strength of acid A : Strength of acid B = 4 : 5 (c) WA SB Type Salt (> 7) (4) Acid A is 10 times stronger than acid B. (d) WA WB Type Salt (Almost 7) Sol. (4) (v) pH of mixture of acid and base. Ex.44 Equal amounts of 0.004 M CaCl2 and 0.002 M Na2SO4 are mixed. Find out, whether CaSO4 will be precipitated or not, if Ksp of CaSO4 is 2.5 × 10–5 ? Sol. When equal amounts of CaCl2 and Na2SO4 are mixed then the concentration becomes half.     and (a) pH of mixture of acids : N1V1 + N2V2 + N3V3 + ....... = NV V = Volume of final solution = V1 + V2 + V3 + .... N = Normality of final solution = [H+] in final solution. 108 (b) pH of mixture of base : N1V1 + N2V2 + N3V3 + ................. = NV V = Volume of final solution Equilibrium A. Simple buffer solution :- Aqueous solution of weak acid-weak base (WA – WB) types of salt. Ex. CH3COONH4 , NH4CN, AgCN etc. = V1 + V2 + V3 + .......... pH = 7 + N = Normality of final solution = [OH–] in final solution. CH3COO– + pKa –  pKb  + H2O   (c) pH of mixture of acids and bases : CH3COOH + NH4OH. For acid : Case 1. N1V1 + N2V2 + N3V3 + ........ = (NV)Acid When mixing of acid [H+] For base : CH3COO– + H+    CH3COOH NH4OH + H+    N1V1 + N2V2 + N3V3 + ........ = (NV)Base If (NV)Acid > (NV)Base then solution is acidic. NV = (NV)Acid – (NV)Base and [H+] = N If (NV)Base > (NV)Acid then solution is basic. NV = (NV)Base – (NV)Acid and [OH–] = N  + H2O Case 2. When mixing of base [OH–]  + OH–    NH4OH  CH3COOH + OH–   CH3COO– + H2O B. Mixed buffer solution : BUFFER SOLUTION (i) Acidic buffer solution :- Aqueous solution of INTRODUCTION mixture of weak acid and salt of same weak acid A. Definition : with any type of strong base is called acidic buffer solution. Those solution which have resistive nature for pH Ex. CH3COOH + CH3 COONa change are called Buffer Solution. WA WASB B. Properties of buffer solution : Weak acid + Strong base (i) pH of buffer solution is never depends on their 1. When N1 V1 = N2 V2 Salt (WA SB) concentration and dilution. (water add or escape). 2. When N1 V1 > N2 V2 Acidic buffer solution (ii) When some amount of strong acid and strong base (WA + WASB salt) are mixed in buffer solution then pH of buffer solution [CH3 COOH + CH3 COONa] is almost unchange. [CH3 COOH + CH3 COO–] Special point :Na+ (Spectator ion) Change in pH of a solution due to change in Case 1. concentration of present free H+ or OH– ions. When mixing of acid [H+]  CH3COO– + H+   CH3COOH TYPES OF BUFFER SOLUTION Case 2. (A). Simple buffer solution When mixing of base [OH–] (B). Mixed buffer solution : CH3 COOH + OH–   CH3COO– + H2O (i) Acidic buffer solution ie. in aqueous solution if weak acid and its conjugate (ii) Basic buffer solution are present then it is definitely acidic buffer solution. 109 Equilibrium a. pH of acidic buffer solution : CH3 COOH + CH3 COONa Acid + Salt [CH3 COOH + CH3 COO–] + Na+ Acid Conjugate   CH3 COOH  CH3COO– + H+            [NH4OH +  or  pH range of acidic buffer solution : CH3 COONa pH = pKa + log pH = pKa + 1 10 : + H2O.  + OH–   NH4OH  NH4OH    1    [OH–]  10  When mixing of base [OH–] Case 2. Henderson's equation : : ] + Cl– [Spectator]   NH4OH + H+      pH= pKa + log 1 ] = [NH4Cl] [NH4OH + NH4Cl] pH = pKa – log CH3 COOH  (a) pOH of basic buffer solution : taking –log on both sides b. [ When mixing of acid [H+] Case 1. or    Ex. NH4OH + NH4Cl    or         taking –log on both sides pOH =       Henderson's equation : pH = pKa –1 So pH range (b) pOH range of basic buffer solution : (c) Maximum buffer action condition of acidic buffer solution : CH3 COOH 1 CH3 COONa : 1 NH4 OH : NH4 Cl 1 : 10 10 : 1 pH = pKa + log (c) Maximum buffer action condition of basic buffer solution : (ii) Basic buffer solution : Definition : Aqueous solution of mixture of weak base and salt of same weak base with any type of strong acid is called basic buffer solution. NH4 OH : NH4 Cl 1 : 1 110 Equilibrium Ex. Acidic indicator [HPh] BUFFER CAPACITY Definition :- It is defined as the number of moles of acid (or base) require by one litre of a buffer solution for changing its pH by one unit. Buffer capacity = Case 1.  HPh   H+ + Ph– In acidic medium [H+] [HPh] > [Ph–] Case2.  HPh  + Ph–  H In basic medium [OH–] [Ph–] > [HPh] Colourless + Pink colour (a) pH of acidic indicator Ex.45 When 2 moles of HCl is added to 1 L. of an acidic buffer solution, its pH changes from 3.9 to 3.4. Find its buffer capacity. Sol. B.C. =  n   H+ =   =4 [H+ ] = INDICATOR         + In–    Taking – log on both sides INTRODUCTION A. Definition :- Those compound or substance which indicate the nature of solution by changes their self colour. pH = pKI – log [HIn] + log [In–]  pH = pKI + log      B. Types of indicators : (a) Acidic indicator (HIn) Ex. Phenolphthalein (HPh) HPh    H+ Ph– + colourless pink (b) pOH of basic indicator - (b) Basic indicator (InOH) Ex. Methyl orange (MeOH) MeOH yellow    Me+ =  OH– +    In+ nOH      + OH–   orange red OSTWALD THEORY OF INDICATORS (i) According to this theory, all the indicators are weak organic acid or base. (ii) Every indicator has light colour in its unionised form and dark colour in ionised form. (c) pH range of acidic indicator : [HIn] : [In–] 1 : 10 10 : 1 (iii) Every indicator works in opposite medium (it means if it is acidic then it will work in basic medium and vice versa) it means colour change in opposite medium due to their unionised form is change in ionised form. 111 Equilibrium (d) pOH range of basic indicator : (e) Neutral condition for indicators + [InOH] : [In ] (No working condition) 1 : 10 For acidic indicator 10 : 1 [HIn] : 1 S.No. Name of indicator Colour in acidic medium 1. Methyl orange (MeOH) 2. 3. medium For basic indicators – [In ] : [InOH] : 1 Colour in basic 1 of indicators Yellow 3.1 to 4.5 Methyl red Red Yellow 4.2 to 6.2 Phenol red Yellow Red 6.2 to 8.2 Colourless Pink 8.2 to 10.2 (HPh) 1 Working pH range Orange red 4. : [In+] ACID-BASE TITRATION S.No. Type of titration pH range of titration Suitable indicators 1. SA/SB. 3 – 11 All indicators (MeOH, HPh etc. ) 2. SA/WB 3–7 Methyl orange (MeOH) and methyl red (HPh) 4. WA/WB 6.5 – 7.5 Phenol red Important points of titration : (i). Titration is always possible in two opposite solution. (i.e. one is acidic and other is basic). (ii). For any titration only that indicator is suitable if their working pH range is in pH range of titration. (iii). At the end point of titration equivalents of acids and bases are always same. (iv). At the end point, nature of solution depends on the type of titration. SA/SB Neutral (pH = 7) SA/WB Acidic WA/SB Basic WA/WB Any one possible BENZENOID AND QUINONOID THEORY OF INDICATORS (i). According to this theory, all indicators are aromatic compounds. (ii). Unionised form of indicator is called benzenoid form and ionised form is called quinonoid form.    Benzenoid form Quinonoid form (iii). Every indicator has light colour in its benzenoid form and dark colour in quinonoid form. (iv). Every indicator works in opposite medium due to their benzenoid form (unionised form) is changes in quinonoid form (ionised form). (a) 112 Equilibrium (i) conducts electricity, (ii) reacts with active metals like zinc, magnesium etc. to give hydrogen, (iii) turns red litmus blue, (iv) has a bitter taste       (b) Benzenoid form (MeOH)      Quinonoid form (MeOH) Acidic solution (red) ACID AND BASE INTRODUCTION (v) whose basic properties are destroyed on reaction with an acid. The above definitions of acids and bases are called operational definitions as they are based certain operations (i.e. tests) to be performed on the substances. However, these have been replaced by conceptual definitions (put forward by Arrhenius, Bronsted-Lowry and Lewis) which go into the causes of the observed behaviour, based upon structure and composition of the substances. ARRHENIUS CONCEPT (1884) (a) Acid:- According to this concept, those substances which produce free H+ ions in aqueous solution are called acid. Ex. HCl, HNO3, H2SO4, H3PO4, H2CO3, H2S, CH3COOH etc. HCl + H2O H+(aq.) + Cl–(aq.) H2SO4 + H2O 2H+(aq.) + SO4–2(aq.) The earliest classification of substances into acids CH3COOH + H2O  and bases was bases upon the characteristic CH3COO–(aq.) + H+(aq.) properties possessed by each one of them. For (b) Base :- Those substances which produce free OH– example, ions in aqueous solution are called base. Acid was defined as a substance whose aqueous solution possesses the following characteristic properties: Ex. NaOH, KOH, Cs(OH), Rb(OH), NH4OH, Ba(OH)2, Ca(OH)2, Al(OH)3 etc. (i) conducts electricity, NaOH + H2O Na+ (aq.) + OH– (aq.) (ii) reacts with active metals like zinc, magnesium etc. to give hydrogen, Ba(OH)2 + H2O Ba+2 (aq.) + 2OH– (aq.) (iii) turns blue litmus red. (iv) has a sour taste. (v) whose acidic properties disappear on reaction with a base. Base was defined as a substance whose aqueous solution possessed the following characteristic properties: Al(OH)3 + H2O Al+3 (aq.) + 3OH– (aq.) (c) Nature of water :- According to this concept nature of water is neutral and act as a solvent. (d) Neutralisation Reaction :- Those reactions in which acid and base react together to form water molecule are called neutralisation reactions. i.e. Na+ + OH– + H+ + Cl– NaCl + H2O H+ + OH– H2O 113 Equilibrium (e) Strength of acids and bases :- This concept BRONSTED-LOWRY CONCEPT (1923) explains the strength of acids and bases depending It is based upon the exchange of proton. upon the basis of degree of ionisation. (a) Acid :- According to this concept those substances Ex. For strong electrolytes  ~ 100% which have tendency to donate the proton (H+) by For weak electrolytes  < 100% any method in any solvent are called acids. (f) Advantage :- This concept explains the acids and Ex. bases practically. i.e. To find out the pH, ionisation (i) HCl, HNO3, H2SO4, H2CO3, H2S, CH3COOH, constant, hydrolysis constants, heat of neutralisation H3PO3 etc. etc. (ii) HS–, HCO3–, HSO4–, H2PO4–, HPO4–2, H2O etc. (g) Disadvantage :(iii) NH4+, H3O+, PH4+, CH3COOH2+ etc. (i) It explains the behaviour of acids and bases only (vi) [Al(H2O)6]+3, [Ag(H2O)2]+1, [Fe(H2O)6]+3 etc. in aqueous (water) solvents. (a) HCl (Acid) + H2O (Solvent)  H3O+ + Cl– (ii) It does not explain the stability of proton (H+). (iii) It does not explain the conjugate acid-base theory. (b) HS– (Acid) + H2O (Solvent)  H3O+ + S–2 (iv) It does not explain the acidic and basic behaviour of aprotic acid and base for example SO2, CO2, SiCl4, AlCl3 etc. (c) NH4+ (Acid) + H2O (Solvent)  NH3 + H3O+ (d) [Al(H2O)6]+3 (Acid) + H2O (Solvent)  [Al(H2O)5OH]+2 + H3O+ (b) Base :- Those substances which have tendency to Ex.46 Gaseous hydrogen chloride is a very poor accept the proton by any method in any solvent are conductor of electricity but a solution of hydrogen called the bases. chloride in water is a good conductor. This is due to Ex. the fact that :(i) NaOH, KOH, Rb(OH), Cs(OH), Ba(OH) 2 , (1) Water is a good conductor of electricity Ca(OH)2, NH4OH, Al(OH)3 etc. (2) Hydrogen chloride ionises in water – – – – –2 (3) A gas cannot conduct electricity but a liquid can (ii) HS , HCO3 , HSO4 , H2PO4 , HPO4 , H2O etc. (4) HCl does not obey Ohm's law where as the (iii) NH3, RNH 2, R 2NH, R 3N, C 6H 5 NH 2, C 5H 5N, H2N-NH2 etc. solution does Sol. (2) Ex.47 Which is acid in the following pairs according to Arrhenius concept ? (i) HCl(g) and HCl (aq) (ii) CH3COOH(l) and CH3COOH(aq) Sol. (i) HCl(aq.) (ii) CH3COOH(aq.) (iv) O–2, SO4–2, CO3–2, Cl–, Br–, I–, CN– etc. (a) NaOH (Base) + H2O (Solvent)     Na+ + H O OH–  2 (b) HS– (Base) + H2O (Solvent)  H2S + OH– (c) NH3 (Base) + H2O (Solvent)  NH4+ + OH– (d) CO3–2(Base) + H2O (Solvent) HCO3– + OH– 114 Equilibrium + (iii) NH + H O NH + OH–  According to this concept nature   3 2 4 Weak base Weak acid of water is amphoteric or amphiprotic i.e. water can Strong acid Strong base act as both acid and base. – +   Cl + CH3COOH2 (i) HCl (Acid) + H2O (Base) Cl– + H3O+ (iv) HCl + CH3COOH  Weak acid Weak base (ii) NH3 (Base) + H2O (Acid) NH4+ + OH– Strong base Strong acid (d) Neutralisation Reaction :- According to this (v) HCl + NH3  Cl– + NH4+   concept those reaction in which acid and base react Strong acid Strong base Weak base Weak acid together and convert into their respective conjugate – + acid and base are called neutralisation reactions. (vi) CH3COOH + NH3    CH3COO + NH4 Strong acid Strong base Ex. Weak base Weak acid Types of solvent : (a) Protogenic or acidic solvent :- They have a tendency to generate or donate the protons. Ex. CH3COOH, HCl, HNO3, H2SO4 etc. (b) Protophilic or basic solvent :- They have a (e) Strength of acids and bases :- This concept tendency to accept the protons. explain the strength of acid and base depending upon Ex. NH3, CH3OH, RNH2 etc. the basis of relative tendency to accept or donate (c) Amphiprotic or amphoteric solvents :- They have the proton. the tendency to accept or donate the protons or they (i) HClO4 (ii) HI are either protophilic or protogenic. (iii) HBr (iv) H2SO 4 Ex. H2O, HS–, C2H5OH, HSO4–, HCO3– etc. (v) HCl (vi) HNO3 (d) Aprotic solvents :- They neither donate nor accept (vii) H3O+ (viii) HSO4– the protons. (ix) H3PO 4 (x) HF Ex. C6H6, CCl4, CHCl3, BrF3, NO2, COCl2 etc. (xi) CH3COOH (xii) H2CO3 (e) Ampholyte solvents:-Those solvents in which the (xiii) H2S (xiv) NH4+ same molecule acts as proton donor and proton (xv) HCN (xvi) C6H5OH acceptor. (xvii) H – OH (xviii) C2H5–OH Ex. (i) (xix) C6H5NH2 (xx) NH3 (xxi) R–NH2 (xxii) CH4 (xxiii) H2 Ex. (i) HCl + H2O    Strong acid Weak base (ii) CH3COOH + H2O Weak acid Strong base Cl– + H3O+ Strong base Weak acid – +    CH3COO + H3O Weak base Strong acid (ii) 115 Equilibrium Special point :- Every ampholyte is amphoteric but every amphoteric is not ampholyte. Ex. H2O (amphoteric) (f) Advantage :(i) It explain the behaviour of acids and bases in any type of solvent. Classification of Lewis acids :(i) Compounds whose central atom have an incomplete octect (electron deficient) Ex. BF3, BBr3, BCl3, BI3, B(CH3)3, B(OH)3, AlCl3, GeCl3 etc. (ii) It explain the stability of proton (H+). (g) Disadvantage :(i) It does not explain the acid and base practically. (ii) Compounds whose central atom have vacant dorbitals and can accept one or more pair of lone (ii) According to this concept every acid and base electrons. reaction is a neutralisation reaction which is not Ex. SF4, SF6, SnCl2, SnCl4, PX3, PX5, GeX4, true. TeX4, IF5, IF7, etc. (iii) It fails to explain the acidic and basic behaviour of aprotic acids and bases also. For ex. SO2, SO3, CO2, AlCl3, SiCl4, etc. – + Ex.48 In the process : NH3 + NH3    NH2 + NH4 , The nature of ammonia is:- (1) Acidic (2) Basic (3) Amphoteric (4) None SF6 IF5 (iii) Molecules with a multiple bond between atoms of different electronegativities. Ex. CO2, SO2, SO3 etc. Sol. (3) Ex.49 Which of the following behave both as Bronsted acid as well as Bronsted bases ? H2O, HCO3–, H2SO4, H3PO4, HS–, NH3 Sol. H2O, HCO3–, HS–, NH3 Ex.50 Assertion : Cl– is weaker base than OC2H5. Reason : Cl– is conjugate base of stronger acid, HCl. Sol. (A) + OH– –O—C=O HCO3– Lewis base  OH Lewis acid SO3 + Lewis acid H2O H2SO4 Lewis base (iv) Cations :LEWIS CONCEPT (1939) (a) Lewis acid :- According to this concept those species which have self tendency to accept the lone pair of electrons are called acids. i.e. Lewis acid is an electron pair acceptor (electrophilic). Ex. Ag+, Na+, Li+, Al+3, Be+2, Mg+2, I+, Cl+, H+ etc. False cations :Ex. NH4+, H3O+, CH3COOH2+, PH4+ etc. + Ex. Ag+ + 2 : NH3    [Ag (:NH3)2] Cu+2 + 2 (:H2O)2 [Cu (:H2O)4]+2 116 Equilibrium (v) Elements which have six electrons in their outermost (e) Strength of acids and bases :- This concept shell or valence shell. explain the strength of acids and bases depending O–Famil y (O, S, Se, Te) upon the basis of self tendency. To accept or donate the lone pair of electrons (e–). Po – Radioactive element Ex. SO 2 < SO3 (Strong acid) (f) Advantage :(b) Lewis base :- Those species which have self (a) This concept explains the acid and base without tendency to donate the lone pair of electrons are any solvent. called bases. i.e. a base is an electron pair (lone (b) This concept explains those acids and bases pair) donor (nucleophile). which are not explained by other concepts. Classification of Lewis bases :(g) Disadvantage :(i) Those species whose central atom have lone pair of (a) This concept fails to explain the acid and base electrons and have self tendency to donate them – practically.     (b) According to this concept forming of coEx. ,  ,  , ,  ordinate bond in neutralisation reaction of acid and    base is a slow process, whereas neutralisation ,  ,  reaction of acids and bases is actually fast process.     (c) This concept fails to explain the real acids and   ,   ,   ,   ,     bases  Ex. HCl, HNO3, H2SO4, NaOH, KOH, etc.   , etc.  Special point :- All the lewis bases are bronsted bases but all the lewis acids are not bronsted acids. Explain :(Electron density on O, increases due to +I effect) (ii) Anions – Ex. O–2, SO4–2, CO3–2, Cl–, Br–, F–, I–, N–3, P–3, NO3–, S–2, CN–, etc. (c) Nature of water :- According to this concept nature of water is basic i.e. water is a base. (d) Neutralisation Reaction :- According to this concept those reactions in which acid and base react together to form a co-ordinate bond between them are called neutralisation reactions. Ex. Ex.51 Lewis base is :(1) NH3 (3) BCl3 Sol. (1) (2) CO2 (4) K+ Ex.52 How does Lewis theory explain the acidic character of CO2 ? Sol. According to Lewis concept, CO2 is Lewis acid because carbon atom is attached to more electronegative oxygen atoms on both sides by double bonds. It can easily accept an electron pair from the Lewis base such as OH– ion. 117 Equilibrium Ex. (i) HF < HCl < HBr < HI CLASSIFICATION OFACIDS AND BASES (ii) NH3 < PH3 < AsH3 (iii)CH4 < NH3 < H2O < HF A. Classification of all the acids :(a) Oxy acids :- Nonmetal + O – H Nonmetal O (c) Acidic anhydrides :- H Ex. (i) When nonmetals are different then – Acidic character  (E.N.) H2SO3  SO2 H2SO4  SO3 2HNO2  N2O3 2HNO3  N2O5 2HClO4  Cl2O7 2H3PO4  P2O5 (HPO3) 2 × CH3COOH  (CH3CO)2O (a) H3BO3 < H2CO3 < HNO3 < H2SO4 < HClO4 (b) HIO4 < HBrO4 < HClO4 (c) HAsO3 < HPO3 < HNO3 (ii) When non metals are same then : Acidic character  (O.N.) (a) (b) Special point :-Acidic order of acidic anhydrides is same as the acidic order of their oxy acids.   Ex. N2O3 < N2O5 (Acidic order)   Cl2O < Cl2O3 < Cl2O5 < Cl2O7 (Acidic order) (d) Organic acid :- (c)  (d)   (b) Hydrides :-   (exception) (i) If R shows + I effect  then acidic character decreases. (ii) If R shows – I effect  then acidic character increases. (e) For Cations :Acidic character  (+)  Electronegativity (E.N.) Oxidation number (O.N.) 118 (ii) Li+ > Na+ > K+ > Rb+ > Cs+ Equilibrium Special point :- Basic order of basic anhydrides is same as the basic order of their metal hydroxides. (iii) Be+2 > Mg+2 > Ca+2 > Sr+2 > Ba+2 (c) Organic Base :- Ex. (i) + +2 Li < Ca < Al +3 (f) Acidic character in terms of H+, OH–, Ka and Kb :Acidic character[H+]  Acidic character if [H+]  or or pOH  [OH–]  pKa or or if Ka  pKb  or pH  or Kb  B. Classification of all the bases :- (i) If R shows + I effect then basic characters are increases. (ii) If R shows – I effect  then basic characters are decreases. Ex. C6H5NH2 < NH3 < CH3NH2 < C2H5NH2 < C3H7NH2 (Basic order) (d) For anions :- (a) Hydroxides :M Basic character  (–) O  H  (Metal)  Basic character  Ex. (i) (ii) F– < O2– < N–3 F– > Cl– > Br– > I– (e) Basic characters in terms of OH–, H+, Kb and Ka :[OH–]  i.e. CsOH is the strongest base. pOH  or  if [OH–]  pH  or or [H+]  or  if Kb  pKa  or Ka  Ex. (i) LiOH < NaOH < KOH < RbOH < CsOH pKb  or (ii) Be(OH)2 < Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2 Ex.53 Least pKa value is for this acid :- (b) Basic anhydrides :Ex. 2NaOH 2LiOH Ba(OH)2 2Al(OH)3 Na2O  Li2O    BaO Al2O3 (1) F3CCOOH (2) CH3COOH (3) (CH3)2CHCOOH (4) ClCH2.COOH Sol. (1) (Amphoteric oxide) 119 Equilibrium Ex.54 Arrange the following in required order : (i) NH3, C6H5NH2, C2H5NH2, CH3NH2 (increasing basic strength) (ii) KOH, Ga(OH)3, Ca(OH)2   H+ + HA– Ex.56 In the dissociation , H2A    H+ + A–2 HA–  (increasing basic strength) Sol. (i) C6H5NH2 < NH3 < CH3NH2 < C2H5NH2 (ii) Ga(OH)3 < Ca(OH)2 < KOH (1) K1 is equal to K2 (2) K1 is smaller than K2 Ex.55 Assertion :– HBr is a stronger acid than HI. (3) K1 is greater than K2 (4) K1 is negligible Reason :– Bond energy of HI is smaller than that of Ans. (3) HBr. Ex.57 Calculate the pH of 2.0 × 10–4 M H3X Solution Sol. (D) assuming first dissociation to be 100%, second to be 50% and third to be neglegible : POLYACIDIC BASE AND Sol. First dissociation : POLYBASIC ACIDS  H3X   H2X– + H+ A. Polyacidic base :  H+ ion concentration = 2 × 10–4 M  Al(OH)3  Al+3 + 3OH– Kb  Second dissociation :  Al(OH)3  [Al(OH)2]+1 + OH–    H2X–   HX– + H+  [Al(OH)2]+1   [Al(OH)]+2 + OH–  H+ ion concentration  [Al(OH)]+2   Kb =  = C = 2 × 10–4 × 0.5 Al+3 + OH–   Third dissociation is negligible   [H+] = 2 × 10–4 + 10–4 = 3 × 10–4 taking –log on both sides    i.e. pH = 4 – log3 = 3.52     B. Polybasic acid : H3PO4    3H+ + PO4–3 Ka H3PO4    H2PO4–1 + H+   H2PO4–1   HPO4–2 + H+   HPO4–2   PO4–3 + H+  taking –log on both sides        120