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a5, , Jonic Equilibria, ee ee, ions on each side of equation (3.51), applying the law at, , By cancelling the common Cl, equilibrium and taking the water concentration constant, since it is present in large EXCESS,, , we get equation as:, , _ [NH4OH][H*] _ [free base] [free acid] eae, [NH] [unhydrolysed salt] Pag, , , , Where, K;,, is the hydrolysis constant of salt, which is the characteristic constant and varies, only with temperature. Further, to maintain hydrolytic equilibrium of salt, two more, , equilibria should have to be satisfied:, , Ionisation of weak base:, | NH,OH —= NH,’ + OH™ (3.53), By the law it gives, the ionisation constant of weak base as:, + [NHj][OH™] | :, ee INH,OH) | (3.54), Ionisation of pure water: HOy = H’ + OH” 3.5 5), By the law it gives, ionic product of water as: |, Ky = [H] [OH] : (3.56), Dividing equation (3.56) by (3.54), we get |, Kw — JH*}(OH7] _(NHgOH][H"] _,, Kp . [NHj] [OH] [NH3], [NH40OH] |, ie Ka = ae ae a, Kp, , Hence, hydrolysis constant of salt of strong acid and ‘weak base is inversely, , proportional to the ionisation constant of weak base (Ky), , Degree of Hydrolysis: :, Consider the hydrolysis reaction of salt, let is the initial molar concentration of salt in aqueous solution and ‘h’ is the degree of hydrolysis of salt., , Rewriting the actual equilibrium hydrolysis reaction of salt, equation (3.51) as:, NH,CI + H,O =— NH,OH + HCl —, 0 0, , ch ch, , tA?, , Cc :, c(1-h), hxch oxi, Therefore, h [NH,OH] [HCl] _chxch, hydrolysis constant, Ky = a eS., , npare 1, than 1—h is approximately equals to |, , Initial molar Pacieaton,, 3 Concentration at equilibrium,