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Derive the hydrolysis constant for the hydrolysis of salt of strong base and weak acid. Deduce its pH. |
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Answer» Solution :Let us find a RELATION between the equilibrium constant for the hydrolysis reaction (hydrolysis constant) and the dissociation constant of the acid. `K_h=([CH_3COOH][OH^-])/([CH_3COO^-][H_2O])` `K_h=([CH_3 COOH][OH^-])/([CH_3 COO^-])""...(1)` `CH_3 COOH_((aq)) hArr CH_3 COO_((aq))+H_((aq))^+` `K-a=([CH_3 COO^-][H^+])/([CH_3 COOH]) ""...(2)` `(1)xx(2)` `rArr K_b. K_a=[H^+][OH^-]` we know that `[H^+][OH^-]=K_w` `K_h.K_a=K_w` `K_b` VALUE in TERM of degree of hydrolysis (h) and the concentration of salt (C) for the equilibrium can be obtained as in the case of Ostwald's dilution law. `K_h=h^2 c` and i.e., `[OH^-]=sqrt(K_h.C)` PH of salt solution in terms of `K_a ` and the concentration of the ELECTROLYTE. `pH+pOH=14` `pH=14-pOH=14-{-log[OH^-]} = 14+log[OH^-]` `therefore pH=14+log(K_h C)^(1/2)` `pH=14+log((K_w C)/(K_a))^(1/2)` `pH=14+((1)/(2)logK_w+(1)/(2)log C-(1)/(2)log K_a) ""[because K_w=10^(-14)]` `pH=14-7+(1)/(2) log C+(1)/(2) pK_a` `(1)/(2)logK_w=(1)/(2)xxlog10^(-14)=(-14)/(2)(I)=-7` `pH=7+(1)/(2)pK_a+(1)/(2)log C` `[-log K_a=pK_a]` |
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