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Explain about the hydrolysis of salt of strong base and weak acid.Derive the value of K_h for that reaction. |
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Answer» Solution :Let us consider the reaction between sodium hydroxide and ACETIC acid to GIVE acetate and water `NaOH_(aq)+CH_3COOH_(aq) leftrightarrow CH_3COONa_(aq)+H_2O_(l)` (II) In aqueous solution `CH_3COONa` is completely dissociated as follows. `CH_3COONa_(aq) to CH_3COO^(-) (aq)+NA^(+) (aq)` (iii) `CH_3COO^-` is a conjugate base of the weak acid `CH_3COOH` and its has a tendency with `H^+` from water to produce unionised acid. But there is no such tendency for `Na^+` to react with `Oh^-` (iv) `CH_3COO^(-) (aq) +H_2O_(l) leftrightarrow CH_3COOH_(aq)+OH^(-) (aq)` and therefore `[OH^-] gt [H^+]` in such case the solution is basic due to the hydrolysis and pH is greater than 7. Relationship between equilibrium basic to the hydrolysis constant and the dissociation constant of acid is derived as follows : `K_h=([CH_3COOH[OH^-]])/([CH_3COO^-][H_2O])` `K_h=([CH_3COOH[OH^-]])/([CH_3COO^-])`....(i) `CH_3COOH (aq) leftrightarrow CH_3COO^(-) (aq) +H^(+) (aq)` `K_h=([CH_3COO^-][H^+])/([CH_3COOH])` ...........(ii) Equation (1) `times `(2) `K_h.K_a=[H^+][OH^-]` `[H^+]=[OH^-]=K_w` `therefore K_h.K_a=K_w` `K_h` VALUE in terms of degree of hydrolysis (H) and the concentration of salt (c) for the equilibrium can be obtained as in the case of Ostwaid.s dilution law `K_w=h^2C` and `[OH^-]=sqrt(k_h.C)` |
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