The dissociation constant of a substituted be | vedclass

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(C) The sodium salt of a weak acid and a strong base undergoes hydrolysis. The $pH$ of such a salt solution is calculated using the formula: $pH = \fr

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) The sodium salt of a weak acid and a strong base undergoes hydrolysis. The $pH$ of such a salt solution is calculated using the formula: $pH = \frac{1}{2} (pK_w + pK_a + \log C)$ Given: $K_a = 1.0 	imes 10^{-4} \implies pK_a = -\log(1.0 	imes 10^{-4}) = 4$ $K_w = 1.0 	imes 10^{-14} \implies pK_w = 14$ $C = 0.01 \ M = 10^{-2} \ M \implies \log C = -2$ Substituting these values: $pH = \frac{1}{2} (14 + 4 + (-2)) = \frac{1}{2} (16) = 8$

The dissociation constant of a substituted benzoic acid at $25^{\circ}C$ is $1.0 \times 10^{-4}$. The $pH$ of a $0.01 \ M$ solution of its sodium salt is

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