Assuming that the degree of hydrolysis is small, the pH of 0.1 M solution of sodium acetate Ka∘=1.0×10−5 will be

A
5.0
B
6.0
C
8.0
D
9.0

Solution:

The acetate ion undergoes hydrolysis as ${CH}_{3} {COO}^{-} + H_{2} O ⇌ {CH}_{3} COOH + {OH}^{-}$

$K_{h} = \frac{[{CH}_{3} COOH] [{OH}^{-}]}{[{CH}_{3} {COO}^{-}]} \equiv \frac{[H^{+}] [{OH}^{-}]}{[{CH}_{3} {COO}^{-}] [H^{+}] / [{CH}_{3} COOH]} = \frac{K_{w}}{K_{a}} = \frac{1.0 \times 10^{- 14} M^{2}}{1.0 \times 10^{- 5} M} = 1.0 \times 10^{- 9} M$ also, $[{CH}_{3} COOH] = [{OH}^{-}]$ Hence, $K_{h} = \frac{{[{OH}^{-}]}^{2}}{{[{CH}_{3} COOH]}_{0}} or [{OH}^{-}] = {(K_{h} {[{CH}_{3} COOH]}_{0})}^{1 / 2}$ O $\begin{aligned} [{OH}^{-}] = {[(1.0 \times 10^{- 9} M) (0.1 M)]}^{1 / 2} = 1.0 \times 10^{- 5} M \\ [H^{+}] = K_{w} / [{OH}^{-}] = 1.0 \times 10^{- 14} M^{2} / (1.0 \times 10^{- 5} M) = 1.0 \times 10^{- 9} M \\ pH = - \log ([H^{+}] / M) = 9 \end{aligned}$

Assuming that the degree of hydrolysis is small, the pH of 0.1 M solution of sodium acetate $(K_{a}^{\circ} = 1.0 \times 10^{- 5})$ will be

Solution:

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