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A 0.1 M sodium acetate solution was prepared. The $k_{h} = 5.6 \times 10^{- 10}$

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The degree of hydrolysis is $7.48 \times 10^{- 5}$

The $[{OH}^{-}]$ concentration is ${7.48×10}^{-3} M$

The $[{OH}^{-}]$ concentration is $7.48 \times 10^{- 6} M$

The pH is approximately 8.88

answer is A.

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Detailed Solution

$α_{h} = \sqrt{\frac{k_{h}}{C_{0}}} = \sqrt{\frac{5.6 \times 10^{- 1}}{10^{- 1}}} = 7.48 \times 10^{- 5}$

$\begin{array}{l} C H_{3} C O O^{-} (a q) + H_{2} O (l) \underset{}{⇄} C H_{3} C O O H + O H^{-} (a q) \\ t = 0 C_{0} \\ t = t_{e q} C_{0} - C_{0} α_{h} C_{0} α_{h} C_{0} α_{w} \end{array}$

$[O H^{-}] = C_{0} α_{w} = 10^{- 1} \times 7.48 \times 10^{- 5}$

$= 7.48 \times 10^{- 6}$

$∴ p O H = 5.12$ $p^{n} = 8.88$

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