If $20 m L$ of $0.1 M N a O H$ is added to $30 m L$ of $0.2 M C H_{3} C O O H (p K_{a} = 4.74)$ , the $p O H$ of the resulting solution is:

4.44

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9.56

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8.96

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9.26

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Solution

The correct option is B $9.56$
$\begin{matrix} H_{3} C C O O H & + & N a O H & \to & H_{3} C C O O N a & + & H_{2} O Initial & 0.006 e q . & + & 0.002 e q . & 0 & 0 Final & 0.004 e q . & + & 0 & 0.002 e q . \end{matrix}$

$p H = p K_{a} + log \frac{[C H_{3} C O O^{-}]}{[C H_{3} C O O H]}$
$= 4.74 + log \frac{2}{4}$
$= 4.74 - 0.30$
$= 4.44$
$p O H = 14 - 4.44$
$= 9.56$

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