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The $P^{H}$ of 1.2% (w/v) aqueous solution of $C H_{3} C O O H$ at $25^{0} C$ is ( $P^{K_{a}}$ = 4.7)

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9.26

2.7

2.9

5.0

answer is C.

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Detailed Solution

$[C H_{3} C O O H] = \frac{1.2}{60} \times \frac{1000}{100} = 0.2 M$

$P^{H} = \frac{1}{2} P^{K_{a}} - \frac{1}{2} \log c$

$= \frac{1}{2} (4.7) - \frac{1}{2} \log (0.2)$

$= 2.7$

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