Solution of 0.1 N ${NH}_{4} OH$ and 0.1 N ${NH}_{4} C l$ has $P^{H}$ 9.25. Then $P^{Kb}$ of ${NH}_{4} OH$ is

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8.25

4.75

9.25

3.75

answer is B.

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Detailed Solution

$\begin{array}{r} P^{H} = 14 - [P^{K_{b}} + l o g \frac{[salt]}{[Base]}] = 14 - [P^{K_{b}} + l o g [\frac{0 .1}{0 .1}]] P^{H} = 14 - p^{k_{b}} \\ \Rightarrow; p^{k_{b}} = 14 - p^{H} = 14 - 9 .25 \\ = 4 .75 \end{array}$

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