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Salt of Strong Acid and Weak Base

An aqueous so...

Question

An aqueous solution of $0.24 M$ aniline ( $K_{b} = 4.166 \times 10^{- 10}$ ) is mixed with $N a O H$ solution to maintain anilinium ion concentration to $1 \times 10^{- 8} M$ .
The $p O H$ of $N a O H$ solution used was:

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Solution

When NaOH is added,
$C_{6} H_{5} N H_{2} + H O H ⇌ C_{6} H_{5} N H_{3}^{+} + O H^{-}$
0.24 $10^{- 8}$ y
$\frac{[C_{6} H_{5} N H_{3}^{+}] [O H^{-}]}{[C_{6} H_{5} N H_{2}]} = K_{b}$

$\frac{y \times 10^{- 8}}{0.24} = 4.166 \times 10^{- 10}$ $⟹ y = 1 \times 10^{- 2} = [O H^{-}]$

$p O H = - l o g [O H^{-}] = - l o g (1 \times 10^{- 2}) = 2$

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0.24 M

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