An aqueous solution of aniline of concentration $0.24 M$ is prepared. What concentration of sodium hydroxide is needed in this solution so that anilinium ion concentration remains at $1 \times 10^{- 8} M$ ?
( $K_{a}$ for $C_{6} H_{5} {N H}_{3}^{+} = 2.4 \times 10^{- 5} M$ )

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Solution

$C_{6} H_{5} {N H}_{2} + H_{2} O ⇌ C_{6} H_{5} N H_{3}^{+} + {O H}^{-}$
$K_{b} = \frac{[C_{6} H_{5} N H_{3}^{+}] [O H^{-}]}{[C_{6} H_{5} {N H}_{2}]}$
$\frac{K_{w}}{K_{a}} = \frac{10^{- 8} [O H^{-}]}{0.24}$
$\frac{10^{- 14}}{2.4 \times 10^{- 5}} = \frac{10^{- 8} [O H^{-}]}{0.24}$
$[O H^{-}] = 0.01$
As $N a O H$ is strong base so all $O H^{-}$ will come from $N a O H$ .
$N a O H = [O H^{-}] = 0.01 M$

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