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

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$10^{- 3} M$

$10^{- 1} M$

$10^{- 2} M$

$10^{- 4} M$

answer is A.

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Detailed Solution

$\begin{array}{l} C_{6} H_{5} N + H_{2} O \underset{}{⇄} C_{6} H_{5} N H_{3}^{+} + O H^{-} \\ 1 0 0 \\ 1- α α α \end{array}$

( $∵$ Dissociation occurs in presence of NaOH and thus dissociation of $C_{6} H_{5} N H_{2}$ will suppress)
$\begin{array}{l} K_{b} = \frac{[C_{6} H_{5} N H_{3}^{+}]}{[C_{6} H_{5} N H_{2}]} = \frac{10^{- 8} \times [O H^{-}]}{[0.24]} \\ ∵ K_{b} f o r C_{6} H_{5} N H_{2} = \frac{K_{w}}{K_{a} f o r C_{6} H_{5} N H_{3}^{+}} = \frac{10^{- 14}}{2.4 \times 10^{- 5}} \\ ∴ [O H^{-}] = \frac{10^{- 14} \times 0.24}{2.4 \times 10^{- 5} \times 10^{- 8}} = 0.01 \\ ∴ N a O H = 10^{- 2} M \end{array}$