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Percentage of ionization of 0.01M aqueous aniline is ( $K_{b}$ of aniline = $3 \times 10^{- 10}$ )

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0.035 %

0.0085 %

0.0234%

0.0173 %

answer is C.

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Detailed Solution

$\begin{array}{l} K_{b} = C α^{2} \\ α = \sqrt{\frac{K_{b}}{C}} \\ α = \sqrt{\frac{3 \times 10^{- 10}}{10^{- 2}}} \\ α = 1.732 \times 10^{- 4} \end{array}$

% of ionizatioin =

$1.732 \times 10^{- 4} \times 100 = 1.732 \times 10^{- 2} = 0.01732 %$

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