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Q.

If $K_{h}$ (hydrolysis constant) for anilinium ion is $2.4 \times 10^{- 5} M,$ then $K_{b}$ for aniline will be

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a

$2.4 \times 10^{- 19}$

b

$2.4 \times 10^{9}$

c

$4.1 \times 10^{10}$

d

$4.1 \times 10^{- 10}$

answer is B.

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Detailed Solution

$K_{h}$ for anilinium ion = $2.4 \times 10^{- 5} M$

$K_{h} = \frac{K_{w}}{K_{b}}$

$K_{b} = \frac{10^{- 14}}{2.4 \times 10^{- 5}} = 4.1 \times 10^{- 10}$

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