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Q.

If $K_{a} = 10^{- 5}$ for a weak acid, then ${pK}_{b}$ for its conjugate base would be

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a

9

b

$10^{- 9}$

c

$10^{- 10}$

d

5

answer is B.

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Detailed Solution

$K_{a} \times K_{b} = K_{w}$

$K_{b} = \frac{K_{w}}{K_{a}} = \frac{10^{- 14}}{10^{- 5}} = 10^{- 9}$

${pK}_{b} = - \log K_{b} = 9$

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