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Q.

The ionization constant of HCN at 298 K is $4.8 x 10^{- 9}$ then the ionization constant of its conjugate base is

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a

$2.08 \times 10^{- 6}$

b

$4.8 \times 10^{- 9}$

c

$2.08 \times 10^{- 9}$

d

$4.8 \times 10^{- 7}$

answer is A.

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Detailed Solution

For a conjugate acid – base pair $K_{a} x K_{b} = K_{w}$

At 298K, $K_{a} \cdot K_{b} = 10^{- 14}$

$K_{a} = 4.8 \times 10^{- 9};$

$K_{b} = \frac{10^{- 14}}{4.8 \times 10^{- 9}} = 2.08 \times 10^{- 6}$

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