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Q.

If ${pK}_{b}$ for fluoride ion at $25^{0} C$ is 10.83, the ionization constant of hydrofluoric acid in water at this temperature is

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a

$3.74 \times 10^{- 3}$

b

$5.38 \times 10^{- 2}$

c

$6.75 \times 10^{- 4}$

d

$1.74 \times 10^{- 5}$

answer is C.

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Detailed Solution

$HF + H_{2} O ⇌ F^{-} + H_{3} O^{+}$

$K_{a} \times K_{b} = K_{w} {pK}_{a} + {pK}_{b} = {pK}_{w}$

${pK}_{a} = 14 - 10.83 \Rightarrow {pK}_{a} = 3.17$

$K_{a} = 6.75 \times 10^{- 4}$

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If pKb for fluoride ion at 250C is 10.83, the ionization constant of hydrofluoric acid in water at this temperature is