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0.5 M ammonium benzoate is hydrolysed to 0.25 percent, hence its hydrolysis constant is

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$2.5 \times 10^{- 5}$

$1.5 \times 10^{- 4}$

$3.125 \times 10^{- 6}$

$6.25 \times 10^{- 4}$

answer is C.

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Detailed Solution

$K_{h} = {Cα}^{2} = \frac{0.25 \times 0.25 \times 0.5}{100 \times 100} = 3125 \times 10^{- 9} = 3.125 \times 10^{- 6}$

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