Question

$0.5M$ ammonium benzoate is hydrolysed to $0.25$%, hence its hydrolysis constant is:

• A. $6.25\times {10}^{-4}$
• B. $3.125\times {10}^{-6}$
• C. $1.5\times {10}^{-4}$
• D. $2.5\times {10}^{-5}$

Answer 1

The correct option is A $6.25\times {10}^{-4}$

Hydrolysis constant

$K_h=\frac{h^2}{V}=C{h}^2$

Given $h=0.25$%$=\frac{0.25}{100}=25\times {10}^{-4}$

$C=0.5M$

$\therefore K_h=0.5(25\times {10}^{-4}{)}^2$