Q. 0.5 M ammonium benzoate is hydrolysed to 0.25 percent, hence, its hydrolysis constant is
A) $2.5 \times 10^{- 5}$
B) $1.5 \times 10^{- 4}$
C) $3.125 \times 10^{- 6}$
D) $6.25 \times 10^{- 6}$
Explanation
(c)
$h = \sqrt{\frac{K_{h}}{C}}$
where, $K_{h} = hydrolysis\ constant = ?$
$h = degree\ of\ hydrolysis = 0.25\% = \frac{0.25}{100}$
$C = concentration = 0.5\ M$
$\therefore$ $\frac{0.25}{100} = \sqrt{\frac{K_{h}}{0.5}}$
or $\left( \frac{0.25}{100} \right)^{2} = \frac{K_{h}}{0.5}$
$\therefore$ $K_{h} = \frac{0.25 \times 0.25 \times 0.5}{100 \times 100}$
$= 3.125 \times 10^{- 6}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.