Question

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. $6.25\times {10}^{-6}$

Answer 1

Answer: (C)

$3.125\times {10}^{-6}$

$h=\sqrt{\frac{K_h}{C}}$
where, $K_h=$ hydrolysis constant =?
$h=$ degree of hydrolysis $=0.25$
$C=$ concentration $=0.5M$
$\therefore \frac{0.25}{100}=\sqrt{\frac{K_h}{0.5}}$
or ${\left(\frac{0.25}{100}\right)}^2=\frac{K_h}{0.5}$
$\therefore K_h=\frac{0.25\times 0.25\times 0.5}{100\times 100}$
$=3.125\times {10}^{-6}$