Question

$HX$ is a weak acid $(K_a={10}^{-5}).$ It forms a salt $NaX$ having concentration of $0.1M$, on reacting with caustic soda. What is the degree of hydrolysis ($h$) of the salt?

• A. ${10}^{-4}$
• B. ${10}^{-2}$
• C. ${10}^{-6}$
• D. ${10}^{-5}$

Answer 1

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

Given, $K_a={10}^{-5},\left[NaX\right]=0.1M$
According to question, $NaX$ is a salt of weak acid and strong base.
Therefore $X^{-}$ only undergoes hydrolysis.
at equilibrium we have,
$\underset{C-Ch}{X^{-}}+H_{2}O\stackrel{K_a}{\rightleftharpoons }\underset{Ch}{HX}+\underset{Ch}{O{H}^{-}}$


'$K_h$' is hydrolysis constant and 'C' is concentration
$K_h=\frac{\left[HX\right]\left[O{H}^{-}\right]}{\left[X^{-}\right]}$
we know,
$\Rightarrow K_h=\frac{K_w}{K_a}=\frac{{10}^{-14}}{{10}^{-5}}$
$\Rightarrow K_h={10}^{-9}$
we know
$h=\sqrt{\frac{K_h}{C}}=\sqrt{\frac{{10}^{-9}}{0.1}}=\sqrt{{10}^{-8}}$
$h={10}^{-4}$
$\Rightarrow \text{degree of hydrolysis}\left(h\right)={10}^{-4}$
and also
$\text{percentage degree of hydrolysis}={10}^{-4}\times 100=0.01$