Question

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

• A. 0.01
• B. 0.1
• C. 0.001
• D. None of the above

Answer 1

The correct option is A 0.01

Given, $K_a={10}^{-5},\left[NaX\right]=0.1M$
According to question, $NaX$ is a mixture of weak acid and strong base.
Therefore $X^{-}-$ only undergoes hydrolysis
at eqb. $\underset{C-Ch}{X^{-}}+H_{2}O\stackrel{K_a}{\rightleftharpoons }\underset{Ch}{HX}+\underset{Ch}{O{H}^{-}}$
'h' is hydrolysis constant and 'C' is concentration
$K_h=\frac{\left[HX\right]\left[O{H}^{-}\right]}{\left[X^{-}\right]\left[H_{2}O\right]}$
we know,
$\Rightarrow K_h=\frac{K_w}{K_a}=\frac{{10}^{-14}}{{10}^{-5}}$
$\Rightarrow K_h={10}^{-9}$
we know
$h=\sqrt{\frac{K_h}{C}}=\sqrt{\frac{{10}^{-9}}{0.1}}=\sqrt{{10}^{-8}}$
$h={10}^{-4}$
$\Rightarrow \text{degree of hydrolysis}\left(h\right)={10}^{-4}$
$\text{percentage degree of hydrolysis}={10}^{-4}\times 100=0.01$