Question

A weak acid HX (${K_a} = 1 \times {10^{ - 5}}$) on reaction with NaOH gives NaX. For 0.1 M aqueous solution of NaX, the % hydrolysis is:
(A) 0.001%
(B) 0.01%
(C) 0.15%
(D) 1%

Answer 1

Hint: The dissociation constant of the acid be found from the formula
\[h = \sqrt {\dfrac{{{K_H}}}{C}} {\text{ }}\]
Where
\[{K_H} \cdot {K_a} = {K_w}{\text{ }}\]

Complete step by step solution:
We are given the dissociation constant (${K_a}$) of the weak acid. Now, we need to find the % hydrolysis of the for the 0.1M solution of NaX.
- Dissociation constant of the acid is the equilibrium constant in the case of dissociation reaction of the acid. We can say that larger the value of the dissociation constant, stronger the acid.
Let’s write the hydrolysis reaction of NaX.
\[NaX + {H_2}O \to NaOH + HX\]
Now, we need to find the % hydrolysis of NaX. Its formula can be given by
\[h = \sqrt {\dfrac{{{K_H}}}{C}} {\text{ }}....{\text{(1)}}\]
Here, C is the concentration and ${K_H}$ can be given as follows.
\[{K_H} \cdot {K_a} = {K_w}{\text{ }}..{\text{(2)}}\]
Putting the value of equation (2) in the equation (1), we get
\[h = \sqrt {\dfrac{{{K_w}}}{{{K_a} \cdot C}}} \]
Here, ${K_w}$ is the ionic constant of the water and its value is always equal to ${10^{ - 14}}$.
Now, we know that value of ${K_w}$ is ${10^{ - 14}}$ and ${K_a}$ is given as ${10^{ - 5}}$. C is given as 0.1M. So, putting this in the above equation give
\[h = \sqrt {\dfrac{{{{10}^{ - 14}}}}{{{{10}^{ - 5}} \times 0.1}}} = \sqrt {{{10}^{ - 8}}} = {10^{ - 4}}\]
Now, we obtained that the degree of hydrolysis is ${10^{ - 4}}$. But we need that % of hydrolysis. So, we will obtain % hydrolysis by following formula
\[\% h = h \times 100\]
So, we can write that
\[\% h = {10^{ - 4}} \times 100 = {10^{ - 2}} = 0.01\% \]
Thus, we can say that the % hydrolysis of NaX will be 0.01%.

Therefore, the correct answer is (B).

Note: Note that ‘h’ here is the degree of the hydrolysis of the salt NaX. It is just the ratio of hydrolysis. Whenever we are asked the % of hydrolysis, then we must multiply the value of degree f of hydrolysis (h) by 100.