A weak acid `HX(K_(a) = 10^(-5))` on reaction with `NaOH` gives `NaX`. For `0.1M` aqueous solution of `NaX`, the `%` hydrolysis is

Options

A. `1%`
B. `0.01%`
C. `0.001%`
D. `0.15%`

Explanation

To solve the problem of finding the percentage of hydrolysis of the salt NaX formed from the weak acid HX (with \( K_a = 10^{-5} \)) and NaOH, we can follow these steps:

### Step 1: Understand the Hydrolysis of the Salt
When the salt NaX is dissolved in water, it undergoes hydrolysis. The weak acid HX will partially dissociate in water, leading to the formation of \( H^+ \) ions and \( X^- \) ions. The hydrolysis reaction can be represented as:
\[ X^- + H_2O \rightleftharpoons HX + OH^- \]

### Step 2: Calculate the Hydrolysis Constant \( K_h \)
The hydrolysis constant \( K_h \) can be calculated using the relationship:
\[ K_h = \frac{K_w}{K_a} \]
Where:
- \( K_w = 10^{-14} \) (the ion product of water at 25°C)
- \( K_a = 10^{-5} \) (given in the problem)

Substituting the values:
\[ K_h = \frac{10^{-14}}{10^{-5}} = 10^{-9} \]

### Step 3: Determine the Concentration of the Salt
The concentration of the salt NaX is given as \( 0.1 \, M \).

### Step 4: Calculate the Hydrolysis Constant in Terms of Concentration
Using the formula for hydrolysis constant:
\[ K_h = C \cdot \alpha^2 \]
Where:
- \( C \) is the concentration of the salt (0.1 M)
- \( \alpha \) is the degree of hydrolysis (the fraction of the salt that hydrolyzes)

Rearranging gives:
\[ \alpha^2 = \frac{K_h}{C} \]
Substituting the values:
\[ \alpha^2 = \frac{10^{-9}}{0.1} = 10^{-8} \]

### Step 5: Solve for \( \alpha \)
Taking the square root of both sides:
\[ \alpha = \sqrt{10^{-8}} = 10^{-4} \]

### Step 6: Calculate the Percentage of Hydrolysis
The percentage of hydrolysis can be calculated as:
\[ \text{Percentage of Hydrolysis} = \alpha \times 100\% \]
Substituting the value of \( \alpha \):
\[ \text{Percentage of Hydrolysis} = 10^{-4} \times 100 = 0.01\% \]

### Final Answer
The percentage of hydrolysis of the 0.1 M aqueous solution of NaX is \( 0.01\% \).

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