A weak acid `HX` has the dissociation constant `1 xx 10^(-5)M`. It forms a salt `NaX` on reaction with alkali. The percentage hydrolysis of `0.1M` solution of `NaX` is

Options

A. `0.001 %`
B. `0.01%`
C. `0.1%`
D. `0.15%`

Explanation

To solve the problem of finding the percentage hydrolysis of a 0.1 M solution of the salt NaX formed from the weak acid HX, we can follow these steps:

### Step 1: Understand the Hydrolysis of NaX
NaX is a salt formed from a weak acid (HX) and a strong base (NaOH). When NaX is dissolved in water, it undergoes hydrolysis, which can be represented as follows:
\[ \text{NaX} \rightleftharpoons \text{Na}^+ + \text{X}^- \]
The anion \( \text{X}^- \) can react with water:
\[ \text{X}^- + \text{H}_2\text{O} \rightleftharpoons \text{HX} + \text{OH}^- \]

### Step 2: Calculate the Hydrolysis Constant (Kh)
The hydrolysis constant \( K_h \) can be calculated using the relationship:
\[ K_h = \frac{K_w}{K_a} \]
Where:
- \( K_w = 1 \times 10^{-14} \) (the ion product of water)
- \( K_a = 1 \times 10^{-5} \) (the dissociation constant of the weak acid HX)

Substituting the values:
\[ K_h = \frac{1 \times 10^{-14}}{1 \times 10^{-5}} = 1 \times 10^{-9} \]

### Step 3: Set Up the Hydrolysis Equation
Let \( x \) be the degree of hydrolysis. The concentration of \( \text{X}^- \) at equilibrium will be:
\[ [\text{X}^-] = 0.1 - x \]
The concentration of \( \text{HX} \) and \( \text{OH}^- \) at equilibrium will both be \( x \).

### Step 4: Write the Expression for Kh
Using the equilibrium concentrations, we can write the expression for \( K_h \):
\[ K_h = \frac{[\text{HX}][\text{OH}^-]}{[\text{X}^-]} = \frac{x \cdot x}{0.1 - x} \]
Assuming \( x \) is small compared to 0.1, we can approximate:
\[ K_h \approx \frac{x^2}{0.1} \]

### Step 5: Solve for x
Substituting the value of \( K_h \):
\[ 1 \times 10^{-9} = \frac{x^2}{0.1} \]
\[ x^2 = 1 \times 10^{-9} \times 0.1 = 1 \times 10^{-10} \]
\[ x = \sqrt{1 \times 10^{-10}} = 1 \times 10^{-5} \]

### Step 6: Calculate the Percentage Hydrolysis
The percentage hydrolysis can be calculated as:
\[ \text{Percentage Hydrolysis} = \left( \frac{x}{C} \right) \times 100 \]
Where \( C = 0.1 \, \text{M} \):
\[ \text{Percentage Hydrolysis} = \left( \frac{1 \times 10^{-5}}{0.1} \right) \times 100 = 0.01\% \]

### Final Answer
The percentage hydrolysis of the 0.1 M solution of NaX is **0.01%**.

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