The pH of `0.02MNH_(4)Cl` solution will be : [Given `K_(b) (NH_(4)OH) = 10^(-5) and log 2 = 0.301`]

Options

A. `4.65`
B. `5.35`
C. `2.65`
D. `4.35`

Explanation

To find the pH of a 0.02 M solution of NH₄Cl, we can follow these steps:

### Step 1: Understand the nature of NH₄Cl
NH₄Cl is a salt formed from the weak base NH₄OH (ammonium hydroxide) and a strong acid HCl. In solution, NH₄Cl dissociates into NH₄⁺ and Cl⁻ ions. The NH₄⁺ ion can hydrolyze to produce H⁺ ions, which will affect the pH of the solution.

### Step 2: Write the hydrolysis reaction
The hydrolysis of NH₄⁺ can be represented as:
\[ \text{NH}_4^+ + \text{H}_2O \rightleftharpoons \text{NH}_4\text{OH} + \text{H}^+ \]

### Step 3: Use the Kb value
Given that \( K_b \) for NH₄OH is \( 10^{-5} \), we can find the \( K_a \) for NH₄⁺ using the relation:
\[ K_w = K_a \cdot K_b \]
Where \( K_w = 1.0 \times 10^{-14} \) at 25°C.

Thus,
\[ K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{10^{-5}} = 1.0 \times 10^{-9} \]

### Step 4: Calculate the concentration of H⁺ ions
Using the formula for \( K_a \):
\[ K_a = \frac{[H^+][NH_4OH]}{[NH_4^+]} \]

Assuming that \( x \) is the concentration of \( H^+ \) produced:
\[ K_a = \frac{x^2}{0.02 - x} \approx \frac{x^2}{0.02} \] (since \( x \) is small compared to 0.02)

Setting up the equation:
\[ 1.0 \times 10^{-9} = \frac{x^2}{0.02} \]

### Step 5: Solve for x
Rearranging gives:
\[ x^2 = 1.0 \times 10^{-9} \times 0.02 \]
\[ x^2 = 2.0 \times 10^{-11} \]
\[ x = \sqrt{2.0 \times 10^{-11}} \]
\[ x \approx 4.47 \times 10^{-6} \]

### Step 6: Calculate the pH
Since \( x \) represents the concentration of \( H^+ \):
\[ \text{pH} = -\log[H^+] = -\log(4.47 \times 10^{-6}) \]

Using the logarithm property:
\[ \text{pH} \approx 5.35 \]

### Final Answer
The pH of the 0.02 M NH₄Cl solution is approximately **5.35**.

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