The `pH` of `0.02 M NH_(4)Cl (aq) (pK_(b)=4.73)` is equal to

Options

A. `3.78`
B. `4.73`
C. `5.48`
D. `7.00`

Explanation

To find the pH of a 0.02 M solution of NH₄Cl, we need to consider the hydrolysis of the salt. NH₄Cl is formed from the weak base NH₃ (ammonia) and the strong acid HCl (hydrochloric acid). The hydrolysis of NH₄Cl in water will produce NH₄⁺ ions, which can donate protons (H⁺) to the solution, making it acidic.

### Step-by-Step Solution:

1. **Identify the components**:
NH₄Cl is a salt derived from a weak base (NH₃) and a strong acid (HCl). This means that the solution will be acidic.

2. **Determine the pKₐ of NH₄⁺**:
The relationship between pKₐ and pKₑ is given by:
\[ pK_a + pK_b = pK_w \]
where \( pK_w \) is 14 at 25°C. Given \( pK_b \) of NH₃ is 4.73, we can find \( pK_a \):
\[ pK_a = pK_w - pK_b = 14 - 4.73 = 9.27 \]

3. **Calculate the concentration of NH₄⁺**:
The concentration of NH₄⁺ in the solution is equal to the concentration of NH₄Cl, which is 0.02 M.

4. **Set up the hydrolysis equilibrium**:
The hydrolysis of NH₄⁺ can be represented as:
\[ NH₄^+ + H₂O \rightleftharpoons NH₃ + H₃O^+ \]
The equilibrium constant \( K_a \) for this reaction can be expressed as:
\[ K_a = \frac{[NH₃][H₃O^+]}{[NH₄^+]} \]
Since we are starting with 0.02 M NH₄⁺ and assuming \( x \) is the amount that dissociates:
\[ [NH₄^+] = 0.02 - x \approx 0.02 \quad (\text{since } x \text{ is small}) \]
\[ [NH₃] = x \quad \text{and} \quad [H₃O^+] = x \]

5. **Substituting into the Ka expression**:
The \( K_a \) can be calculated using:
\[ K_a = 10^{-pK_a} = 10^{-9.27} \approx 5.37 \times 10^{-10} \]
Substituting into the equilibrium expression:
\[ 5.37 \times 10^{-10} = \frac{x^2}{0.02} \]

6. **Solving for x**:
Rearranging gives:
\[ x^2 = 5.37 \times 10^{-10} \times 0.02 = 1.074 \times 10^{-11} \]
\[ x = \sqrt{1.074 \times 10^{-11}} \approx 1.037 \times 10^{-6} \]

7. **Calculating pH**:
Since \( [H₃O^+] = x \), we can find the pH:
\[ pH = -\log[H₃O^+] = -\log(1.037 \times 10^{-6}) \approx 5.98 \]

### Final Answer:
The pH of the 0.02 M NH₄Cl solution is approximately **5.98**.