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 can follow these steps:

### Step 1: Identify the nature of NH₄Cl
NH₄Cl is a salt formed from the weak base NH₄OH (ammonium hydroxide) and the strong acid HCl. When dissolved in water, it dissociates into NH₄⁺ and Cl⁻ ions. The NH₄⁺ ion can react with water to form NH₄OH and H⁺ ions, making the solution acidic.

### Step 2: Determine the pKₐ of NH₄⁺
Since we are given the pKₐ of the conjugate base (pKₑ of NH₄⁺), we can find the pKₐ using the relation:
\[ pK_a + pK_b = 14 \]
Given that \( pK_b = 4.73 \):
\[ pK_a = 14 - pK_b = 14 - 4.73 = 9.27 \]

### Step 3: Use the formula to calculate pH
The pH of a solution containing a weak acid can be calculated using the formula:
\[ pH = \frac{1}{2} (pK_a - \log C) \]
Where \( C \) is the concentration of the acid (NH₄⁺ in this case).

Substituting the values:
- \( pK_a = 9.27 \)
- \( C = 0.02 \, M \)

We can now calculate:
\[ pH = \frac{1}{2} (9.27 - \log(0.02)) \]

### Step 4: Calculate the logarithm
First, we need to calculate \( \log(0.02) \):
\[ \log(0.02) = \log(2 \times 10^{-2}) = \log(2) + \log(10^{-2}) \]
Using \( \log(2) \approx 0.301 \) and \( \log(10^{-2}) = -2 \):
\[ \log(0.02) \approx 0.301 - 2 = -1.699 \]

### Step 5: Substitute back into the pH formula
Now substituting back into the pH formula:
\[ pH = \frac{1}{2} (9.27 - (-1.699)) \]
\[ pH = \frac{1}{2} (9.27 + 1.699) \]
\[ pH = \frac{1}{2} (10.969) \]
\[ pH \approx 5.4845 \]

### Step 6: Finalize the pH value
Rounding to two decimal places, we get:
\[ pH \approx 5.48 \]

### Conclusion
The pH of the 0.02 M NH₄Cl solution is approximately **5.48**.

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