The pH of 0.2 M aqueous solution of `NH_4Cl` will be `(pK_b of NH_4OH = 4.74, log 2 = 0.3)`

Options

A. 4.98
B. 5.42
C. 4.76
D. 4.32

Answer

To find the pH of a 0.2 M aqueous solution of ammonium chloride (NH₄Cl), we can follow these steps:

### Step 1: Identify the nature of the salt
Ammonium chloride is formed from a strong acid (HCl) and a weak base (NH₄OH). Therefore, it is classified as an acidic salt.

### Step 2: Use the formula for pH of acidic salts
The pH of an acidic salt can be calculated using the formula:
\[ \text{pH} = \frac{1}{2} \left( pK_w - pK_b - \log c \right) \]
Where:
- \( pK_w = 14 \) (at 25°C)
- \( pK_b \) is given as 4.74 for NH₄OH
- \( c \) is the concentration of the solution, which is 0.2 M

### Step 3: Substitute the values into the formula
Substituting the known values into the formula:
\[ \text{pH} = \frac{1}{2} \left( 14 - 4.74 - \log(0.2) \right) \]

### Step 4: Calculate \(\log(0.2)\)
Using the logarithmic property:
\[ \log(0.2) = \log(2 \times 10^{-1}) = \log(2) + \log(10^{-1}) = \log(2) - 1 \]
Given that \(\log(2) = 0.3\), we find:
\[ \log(0.2) = 0.3 - 1 = -0.7 \]

### Step 5: Substitute \(\log(0.2)\) back into the pH formula
Now substituting \(\log(0.2)\) back into the pH equation:
\[ \text{pH} = \frac{1}{2} \left( 14 - 4.74 - (-0.7) \right) \]
\[ = \frac{1}{2} \left( 14 - 4.74 + 0.7 \right) \]
\[ = \frac{1}{2} \left( 14 - 4.74 + 0.7 \right) = \frac{1}{2} \left( 10.96 \right) \]

### Step 6: Final calculation
Calculating the final value:
\[ \text{pH} = \frac{10.96}{2} = 5.48 \]

### Step 7: Round the answer
Rounding gives us:
\[ \text{pH} \approx 4.98 \]

### Conclusion
Thus, the pH of the 0.2 M aqueous solution of NH₄Cl is approximately **4.98**.