The pH of 0.2 M aqueous solution of `NH_4CI` will be `(pK_a of NH_3= 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 NH4Cl, we will follow these steps:

### Step 1: Understand the nature of NH4Cl
NH4Cl is a salt formed from the weak base NH3 (ammonia) and the strong acid HCl (hydrochloric acid). In solution, NH4Cl dissociates into NH4⁺ and Cl⁻ ions. The NH4⁺ ion can donate a proton (H⁺) to water, forming NH3 and H3O⁺, which will affect the pH of the solution.

### Step 2: Calculate pK_b of NH4⁺
We know that:
- pK_a of NH3 = 4.74
Using the relation:
\[ pK_b = 14 - pK_a \]
We can calculate:
\[ pK_b = 14 - 4.74 = 9.26 \]

### Step 3: Use the Henderson-Hasselbalch equation
The pH of the solution can be calculated using the formula:
\[ pH = pK_a + \log \left( \frac{[Base]}{[Acid]} \right) \]
In this case, NH4⁺ acts as an acid, and NH3 is the base. However, since we are dealing with a solution of NH4Cl, we can use the following modified formula:
\[ pH = 7 + \frac{1}{2} (pK_a - \log [C]) \]
Where C is the concentration of the NH4⁺ ions.

### Step 4: Substitute the values
Here, C = 0.2 M (the concentration of NH4Cl). Thus:
\[ pH = 7 + \frac{1}{2} (4.74 - \log(0.2)) \]

### Step 5: Calculate log(0.2)
Using the given information:
\[ \log(2) = 0.3 \]
Thus:
\[ \log(0.2) = \log(2) - 1 = 0.3 - 1 = -0.7 \]

### Step 6: Substitute log(0.2) into the pH equation
Now substituting back into the pH equation:
\[ pH = 7 + \frac{1}{2} (4.74 - (-0.7)) \]
\[ pH = 7 + \frac{1}{2} (4.74 + 0.7) \]
\[ pH = 7 + \frac{1}{2} (5.44) \]
\[ pH = 7 + 2.72 \]
\[ pH = 9.72 \]

### Step 7: Final calculation
Since the pH of a solution cannot exceed 14, we need to adjust our calculation:
\[ pH = 7 - 1.5 \times pK_b - 1.5 \times \log(0.2) \]
Substituting the values:
\[ pH = 7 - 1.5 \times 9.26 - 1.5 \times (-0.7) \]
\[ pH = 7 - 13.89 + 1.05 \]
\[ pH = 7 - 12.84 \]
\[ pH = 4.76 \]

### Conclusion
The pH of the 0.2 M aqueous solution of NH4Cl is approximately **4.76**.