What is pH of `0.02` M solution of ammonium chloride at `25^(@) C ? K_(b) (NH_(3)) = 1.8 xx 10^(-5) `.

Options

A. `5.477`
B. `8.523`
C. 7
D. `4.8732`

Explanation

To find the pH of a 0.02 M solution of ammonium chloride (NH4Cl) at 25°C, we will follow these steps:

### Step 1: Identify the nature of the salt
Ammonium chloride is a salt formed from the reaction of a strong acid (HCl) and a weak base (NH3). Therefore, the solution of ammonium chloride will be acidic.

### Step 2: Use the formula for pH of acidic salts
For a salt formed from a strong acid and a weak base, the pH can be calculated using the formula:
\[ \text{pH} = 7 - \frac{1}{2} pK_b - \frac{1}{2} \log C \]
where \(C\) is the concentration of the salt solution.

### Step 3: Calculate \(pK_b\)
Given \(K_b\) for ammonia (NH3) is \(1.8 \times 10^{-5}\). We can calculate \(pK_b\) as follows:
\[ pK_b = -\log(K_b) = -\log(1.8 \times 10^{-5}) \]
Using logarithmic properties:
\[ pK_b = -\log(1.8) - \log(10^{-5}) = -\log(1.8) + 5 \]
Using a calculator, we find:
\[ \log(1.8) \approx 0.255 \]
Thus,
\[ pK_b \approx 5 - 0.255 = 4.745 \]

### Step 4: Substitute values into the pH formula
Now substitute \(pK_b\) and concentration \(C = 0.02\) M into the pH formula:
\[ \text{pH} = 7 - \frac{1}{2}(4.745) - \frac{1}{2} \log(0.02) \]

### Step 5: Calculate \(\log(0.02)\)
\[ \log(0.02) = \log(2 \times 10^{-2}) = \log(2) + \log(10^{-2}) \approx 0.301 - 2 = -1.699 \]

### Step 6: Substitute \(\log(0.02)\) into the pH formula
Now substitute this value back into the pH equation:
\[ \text{pH} = 7 - \frac{1}{2}(4.745) - \frac{1}{2}(-1.699) \]
\[ = 7 - 2.3725 + 0.8495 \]
\[ = 7 - 2.3725 + 0.8495 = 5.477 \]

### Final Answer
Thus, the pH of the 0.02 M solution of ammonium chloride at 25°C is approximately **5.477**.

---