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 can follow these steps:

### Step 1: Identify the nature of the salt
Ammonium chloride is formed from the weak base ammonia (NH3) and the strong acid hydrochloric acid (HCl). Therefore, NH4Cl will behave as an acidic salt in solution.

### Step 2: Use the relationship between Kb and Ka
Since NH4Cl is derived from a weak base (NH3), we can relate the base dissociation constant (Kb) of NH3 to the acid dissociation constant (Ka) of NH4+ using the equation:
\[ K_w = K_a \times K_b \]
Where \( K_w \) (the ion product of water) is \( 1.0 \times 10^{-14} \) at 25°C.

Given:
\[ K_b(NH3) = 1.8 \times 10^{-5} \]

We can calculate \( K_a(NH4^+) \):
\[ K_a(NH4^+) = \frac{K_w}{K_b(NH3)} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \]

### Step 3: Calculate Ka
Calculating \( K_a \):
\[ K_a(NH4^+) = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10} \]

### Step 4: Set up the equilibrium expression
For the dissociation of NH4+ in water:
\[ NH4^+ \rightleftharpoons NH3 + H^+ \]

Let \( x \) be the concentration of \( H^+ \) produced at equilibrium. The initial concentration of \( NH4^+ \) is 0.02 M, and at equilibrium, we have:
- \([NH4^+] = 0.02 - x\)
- \([NH3] = x\)
- \([H^+] = x\)

Using the Ka expression:
\[ K_a = \frac{[NH3][H^+]}{[NH4^+]} \]
\[ 5.56 \times 10^{-10} = \frac{x \cdot x}{0.02 - x} \approx \frac{x^2}{0.02} \]

### Step 5: Solve for x
Assuming \( x \) is small compared to 0.02, we can simplify:
\[ 5.56 \times 10^{-10} \cdot 0.02 = x^2 \]
\[ x^2 = 1.112 \times 10^{-11} \]
\[ x = \sqrt{1.112 \times 10^{-11}} \approx 1.05 \times 10^{-6} \]

### Step 6: Calculate pH
Since \( x \) represents the concentration of \( H^+ \):
\[ pH = -\log[H^+] = -\log(1.05 \times 10^{-6}) \]

Calculating the pH:
\[ pH \approx 5.98 \]

### Final Answer
The pH of a 0.02 M solution of ammonium chloride at 25°C is approximately **5.98**.