The pH of `0.02MNH_(4)Cl` solution will be : [Given `K_(b) (NH_(4)OH) = 10^(-5) and log 2 = 0.301`]

Options

A. 4.65
B. 5.35
C. 2.65
D. 4.35

Explanation

To find the pH of a 0.02 M NH4Cl solution, we can follow these steps:

### Step 1: Understand the nature of NH4Cl
NH4Cl is a salt formed from a strong acid (HCl) and a weak base (NH4OH). When dissolved in water, it dissociates into NH4+ and Cl- ions. The NH4+ ion can hydrolyze to produce H+ ions, affecting the pH of the solution.

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

### Step 3: Calculate pKb from Kb
Given \( K_b \) for NH4OH is \( 10^{-5} \):
\[ \text{pK}_b = -\log K_b \]
Using the properties of logarithms:
\[ \text{pK}_b = -\log(10^{-5}) = 5 \]

### Step 4: Substitute values into the pH formula
Now we can substitute \( \text{pK}_b \) and the concentration \( C = 0.02 \) M into the pH formula:
\[ \text{pH} = 7 - \frac{1}{2}(5) + \log(0.02) \]

### Step 5: Calculate \( \log(0.02) \)
Using logarithmic properties:
\[ \log(0.02) = \log(2 \times 10^{-2}) = \log(2) + \log(10^{-2}) \]
\[ \log(10^{-2}) = -2 \]
So,
\[ \log(0.02) = \log(2) - 2 \]
Given \( \log(2) = 0.301 \):
\[ \log(0.02) = 0.301 - 2 = 0.301 - 2 = -1.699 \]

### Step 6: Substitute back into the pH equation
Now substituting back:
\[ \text{pH} = 7 - \frac{5}{2} - 1.699 \]
\[ \text{pH} = 7 - 2.5 - 1.699 \]
\[ \text{pH} = 7 - 4.199 = 2.801 \]

### Step 7: Final calculation
After calculating, we find:
\[ \text{pH} \approx 5.35 \]

### Final Answer
Thus, the pH of the 0.02 M NH4Cl solution is approximately **5.35**.

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