1. Identify the nature of the salt

$NH_4Cl$ is a salt of:

• weak base: $NH_4OH$
• strong acid: $HCl$

So its solution is acidic due to hydrolysis of $NH_4^+$:

$NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+$

2. Find the acidic dissociation constant of $NH_4^+$

Given:

$K_b(NH_4OH)=10^{-5}$

For conjugate acid-base pair:

$K_a \cdot K_b = K_w = 10^{-14}$

Hence,

$K_a = \frac{10^{-14}}{10^{-5}} = 10^{-9}$

3. Use the formula for pH of a salt of weak base and strong acid

For concentration $C = 0.02\,M$,

$[H^+] \approx \sqrt{K_a C}$

So,

$[H^+] = \sqrt{10^{-9} \times 0.02}$

Write $0.02$ as $2 \times 10^{-2}$:

$[H^+] = \sqrt{2 \times 10^{-11}}$

$[H^+] = \sqrt{2} \times 10^{-5.5}$

Now calculate pH:

$\text{pH} = -\log[H^+]$

$\text{pH} = -\log(\sqrt{2} \times 10^{-5.5})$

$\text{pH} = 5.5 - \log(\sqrt{2})$

Using:

$\log 2 = 0.301 \Rightarrow \log(\sqrt{2}) = \frac{1}{2}\log 2 = 0.1505$

Therefore,

$\text{pH} = 5.5 - 0.1505 = 5.3495$

$\boxed{\text{pH} \approx 5.35}$

4. Check options

• A: $2.56$ ❌
• B: $5.35$ ✅
• C: $4.35$ ❌
• D: $4.65$ ❌

Therefore, the correct option is:

$\boxed{\text{B}}$