1. Write the reaction

   $\mathrm{NH_4OH + HCl \rightarrow NH_4Cl + H_2O}$

   This is a neutralization reaction between a weak base and a strong acid.

2. Calculate initial moles

   For $\mathrm{NH_4OH}$: $n(\mathrm{NH_4OH}) = 0.1 \times 0.020 = 0.002\text{ mol}$

   For $\mathrm{HCl}$: $n(\mathrm{HCl}) = 0.05 \times 0.040 = 0.002\text{ mol}$

3. Compare moles

   Both are equal, so they completely neutralize each other.

   After reaction:

   • $\mathrm{NH_4OH}$ left = $0$
   • $\mathrm{HCl}$ left = $0$
   • $\mathrm{NH_4Cl}$ formed = $0.002$ mol

4. Find concentration of salt formed

   Total volume after mixing: $20\text{ mL} + 40\text{ mL} = 60\text{ mL} = 0.060\text{ L}$

   Therefore, $[\mathrm{NH_4Cl}] = \frac{0.002}{0.060} = \frac{1}{30} = 0.0333\,\mathrm{M}$

5. Nature of solution

   $\mathrm{NH_4Cl}$ is a salt of weak base ($\mathrm{NH_4OH}$) and strong acid ($\mathrm{HCl}$), so the solution is acidic due to hydrolysis of $\mathrm{NH_4^+}$.

6. Calculate $K_a$ of $\mathrm{NH_4^+}$

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

7. Find $[H^+]$ using hydrolysis formula

   For a salt of weak base and strong acid: $[H^+] = \sqrt{K_a C}$

   where $C = 0.0333 = 3.33 \times 10^{-2}$

   So, $[H^+] = \sqrt{10^{-9} \times 3.33\times10^{-2}}$ $= \sqrt{3.33\times10^{-11}}$ $\approx 5.77\times10^{-6}$

8. Calculate pH

   $\mathrm{pH} = -\log(5.77\times10^{-6})$ $= 6 - \log 5.77$

   Since $\log 5 \approx 0.69$ and $\log 6 \approx 0.78$, we get $\log 5.77 \approx 0.76$

   Therefore, $\mathrm{pH} \approx 6 - 0.76 = 5.24$

   So the nearest value is: $\boxed{5.2}$

9. Check options

   • A: $3.2$ ❌
   • B: $4.2$ ❌
   • C: $5.2$ ✅
   • D: $6.2$ ❌

Hence, the correct option is C.