1. Write the reaction

Ammonia reacts with hydrochloric acid as:

$\mathrm{NH_3 + HCl \rightarrow NH_4^+ + Cl^-}$

Since HCl is a strong acid, it will neutralize ammonia completely.

2. Calculate initial moles

For ammonia:

$n(\mathrm{NH_3}) = 0.2 \times 0.050 = 0.010\ \text{mol}$

For HCl:

$n(\mathrm{HCl}) = 0.2 \times 0.025 = 0.005\ \text{mol}$

3. Find moles after reaction

Reaction is in $1:1$ ratio.

• HCl is limiting, so it will consume $0.005$ mol of $\mathrm{NH_3}$.
• Remaining $\mathrm{NH_3}$:

$0.010 - 0.005 = 0.005\ \text{mol}$

• Formed $\mathrm{NH_4^+}$:

$0.005\ \text{mol}$

So after mixing, we have a buffer containing equal moles of $\mathrm{NH_3}$ and $\mathrm{NH_4^+}$.

4. Use buffer relation for basic buffer

For a basic buffer:

$\mathrm{pOH} = \mathrm{p}K_b + \log \frac{[\mathrm{salt}]}{[\mathrm{base}]}$

Here,

$[\mathrm{salt}] = [\mathrm{NH_4^+}], \quad [\mathrm{base}] = [\mathrm{NH_3}]$

Since moles are equal, their concentrations are equal in the same total volume, so:

$\frac{[\mathrm{salt}]}{[\mathrm{base}]} = 1$

Thus,

$\mathrm{pOH} = \mathrm{p}K_b + \log 1 = 4.75$

Now,

$\mathrm{pH} = 14 - 4.75 = 9.25$

5. Final answer

$\boxed{\mathrm{pH} = 9.25}$

So the correct option is D.