20 ml of 0.1 M $$H_2SO_4$$ solution is added to 30 mL of 0.2 M $$NH_4OH$$ solution. The pH of the resultant mixture is: ($$pK_b$$ of $$NH_4OH = 4.7$$)

First we calculate the millimoles (mmol) of each reagent, using the relation $$\text{millimoles} = \text{molarity (M)} \times \text{volume (mL)}\;.$$

For the acid, $$H_2SO_4$$ is diprotic, so every mole gives two moles of $$H^+$$. We have

$$20\ \text{mL}\times 0.1\ \text{M}=2.0\ \text{mmol of }H_2SO_4$$

which produces $$2 \times 2.0 = 4.0\ \text{mmol of }H^+\;.$$

For the base, $$NH_4OH$$ (a weak base) gives

$$30\ \text{mL}\times 0.2\ \text{M}=6.0\ \text{mmol of }NH_4OH\;.$$

The neutralisation reaction is

$$\displaystyle NH_4OH + H^+ \rightarrow NH_4^+ + H_2O\;.$$

4.0 mmol of $$H^+$$ react with 4.0 mmol of $$NH_4OH$$ (one-to-one stoichiometry), leaving

$$6.0-4.0 = 2.0\ \text{mmol of unreacted }NH_4OH$$

and producing

$$4.0\ \text{mmol of }NH_4^+\;.$$

The total volume after mixing is

$$20\ \text{mL}+30\ \text{mL}=50\ \text{mL}=0.050\ \text{L}\;.$$

Hence the concentrations in the final 50 mL are

$$[NH_4OH]=\frac{2.0\ \text{mmol}}{50\ \text{mL}}=0.04\ \text{M},\qquad [NH_4^+]=\frac{4.0\ \text{mmol}}{50\ \text{mL}}=0.08\ \text{M}\;.$$

The mixture is now a buffer containing a weak base and its conjugate acid. For such a buffer we use the Henderson-Hasselbalch equation written for bases:

$$\boxed{\mathrm{pOH}=pK_b+\log\frac{[\text{salt}]}{[\text{base}]}}\;.$$

Substituting the values, with $$pK_b=4.7$$ and $$\dfrac{[\text{salt}]}{[\text{base}]}=\dfrac{0.08}{0.04}=2$$, we obtain

$$\mathrm{pOH}=4.7+\log 2=4.7+0.3=5.0\;.$$

The relation between pH and pOH for water at 298 K is $$\mathrm{pH}=14-\mathrm{pOH}\;,$$ so

$$\mathrm{pH}=14-5.0=9.0\;.$$

Hence, the correct answer is Option D.