$$20 \text{ mL}$$ of $$0.1 \text{ M } NH_4OH$$ is mixed with $$40 \text{ mL}$$ of $$0.05 \text{ M HCl}$$. The pH of the mixture is nearest to:
(Given: $$K_b(NH_4OH) = 1 \times 10^{-5}$$, $$\log 2 = 0.30$$, $$\log 3 = 0.48$$, $$\log 5 = 0.69$$, $$\log 7 = 0.84$$, $$\log 11 = 1.04$$)

Given: 20 mL of 0.1 M $$NH_4OH$$ and 40 mL of 0.05 M $$HCl$$ with $$K_b(NH_4OH) = 1 \times 10^{-5}$$.

The moles of $$NH_4OH$$ equal 20 × 0.1 = 2 mmol and the moles of $$HCl$$ equal 40 × 0.05 = 2 mmol.

They react according to $$NH_4OH + HCl \rightarrow NH_4Cl + H_2O$$ and, since the moles are equal, complete neutralization occurs, forming 2 mmol of $$NH_4Cl$$ in 60 mL of solution.

The $$NH_4^+$$ ion undergoes hydrolysis according to $$NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+$$.

The concentration of $$NH_4^+$$ is $$C = \dfrac{2}{60} = \dfrac{1}{30} \text{ M}$$.

The acid dissociation constant of $$NH_4^+$$ is $$K_a = \dfrac{K_w}{K_b} = \dfrac{10^{-14}}{10^{-5}} = 10^{-9}$$.

The hydrogen ion concentration is $$[H^+] = \sqrt{K_a \times C} = \sqrt{10^{-9} \times \dfrac{1}{30}} = \sqrt{\dfrac{10^{-9}}{30}}$$.

Taking the negative logarithm gives $$pH = -\log[H^+] = -\dfrac{1}{2}\log(3.33 \times 10^{-11}) = \dfrac{1}{2}[11 - \log 3.33]$$.

Since $$\log 3.33 = \log\dfrac{10}{3} = \log 10 - \log 3 = 1 - 0.48 = 0.52$$, it follows that $$pH = \dfrac{10.48}{2} = 5.24$$.

Therefore, the correct answer is Option C: $$5.2$$.