An aqueous solution of a weak monoprotic acid, HA , has a...

Given concentration of the weak monoprotic acid, $C = 0.02 \mathrm{~M}$

Degree of dissociation, $\alpha = 0.05\% = \dfrac{0.05}{100} = 5 \times 10^{-4}$

For a weak monoprotic acid, the concentration of hydrogen ions is given by:

$[H^+] = C\alpha$

Substituting the given values:

$[H^+] = 0.02 \times 5 \times 10^{-4} = 10^{-5} \mathrm{~M}$

The $\mathrm{pH}$ of the solution is calculated as:

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

$\mathrm{pH} = -\log(10^{-5}) = 5$

Answer: 5