\((ii)\,{\text{pH}} = \frac{{\text{1}}}{{\text{2}}}p{K_a} - \frac{1}{2}\log c\,\,\,\,\,\,\,\,\because \,\,{K_a} = {10^{ - 5}}\)
\(pH = \frac{1}{2} \times 5 - \frac{1}{2}\log \,{10^{ - 3}}\,\,\,\,\,\,\,\,\,\,\,\therefore \,p{K_a} = 5\)
\( = \frac{5}{2} + \frac{3}{2} = \frac{8}{2} = 4\)
(આપેલ છે : ${Zn}({OH})_{2}$નો દ્રાવ્યતા ગુણાકાર $2 \times 10^{-20}$ છે.)