Calculate the pH of 0.01 M sulphuric acid.

\(\begin{aligned} & \mathrm{H}_2 \mathrm{SO}_{4(\mathrm{aq})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \longrightarrow 2 \mathrm{H}_3 \mathrm{O}_{(\mathrm{aq})}^{+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-} \\ & \text { Hence, }\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=2 \times \mathrm{c}=2 \times 0.01 \mathrm{M} \\ & \\ & \quad=2 \times 10^{-2} \mathrm{M} \\ & \begin{aligned} \mathrm{pH} & =-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \\ \quad & =-\log _{10}\left(2 \times 10^{-2}\right) \\ & =-\log _{10} 2-\log _{10} 10^{-2} \\ & =-\log _{10} 2+2 \\ & =2-0.3010\end{aligned} \\ & \begin{aligned} \mathrm{pH} & =1.699\end{aligned}\end{aligned}\)