Calculate the $\mathrm{pH}$ of $0.01 \mathrm{M}$ strong dibasic acid.

Calculate the $\mathrm{pH}$ of $0.01 \mathrm{M}$ strong dibasic acid.
  1. $5.5$
  2. $2.5$
  3. $2.0$
  4. $1.7$

Solution

$\begin{aligned} & {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=2 \times \mathrm{c}=2 \times 0.01 \mathrm{M}=2 \times 10^{-2} \mathrm{M}} \\ & \mathrm{pH}=-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \\ & =-\log _{10}\left[2 \times 10^{-2}\right] \\ & =-\log _{10} 2-\log _{10} 10^{-2} \\ & =-\log _{10} 2+2 \\ & =2-0.3010 \\ & \mathrm{pH}=1.7 \\ & \end{aligned}$

Asked in: MHT CET 2023 (14 May Shift 1)

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