Calculate the pH of 0.02 M monobasic acid having $2 \%$ dissociation.

Calculate the pH of 0.02 M monobasic acid having $2 \%$ dissociation.
  1. 3.4
  2. 4.5
  3. 5.1
  4. 5.8

Solution

$\begin{aligned} & \mathrm{HA} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{A}_{(\mathrm{aq})}^{-} \\ & {[\mathrm{H}]^{+}=\alpha \mathrm{C}=\frac{2}{100} \times 0.02 \mathrm{~m}=0.0004 \mathrm{~m}} \\ & \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log (0.0004)=3.4\end{aligned}$

Asked in: MHT CET 2024 (16 May Shift 2)

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