A buffer solution is prepared by mixing 0.01 M HCN and 0.02 MNaCN . If $\mathrm{K}_{\mathrm{a}}$ for HCN is…

A buffer solution is prepared by mixing 0.01 M HCN and 0.02 MNaCN . If $\mathrm{K}_{\mathrm{a}}$ for HCN is $6.6 \times 10^{-10}$, what is the concentration of $\mathrm{H}^{+}$ions in solution?
  1. $3.3 \times 10^{-6} \mathrm{M}$
  2. $3.3 \times 10^{-10} \mathrm{M}$
  3. $1.32 \times 10^{-6} \mathrm{M}$
  4. $1.32 \times 10^{-10} \mathrm{M}$

Solution

For an acidic buffer solution, $\begin{aligned} & \mathrm{pH}=\mathrm{pK}_{\mathrm{a},}+\log _{10} \frac{[\text { Salt }]}{[\text { Acid }]} \\ & \mathrm{pH}=-\log _{10}\left(6.6 \times 10^{-10}\right)+\log _{10} \frac{0.02}{0.01} \\ & \left(\because \mathrm{pK}_{\mathrm{a}}=-\log _{10}\left[\mathrm{~K}_{\mathrm{a}}\right]\right) \\ \therefore \quad & \mathrm{pH}=9.18+\log _{10} 2=9.18+0.3010=9.481 \\ & \mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right] \\ \therefore \quad & {\left[\mathrm{H}^{+}\right]=\operatorname{Antilog}(9.481)=3.3 \times 10^{-10} \mathrm{M} } \end{aligned}$

Asked in: MHT CET 2024 (11 May Shift 1)

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