The dissociation constants for acetic acid and $\mathrm{HCN}$ at $25^{\circ} \mathrm{C}$ are $1.5 \times…

The dissociation constants for acetic acid and $\mathrm{HCN}$ at $25^{\circ} \mathrm{C}$ are $1.5 \times 10^{-5}$ and $4.5 \times 10^{-10}$, respectively. The equilibrium constant for the equilibrium - $\mathrm{CN}^{-}+\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{HCN}+\mathrm{CH}_3 \mathrm{COO}^{-}$ would be :
  1. $3.0 \times 10^4$
  2. $3.0 \times 10^5$
  3. $3.0 \times 10^{-5}$
  4. $3.0 \times 10^{-4}$

Solution

$\begin{aligned} \mathrm{K}_{\mathrm{c}} & =\mathrm{K}_{\mathrm{a}\left(\mathrm{CH}_3 \mathrm{COOH}\right)} \times \frac{1}{\mathrm{~K}_{\mathrm{a}(\mathrm{HCN})}} \\ & =1.5 \times 10^{-5} \times \frac{1}{4.5 \times 10^{-10}} \\ & \cong 3 \times 10^4 \end{aligned}$

Asked in: NEET 2009 (Mains)

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