Question

The dissociation constant 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 reaction \(\mathrm{CN}^{-}+\mathrm{CH}_{3}\) \(\mathrm{COOH} \leftrightharpoons \mathrm{HCN}+\mathrm{CH}_{3} \mathrm{COO}^{-}\)
(a) \(3 \times 10^{5}\)
(b) \(3 \times 10^{-5}\)
(c) \(3 \times 10^{-4}\)
(d) \(3 \times 10^{4}\)

Answer 1

Correct option (d) \(3 \times 10^{4}\)
Explanation:
Dissociation of \(\mathrm{CH}_{3} \mathrm{COOH}\)
$$
\begin{gathered}
\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{H}^{+}+\mathrm{CH}_{3} \mathrm{COO}^{-} \\
\mathrm{k}_{\mathrm{a} 1}=1.5 \times 10^{-3}
\end{gathered}
$$
Dissociation of HCN: \(\mathrm{HCN} \rightleftharpoons \mathrm{H}^{+}+\mathrm{CN}^{-}\)
$$
\mathrm{k}_{\mathrm{a} 2}=4.5 \times 10^{-3}
$$
for a reaction
\(\mathrm{CN}^{-}+\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{HCN}\) is
\(\mathrm{Ka}=\frac{\mathrm{Ka}_{1}}{\mathrm{Ka}_{2}}=\frac{1.5 \times 10^{-3}}{4.5 \times 10^{-10}}=3.33 \times 10^{4}\)

Answer 2

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Answer 3

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