Equilibrium - Result Question 27
####28. The dissociation constants for acetic acid and $HCN$ at $25^{\circ} C$ are $1.5 \times 10^{-5}$ and $4.5 \times 10^{-10}$ respectively. The equilibrium constant for the equilibrium
[2009]
$CN^{-}+CH_3 COOH \rightarrow HCN+CH_3 COO^{-}$ would be:
(a) $3.0 \times 10^{-5}$
(b) $3.0 \times 10^{-4}$
(c) $3.0 \times 10^{4}$
(d) $3.0 \times 10^{5}$
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Solution:
- (c) Given, $CH_3 COOH \rightarrow CH_3 COO^{-}+H^{+}$;
$K _{a_1}=1.5 \times 10^{-5}$ $HCN \rightarrow H^{+}+CN^{-} ; K _{a_2}=4.5 \times 10^{-10}$
or $H^{+}+CN^{-} \rightarrow HCN$;
$K_a_2^{\prime}=\frac{1}{K _{a_2}}=\frac{1}{4.5 \times 10^{-10}}$
$\therefore \quad$ From (i) and (ii), we find that the equilibrium constant $(K_a)$ for the reaction,
$ \begin{aligned} & CN^{-}+CH_3 COOH \rightarrow CH_3 COO^{-}+HCN, \text{ is } \\ & K_a=K _{a_1} \times K_a_2^{\prime} \\ & \quad=\frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}}=\frac{1}{3} \times 10^{5}=3.33 \times 10^{4} \end{aligned} $
