Equilibrium Ques 21
21. 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]
$CN^{-}+CH_3 COOH \rightleftharpoons HCN+CH_3 COO^{-}$ would be:
(a) $3.0 \times 10^{-5}$
(b) $3.0 \times 10^{-4}$
(c) $3.0 \times 10^{4}$
(d) $3.0 \times 10^{5}$
Show Answer
Answer:
Correct Answer: 21.(c)
Solution:
- (c) Given, $CH_3 COOH \rightleftharpoons CH_3 COO^{-}+H^{+}$;
$K _{a_1}=1.5 \times 10^{-5}$ $\quad$ ……..(i)
$HCN \rightleftharpoons H^{+}+CN^{-} ; K _{a_2}=4.5 \times 10^{-10}$
or $H^{+}+CN^{-} \rightleftharpoons HCN$;
$K_{a_2}^{\prime}=\frac{1}{K _{a_2}}$ $=\frac{1}{4.5 \times 10^{-10}}$ $\quad$ ……..(ii)
$\therefore \quad$ From (i) and (ii), we find that the equilibrium constant $(K_a)$ for the reaction,
$ CN^{-}+CH_3 COOH \rightleftharpoons CH_3 COO^{-}+HCN, \text{ is } $
$ K_a=K_{a_1} \times $ $K_{a_2}^{\prime} $
$ =\frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}}=\frac{1}{3} \times 10^{5}=3.33 \times 10^{4}$
BETA
