Equilibrium Ques 64
64. Ionisation constant of $\mathrm{CH}_3 \mathrm{COOH}$ is $1.7 \times 10^{-5}$. If concentration of $\mathrm{H}^{+}$ ions is $3.4 \times 10^{-4} \mathrm{M}$, then find out initial concentration of $\mathrm{CH}_3 \mathrm{COOH}$ molecules
$[2001]$
(a) $3.4 \times 10^{-4} \mathrm{M}$
(b) $3.4 \times 10^{-3} \mathrm{M}$
(c) $6.8 \times 10^{-3} \mathrm{M}$
(d) $6.8 \times 10^{-4} \mathrm{M}$
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Answer:
Correct Answer: 64.(c)
Solution: (c) $\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}$
$ K_a=\frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} $
Given that,
$ \begin{aligned} & {\left[\mathrm{CH} _3 \mathrm{COO}^{-}\right]=\left[\mathrm{H}^{+}\right]=3.4 \times 10^{-4} \mathrm{M}} \\ & \mathrm{K} _{\mathrm{a}} \text { for } \mathrm{CH} _3 \mathrm{COOH}=1.7 \times 10^{-5} \end{aligned} $
$\mathrm{CH}_3 \mathrm{COOH}$ is weak acid, so in it $\left[\mathrm{CH}_3 \mathrm{COOH}\right]$ is equal to initial concentration. Hence
$ \begin{aligned} & 1.7 \times 10^{-5}=\frac{\left(3.4 \times 10^{-4}\right)\left(3.4 \times 10^{-4}\right)}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]} \\ & {\left[\mathrm{CH}_3 \mathrm{COOH}\right]=\frac{3.4 \times 10^{-4} \times 3.4 \times 10^{-4}}{1.7 \times 10^{-5}}} \\ & =6.8 \times 10^{-3} \mathrm{M} \\ & \end{aligned} $
BETA
