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Ionisation constant of $CH_3COOH$ is $1.7 \times 10^{-5}$ and concentration of $H^+$ ions is $3.4 \times 10^{-4}$. Then find out initial concentration of $CH_3COOH$ molecules : -
A
$3.4 \times 10^{-4}$
B
$3.4 \times 10^{-3}$
C
$6.8 \times 10^{-4}$
D
$6.8 \times 10^{-3}$
Detailed Solution
For a weak acid: $K_a = \dfrac{[H^+][CH_3COO^-]}{[CH_3COOH]} \approx \dfrac{[H^+]^2}{C}$ (since the degree of ionisation is small, $[H^+] = [CH_3COO^-]$ and $[CH_3COOH] \approx C$)
$C = \dfrac{[H^+]^2}{K_a} = \dfrac{(3.4 \times 10^{-4})^2}{1.7 \times 10^{-5}}$
$= \dfrac{11.56 \times 10^{-8}}{1.7 \times 10^{-5}} = 6.8 \times 10^{-3}$ M
$C = \dfrac{[H^+]^2}{K_a} = \dfrac{(3.4 \times 10^{-4})^2}{1.7 \times 10^{-5}}$
$= \dfrac{11.56 \times 10^{-8}}{1.7 \times 10^{-5}} = 6.8 \times 10^{-3}$ M