The ionization constant of CH_3COOH is 1.7 ti | vedclass

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(D) For a weak acid $CH_3COOH$,the dissociation equilibrium is $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$. The ionization constant $K_a$ is given b

Vedclass · Accepted Answer

$6.8 	imes 10^{-3} \ M$

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(D) For a weak acid $CH_3COOH$,the dissociation equilibrium is $CH_3COOH ightleftharpoons CH_3COO^- + H^+$. The ionization constant $K_a$ is given by $K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$. Since $[CH_3COO^-] = [H^+] = 3.4 	imes 10^{-4} \ M$,we have $K_a = \frac{[H^+]^2}{C}$,where $C$ is the initial concentration of the acid. Substituting the values: $1.7 	imes 10^{-5} = \frac{(3.4 	imes 10^{-4})^2}{C}$. $C = \frac{11.56 	imes 10^{-8}}{1.7 	imes 10^{-5}} = 6.8 	imes 10^{-3} \ M$.

The ionization constant of $CH_3COOH$ is $1.7 \times 10^{-5}$. In a certain solution of acetic acid,the concentration of $H^+$ is $3.4 \times 10^{-4} \ M$. The concentration of the acetic acid solution is ............ .

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