Ionisation constant of CH_3COOH is 1.7 times | vedclass

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Question

(D) The ionization reaction for $CH_3COOH$ is:$CH_3COOH \rightleftharpoons CH_3COO^{-} + H^{+}$At equilibrium,the concentration of $[CH_3COO^{-}] = [H

Vedclass · Accepted Answer

$6.8 	imes 10^{-3}$

Vedclass · Answer

(D) The ionization reaction for $CH_3COOH$ is:$CH_3COOH ightleftharpoons CH_3COO^{-} + H^{+}$At equilibrium,the concentration of $[CH_3COO^{-}] = [H^{+}] = 3.4 	imes 10^{-4} \ M$.Let the initial concentration of $CH_3COOH$ be $C$. At equilibrium,$[CH_3COOH] = C - 3.4 	imes 10^{-4} \approx C$.The ionization constant $K_a$ is given by:$K_a = \frac{[CH_3COO^{-}][H^{+}]}{[CH_3COOH]}$Substituting the values:$1.7 	imes 10^{-5} = \frac{(3.4 	imes 10^{-4})(3.4 	imes 10^{-4})}{C}$$C = \frac{11.56 	imes 10^{-8}}{1.7 	imes 10^{-5}}$$C = 6.8 	imes 10^{-3} \ M$Hence,option $D$ is correct.

Ionisation constant of $CH_3COOH$ is $1.7 \times 10^{-5}$ and concentration of $H^{+}$ ions is $3.4 \times 10^{-4} \ M$. Find the initial concentration of $CH_3COOH$ molecules.

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