50 mL of 10^{-3} M HCl is mixed with 50 m | vedclass

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Question

$(A)$ The concentration of $H^+$ from $HCl$ is $10^{-3} \ M$. Since $HCl$ is a strong acid, it suppresses the dissociation of the weak acid $CH_3COOH$

Vedclass · Accepted Answer

$1.8 	imes 10^{-13}$

Vedclass · Answer

$(A)$ The concentration of $H^+$ from $HCl$ is $10^{-3} \ M$. Since $HCl$ is a strong acid, it suppresses the dissociation of the weak acid $CH_3COOH$. The total $[H^+]$ in the mixture is dominated by the strong acid $HCl$, which is approximately $10^{-3} \ M$. Using the ionic product of water, $K_w = [H^+][OH^-] = 10^{-14}$, we find $[OH^-] = \frac{10^{-14}}{10^{-3}} = 10^{-11} \ M$. The degree of dissociation of water $(\alpha_w)$ is given by $\alpha_w = \frac{[OH^-]}{[H_2O]_{total}} = \frac{10^{-11}}{1000/18} = 1.8 	imes 10^{-13}$.

$50 \ mL$ of $10^{-3} \ M \ HCl$ is mixed with $50 \ mL$ of $0.05 \ M \ CH_3COOH$ $(pK_a = 4.7)$. The degree of dissociation of water in the resulting solution is

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