The pH of 0.1 M HCN solution is 5.2. Calcula | vedclass

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Question

(A) For a weak acid $HCN$,the dissociation is given by: $HCN \rightleftharpoons H^+ + CN^-$. The concentration of $H^+$ ions is calculated from $pH$ a

Vedclass · Accepted Answer

$3.98 	imes 10^{-10}$

Vedclass · Answer

(A) For a weak acid $HCN$,the dissociation is given by: $HCN ightleftharpoons H^+ + CN^-$. The concentration of $H^+$ ions is calculated from $pH$ as: $[H^+] = 10^{-pH} = 10^{-5.2}$. $[H^+] = 6.31 	imes 10^{-6} \ M$. The dissociation constant $K_a$ is given by the formula: $K_a = \frac{[H^+][CN^-]}{[HCN]}$. Since $[H^+] = [CN^-]$,we have $K_a = \frac{[H^+]^2}{[HCN]}$. Substituting the values: $K_a = \frac{(6.31 	imes 10^{-6})^2}{0.1} = \frac{3.98 	imes 10^{-11}}{0.1} = 3.98 	imes 10^{-10}$.

The $pH$ of $0.1 \ M$ $HCN$ solution is $5.2$. Calculate the dissociation constant $K_a$ of this solution.

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