For a 0.01 M HCN solution,the van't Hoff fa | vedclass

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(B) The van't Hoff factor $i$ for a weak electrolyte $HCN$ is given by $i = 1 + \alpha$,where $\alpha$ is the degree of dissociation.Given $i = 1.002$

Vedclass · Accepted Answer

$4 	imes 10^{-8}$

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(B) The van't Hoff factor $i$ for a weak electrolyte $HCN$ is given by $i = 1 + \alpha$,where $\alpha$ is the degree of dissociation.Given $i = 1.002$,we have $1 + \alpha = 1.002$,which implies $\alpha = 0.002$.The dissociation of $HCN$ is $HCN ightleftharpoons H^+ + CN^-$.The equilibrium concentrations are $[H^+] = c\alpha$,$[CN^-] = c\alpha$,and $[HCN] = c(1 - \alpha)$.The acid dissociation constant $K_a$ is given by $K_a = \frac{[H^+][CN^-]}{[HCN]} = \frac{c^2 \alpha^2}{c(1 - \alpha)} = \frac{c \alpha^2}{1 - \alpha}$.Since $\alpha$ is very small,$1 - \alpha \approx 1$.Thus,$K_a \approx c \alpha^2 = (0.01) 	imes (0.002)^2 = 10^{-2} 	imes (2 	imes 10^{-3})^2 = 10^{-2} 	imes 4 	imes 10^{-6} = 4 	imes 10^{-8}$.

For a $0.01 \ M \ HCN$ solution,the van't Hoff factor is $1.002$. What will be the acid dissociation constant $(K_a)$ for $HCN$?

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