Calculate the ionisation constant of 0.08 mo | vedclass

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(D) For a monobasic acid $HA$,the dissociation is $HA \rightleftharpoons H^+ + A^-$. Given concentration $C = 0.08 \ mol \ dm^{-3}$ and $pH = 2$. We k

Vedclass · Accepted Answer

$1.25 	imes 10^{-3}$

Vedclass · Answer

(D) For a monobasic acid $HA$,the dissociation is $HA ightleftharpoons H^+ + A^-$. Given concentration $C = 0.08 \ mol \ dm^{-3}$ and $pH = 2$. We know that $[H^+] = 10^{-pH} = 10^{-2} = 0.01 \ mol \ dm^{-3}$. The degree of dissociation $\alpha$ is given by $\alpha = \frac{[H^+]}{C} = \frac{0.01}{0.08} = 0.125$. The ionisation constant $K_a$ is given by the formula $K_a = \frac{C\alpha^2}{1-\alpha}$. Substituting the values: $K_a = \frac{0.08 	imes (0.125)^2}{1 - 0.125} = \frac{0.08 	imes 0.015625}{0.875} = \frac{0.00125}{0.875} \approx 1.428 	imes 10^{-3}$. Re-evaluating the approximation for $K_a \approx C\alpha^2$ (if $\alpha$ is small) or using the exact formula,the closest provided option is $D$.

Calculate the ionisation constant of $0.08 \ mol \ dm^{-3}$ of a monobasic acid having $pH = 2$.

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