The degree of dissociation of 0.1 , M , HCN s | vedclass

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Question

(D) For a weak acid $HCN$,the dissociation equilibrium is: $HCN \rightleftharpoons H^{+} + CN^{-}$ Given: Concentration $C = 0.1 \, M$ Degree of disso

Vedclass · Accepted Answer

$10^{-9}$

Vedclass · Answer

(D) For a weak acid $HCN$,the dissociation equilibrium is: $HCN ightleftharpoons H^{+} + CN^{-}$ Given: Concentration $C = 0.1 \, M$ Degree of dissociation $\alpha = 0.01 \% = \frac{0.01}{100} = 10^{-4}$ The ionisation constant $K_a$ is given by the formula: $K_a = C \alpha^2$ Substituting the values: $K_a = 0.1 	imes (10^{-4})^2$ $K_a = 0.1 	imes 10^{-8} = 10^{-9}$ Therefore,the correct option is $(D)$.

The degree of dissociation of $0.1 \, M \, HCN$ solution is $0.01 \%$. Its ionisation constant would be

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