Aniline is a very weak base. For which of the | vedclass

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(C) For a weak base,the degree of dissociation $(\alpha)$ is given by the formula $\alpha = \sqrt{\frac{K_b}{C}}$,where $K_b$ is the dissociation cons

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$0.01$

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(C) For a weak base,the degree of dissociation $(\alpha)$ is given by the formula $\alpha = \sqrt{\frac{K_b}{C}}$,where $K_b$ is the dissociation constant of the base and $C$ is the molar concentration.From the relation $\alpha \propto \frac{1}{\sqrt{C}}$,it is clear that the degree of dissociation increases as the concentration $(C)$ decreases.Comparing the given concentrations: $1 \ M$,$0.1 \ M$,$0.01 \ M$,and $0.02 \ M$.The lowest concentration is $0.01 \ M$.Therefore,the degree of dissociation will be the highest for the $0.01 \ M$ solution.

Aniline is a very weak base. For which of the following concentrations $(M)$ of an aniline solution will the degree of dissociation be the highest?

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