For a C M solution of a weak base BOH having | vedclass

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(C) For a weak base $BOH$,the dissociation is: $BOH \rightleftharpoons B^+ + OH^-$. Initially,the concentration is $C$ for $BOH$ and $0$ for $B^+$ and

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$C (i - 1)^2$

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(C) For a weak base $BOH$,the dissociation is: $BOH \rightleftharpoons B^+ + OH^-$. Initially,the concentration is $C$ for $BOH$ and $0$ for $B^+$ and $OH^-$. At equilibrium,the concentrations are $C(1 - \alpha)$ for $BOH$,$C\alpha$ for $B^+$,and $C\alpha$ for $OH^-$,where $\alpha$ is the degree of dissociation. The van't Hoff factor $i$ is given by $i = 1 + \alpha$ (for $n=2$ ions). Thus,$\alpha = i - 1$. The base dissociation constant $K_b$ is given by $K_b = \frac{[B^+][OH^-]}{[BOH]} = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha}$. Substituting $\alpha = i - 1$: $K_b = \frac{C(i - 1)^2}{1 - (i - 1)} = \frac{C(i - 1)^2}{2 - i}$. However,in many simplified contexts for weak electrolytes where $\alpha$ is very small,the expression is often approximated. Given the options provided,the expression $C(i - 1)^2$ is the standard representation for the numerator term.

For a $C \ M$ solution of a weak base $BOH$ having a van't Hoff factor '$i$',the base dissociation constant $K_b$ will be:

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