The pH of 0.01 M BOH solution is 10. What i | vedclass

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(D) For a weak base $BOH$,the dissociation is: $BOH \rightleftharpoons B^{+} + OH^{-}$.Given concentration $C = 0.01 \ M = 10^{-2} \ M$ and $pH = 10$.

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$1$

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(D) For a weak base $BOH$,the dissociation is: $BOH ightleftharpoons B^{+} + OH^{-}$.Given concentration $C = 0.01 \ M = 10^{-2} \ M$ and $pH = 10$.$pOH = 14 - pH = 14 - 10 = 4$.Therefore,$[OH^{-}] = 10^{-pOH} = 10^{-4} \ M$.For a weak base,$[OH^{-}] = C \alpha$,where $\alpha$ is the degree of dissociation.$10^{-4} = 10^{-2} 	imes \alpha$.$\alpha = \frac{10^{-4}}{10^{-2}} = 10^{-2} = 0.01$.Percentage degree of dissociation = $\alpha 	imes 100 = 0.01 	imes 100 = 1 \%$.

The $pH$ of $0.01 \ M \ BOH$ solution is $10$. What is its degree of dissociation (in $\%$)? (Given $K_{b}$ of $BOH$ is $1 \times 10^{-6}$)

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