The dissociation constant of a base BOH at 25 | vedclass

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Question

(B) For a weak base $BOH$,the dissociation is $BOH \rightleftharpoons B^{+} + OH^{-}$.The dissociation constant $K_b = \frac{[B^{+}][OH^{-}]}{[BOH]} =

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$1.0 	imes 10^{-7} \ M$

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(B) For a weak base $BOH$,the dissociation is $BOH ightleftharpoons B^{+} + OH^{-}$.The dissociation constant $K_b = \frac{[B^{+}][OH^{-}]}{[BOH]} = \frac{c \alpha^2}{1-\alpha} \approx c \alpha^2$.Given $K_b = 1.0 	imes 10^{-12}$ and $c = 0.01 \ M = 10^{-2} \ M$.$1.0 	imes 10^{-12} = 10^{-2} 	imes \alpha^2 \implies \alpha^2 = 10^{-10} \implies \alpha = 10^{-5}$.$[OH^{-}] = c \alpha = 10^{-2} 	imes 10^{-5} = 1.0 	imes 10^{-7} \ M$.

The dissociation constant of a base $BOH$ at $25^{\circ} C$ is $1.0 \times 10^{-12}$. The concentration of $OH^{-}$ in a $0.01 \ M$ aqueous solution is .......

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