At 25 , ^circ C,the equilibrium constant for | vedclass

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

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$1.0 	imes 10^{-7} \, mol \, L^{-1}$

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(B) For a weak base $BOH$,the dissociation is given by: $BOH ightleftharpoons B^+ + OH^-$. The dissociation constant $K_b$ is given by $K_b = \frac{[B^+][OH^-]}{[BOH]}$. Since $K_b = 1.0 	imes 10^{-12}$ is very small,we can assume $[BOH] \approx C = 0.01 \, M$. Let the concentration of $OH^-$ be $x$. Then $[B^+] = [OH^-] = x$. $K_b = \frac{x^2}{C} \implies x^2 = K_b 	imes C$. $x^2 = (1.0 	imes 10^{-12}) 	imes (0.01) = 1.0 	imes 10^{-14}$. $x = \sqrt{1.0 	imes 10^{-14}} = 1.0 	imes 10^{-7} \, mol \, L^{-1}$.

At $25 \, ^\circ C$,the equilibrium constant for a base $BOH$ is $1.0 \times 10^{-12}$. What will be the concentration of hydroxide ions in a $0.01 \, M$ aqueous solution of the base?

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