If the solubility product of a base M(OH)_3 i | vedclass

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(B) The dissociation of the base is given by: $M(OH)_3 \rightleftharpoons M^{3+} + 3OH^-$Let the solubility be $s$. Then $[M^{3+}] = s$ and $[OH^-] =

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$3 	imes 10^{-3}$

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(B) The dissociation of the base is given by: $M(OH)_3 ightleftharpoons M^{3+} + 3OH^-$Let the solubility be $s$. Then $[M^{3+}] = s$ and $[OH^-] = 3s$.The solubility product expression is: $K_{sp} = [M^{3+}][OH^-]^3$Substituting the values: $2.7 	imes 10^{-11} = (s)(3s)^3$$2.7 	imes 10^{-11} = 27s^4$$s^4 = \frac{2.7 	imes 10^{-11}}{27} = 0.1 	imes 10^{-11} = 10^{-12}$$s = (10^{-12})^{1/4} = 10^{-3} \ M$The concentration of $OH^-$ is $[OH^-] = 3s = 3 	imes 10^{-3} \ M$.

If the solubility product of a base $M(OH)_3$ is $2.7 \times 10^{-11}$,find the concentration of $OH^-$.

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