Find out the solubility of Ni(OH)_2 in 0.1 M | vedclass

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(B) $NaOH$ is a strong electrolyte,so it dissociates completely:$NaOH_{(aq)} 	o Na^+_{(aq)} + OH^-_{(aq)}$$[OH^-] = 0.1 \ M$For $Ni(OH)_2$:$Ni(OH)_{2

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$2 	imes 10^{-13} \ M$

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(B) $NaOH$ is a strong electrolyte,so it dissociates completely:$NaOH_{(aq)} 	o Na^+_{(aq)} + OH^-_{(aq)}$$[OH^-] = 0.1 \ M$For $Ni(OH)_2$:$Ni(OH)_{2(s)} ightleftharpoons Ni^{2+}_{(aq)} + 2OH^-_{(aq)}$Let the solubility of $Ni(OH)_2$ in the presence of $0.1 \ M \ NaOH$ be $S' \ M$.Then,$[Ni^{2+}] = S'$ and $[OH^-] = (0.1 + 2S') \ M$.Since $S'$ is very small,we can approximate $(0.1 + 2S') \approx 0.1$.$K_{sp} = [Ni^{2+}][OH^-]^2$$2 	imes 10^{-15} = (S')(0.1)^2$$2 	imes 10^{-15} = S' 	imes 0.01$$S' = \frac{2 	imes 10^{-15}}{10^{-2}} = 2 	imes 10^{-13} \ M$.

Find out the solubility of $Ni(OH)_2$ in $0.1 \ M \ NaOH$.
Given that the solubility product constant $(K_{sp})$ of $Ni(OH)_2$ is $2 \times 10^{-15}$.

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Find out the solubility of $Ni(OH)_2$ in $0.1 \ M \ NaOH$.Given that the solubility product constant $(K_{sp})$ of $Ni(OH)_2$ is $2 \times 10^{-15}$.