Calculate the molar solubility of Ni(OH)_2 in | vedclass

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(A) The dissociation of $Ni(OH)_2$ is given by: $Ni(OH)_2(s) \rightleftharpoons Ni^{2+}(aq) + 2OH^-(aq)$.Let $s$ be the molar solubility of $Ni(OH)_2$

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The dissociation of $Ni(OH)_2$ is given by: $Ni(OH)_2(s) ightleftharpoons Ni^{2+}(aq) + 2OH^-(aq)$.Let $s$ be the molar solubility of $Ni(OH)_2$.In $0.10 \ M \ NaOH$,the concentration of $OH^-$ ions is $0.10 \ M$ from $NaOH$ plus $2s$ from $Ni(OH)_2$. Since $s$ is very small,we can approximate $[OH^-] \approx 0.10 \ M$.The solubility product expression is $K_{sp} = [Ni^{2+}][OH^-]^2$.Substituting the values: $2 	imes 10^{-15} = (s)(0.10)^2$.$2 	imes 10^{-15} = s 	imes 0.01$.$s = \frac{2 	imes 10^{-15}}{0.01} = 2 	imes 10^{-13} \ M$.

Calculate the molar solubility of $Ni(OH)_2$ in $0.10 \ M \ NaOH$. The solubility product of $Ni(OH)_2$ is $2 \times 10^{-15}$.

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