K_{sp} of a salt Ni(OH)_2 is 2 times 10^{-15} | vedclass

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(C) The dissociation of $Ni(OH)_2$ is given by: $Ni(OH)_2(s) \rightleftharpoons Ni^{2+}(aq) + 2OH^-(aq)$.In $0.01 \ M \ NaOH$,the concentration of $OH

Vedclass · Accepted Answer

$2 	imes 10^{-11} \ M$

Vedclass · Answer

(C) The dissociation of $Ni(OH)_2$ is given by: $Ni(OH)_2(s) ightleftharpoons Ni^{2+}(aq) + 2OH^-(aq)$.In $0.01 \ M \ NaOH$,the concentration of $OH^-$ ions is dominated by the strong electrolyte $NaOH$,so $[OH^-] \approx 0.01 \ M = 10^{-2} \ M$.Let the molar solubility of $Ni(OH)_2$ be $S$.Then,$[Ni^{2+}] = S$ and $[OH^-] = (2S + 0.01) \approx 0.01 \ M$.The solubility product expression is $K_{sp} = [Ni^{2+}][OH^-]^2$.Substituting the values: $2 	imes 10^{-15} = S 	imes (10^{-2})^2$.$2 	imes 10^{-15} = S 	imes 10^{-4}$.$S = \frac{2 	imes 10^{-15}}{10^{-4}} = 2 	imes 10^{-11} \ M$.

Metal sulfide	Solubility product
$A$. $PbS$	$I$. $4.0 \times 10^{-53}$
$B$. $HgS$	$II$. $8.0 \times 10^{-28}$
$C$. $MnS$	$III$. $1.6 \times 10^{-24}$
$D$. $ZnS$	$IV$. $2.5 \times 10^{-13}$

$K_{sp}$ of a salt $Ni(OH)_2$ is $2 \times 10^{-15}$,then molar solubility of $Ni(OH)_2$ in $0.01 \ M \ NaOH$ is $:-$

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