The solubility product of NiS is 4.9 times 10 | vedclass

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(B) For a sparingly soluble salt like $NiS$,the dissociation equilibrium is given by: $NiS(s) \rightleftharpoons Ni^{2+}(aq) + S^{2-}(aq)$ Let the sol

Vedclass · Accepted Answer

$7.0 	imes 10^{-3}$

Vedclass · Answer

(B) For a sparingly soluble salt like $NiS$,the dissociation equilibrium is given by: $NiS(s) ightleftharpoons Ni^{2+}(aq) + S^{2-}(aq)$ Let the solubility of $NiS$ be $s \ mol \ dm^{-3}$. Then,$[Ni^{2+}] = s$ and $[S^{2-}] = s$. The solubility product constant $(K_{sp})$ is given by: $K_{sp} = [Ni^{2+}][S^{2-}] = s 	imes s = s^2$ Given $K_{sp} = 4.9 	imes 10^{-5}$. $s^2 = 4.9 	imes 10^{-5} = 49 	imes 10^{-6}$ $s = \sqrt{49 	imes 10^{-6}} = 7.0 	imes 10^{-3} \ mol \ dm^{-3}$ Therefore,the correct option is $B$.

The solubility product of $NiS$ is $4.9 \times 10^{-5}$ at $298 \ K$. Calculate its solubility in $mol \ dm^{-3}$ at the same temperature?

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