If the solubility product of HgSO_4 is 6.4 ti | vedclass

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Question

(A) The dissociation of $HgSO_4$ is given by: $HgSO_4(s) \rightleftharpoons Hg^{2+}(aq) + SO_4^{2-}(aq)$.Let the solubility be $s \ mol \ L^{-1}$.Then

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$8 	imes 10^{-3} \ mol \ L^{-1}$

Vedclass · Answer

(A) The dissociation of $HgSO_4$ is given by: $HgSO_4(s) ightleftharpoons Hg^{2+}(aq) + SO_4^{2-}(aq)$.Let the solubility be $s \ mol \ L^{-1}$.Then,$[Hg^{2+}] = s$ and $[SO_4^{2-}] = s$.The solubility product expression is $K_{sp} = [Hg^{2+}][SO_4^{2-}] = s 	imes s = s^2$.Given $K_{sp} = 6.4 	imes 10^{-5}$.Therefore,$s^2 = 6.4 	imes 10^{-5} = 64 	imes 10^{-6}$.Taking the square root,$s = \sqrt{64 	imes 10^{-6}} = 8 	imes 10^{-3} \ mol \ L^{-1}$.

If the solubility product of $HgSO_4$ is $6.4 \times 10^{-5}$,then its solubility is:

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