The solubility of Sb_2S_3 in water is 1.0 tim | vedclass

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(A) The dissociation of $Sb_2S_3$ is given by: $Sb_2S_3(s) ightleftharpoons 2Sb^{3+}(aq) + 3S^{2-}(aq)$.Let the solubility be $s = 1.0 	imes 10^{-5

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$108 	imes 10^{-25}$

Vedclass · Answer

(A) The dissociation of $Sb_2S_3$ is given by: $Sb_2S_3(s) ightleftharpoons 2Sb^{3+}(aq) + 3S^{2-}(aq)$.Let the solubility be $s = 1.0 	imes 10^{-5} \ mol/L$.The concentrations of the ions are $[Sb^{3+}] = 2s$ and $[S^{2-}] = 3s$.The solubility product expression is $K_{sp} = [Sb^{3+}]^2 [S^{2-}]^3$.Substituting the values: $K_{sp} = (2s)^2 (3s)^3 = 4s^2 \cdot 27s^3 = 108s^5$.Calculating $K_{sp}$: $K_{sp} = 108 	imes (1.0 	imes 10^{-5})^5 = 108 	imes 10^{-25}$.

The solubility of $Sb_2S_3$ in water is $1.0 \times 10^{-5} \ mol/L$ at $298 \ K$. What will be its solubility product?

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