The solubility product of Hg_2Cl_2 in water a | vedclass

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(C) The dissociation of $Hg_2Cl_2$ is given by: $Hg_2Cl_2 \rightleftharpoons Hg_2^{2+} + 2Cl^-$Let the solubility of $Hg_2Cl_2$ be $x \, mol/L$.Then,$

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$2.0 	imes 10^{-6} \, M$

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(C) The dissociation of $Hg_2Cl_2$ is given by: $Hg_2Cl_2 ightleftharpoons Hg_2^{2+} + 2Cl^-$Let the solubility of $Hg_2Cl_2$ be $x \, mol/L$.Then,$[Hg_2^{2+}] = x$ and $[Cl^-] = 2x$.The solubility product expression is: $K_{sp} = [Hg_2^{2+}][Cl^-]^2$.Substituting the values: $K_{sp} = (x)(2x)^2 = 4x^3$.Given $K_{sp} = 3.2 	imes 10^{-17}$,we have:$4x^3 = 3.2 	imes 10^{-17}$$x^3 = 0.8 	imes 10^{-17} = 8 	imes 10^{-18}$$x = \sqrt[3]{8 	imes 10^{-18}} = 2 	imes 10^{-6} \, M$.

The solubility product of $Hg_2Cl_2$ in water at $25 \, ^\circ C$ is $3.2 \times 10^{-17} \, mol^3 \, dm^{-9}$. The solubility of $Hg_2Cl_2$ at $25 \, ^\circ C$ is:

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