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The solubility of Hg2I2\mathrm{Hg}_2 \mathrm{I}_2 in H2O\mathrm{H}_2 \mathrm{O} is 3.04×107 g/L3.04 \times 10^{-7} \mathrm{~g} / \mathrm{L}. The reaction Hg2I2Hg22++2 I\mathrm{Hg}_2 \mathrm{I}_2 \rightleftharpoons \mathrm{Hg}_2^{2+}+2 \mathrm{~I}^{-}represents the equilibrium. Calculate the Ksp K_{\text {sp }}.

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In this task, we need to calculate KspK_{sp}.

The solubility product constant is the product of the equilibrium concentrations of the precipitated ions raised to the potency of its stoichiometric factor in the equilibrium equation. For any compound of general formula:

AaBb(s)aA+(aq)+bB(aq)\text{A$_a$B$_b$(s)} \rightleftharpoons \text{aA$^+$(aq)} + \text{bB$^-$(aq)}

the constant of the solubility product is:

Ksp=[A+]a[B]bK_{sp}=[\text{A$^+$}]^a \cdot [\text{B$^-$}]^b

First we need to convert solubility to molar solubility using the solubility and molar mass:

c=γMc=\dfrac{\gamma}{M}

Then, using the mole ratio, we calculate concentration of ions. When we know the concentration of ion, we can calculate KspK_{sp}.

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