15.1 Precipitation and dissolution  (Page 4/17)

Calculating the solubility of hg ₂ Cl ₂

Calomel, Hg ₂ Cl ₂ , is a compound composed of the diatomic ion of mercury(I), ${\text{Hg}}_2{}^{\text{2+}},$ and chloride ions, Cl ^– . Although most mercury compounds are now known to be poisonous, eighteenth-century physicians used calomel as a medication. Their patients rarely suffered any mercury poisoning from the treatments because calomel is quite insoluble:

Calculate the molar solubility of Hg ₂ Cl ₂ .

Solution

The molar solubility of Hg ₂ Cl ₂ is equal to the concentration of ${\text{Hg}}_2{}^{\text{2+}}$ ions because for each 1 mol of Hg ₂ Cl ₂ that dissolves, 1 mol of ${\text{Hg}}_2{}^{\text{2+}}$ forms:

1. Determine the direction of change. Before any Hg ₂ Cl ₂ dissolves, Q is zero, and the reaction will shift to the right to reach equilibrium.

2. Determine x and equilibrium concentrations. Concentrations and changes are given in the following ICE table:

   

   Note that the change in the concentration of Cl ^– (2 x ) is twice as large as the change in the concentration of ${\text{Hg}}_2{}^{\text{2+}}$ ( x ) because 2 mol of Cl ^– forms for each 1 mol of ${\text{Hg}}_2{}^{\text{2+}}$ that forms. Hg ₂ Cl ₂ is a pure solid, so it does not appear in the calculation.

3. Solve for x and the equilibrium concentrations. We substitute the equilibrium concentrations into the expression for K _sp and calculate the value of x :

   $K_{\text{sp}}=[{\text{Hg}}_{\text{2}}{}^{\text{2+}}]{[{\text{Cl}}^{\text{−}}]}^2$
   $1.1\times {10}^{-18}=(x)(2x{)}^2$
   $4x^3=\text{1.1}\times {10}^{-18}$
   $x=\sqrt[3]{\left(\frac{1.1\times {10}^{-18}}{4}\right)}=\text{6.5}\times {10}^{-7}M$
   $[{\text{Hg}}_2{}^{\text{2+}}]=\text{6.5}\times {10}^{-7}M=\text{6.5}\times {10}^{-7}M$
   $[{\text{Cl}}^{\text{−}}]=2x=\text{2(6.5}\times {10}^{-7})=1.3\times {10}^{-6}M$

   The molar solubility of Hg ₂ Cl ₂ is equal to $[{\text{Hg}}_{\text{2}}{}^{\text{2+}}],$ or 6.5 $\times$ 10 ^–7 M .

4. Check the work. At equilibrium, Q = K _sp :

   $Q=[{\text{Hg}}_{\text{2}}{}^{\text{2+}}]{[{\text{Cl}}^{\text{−}}]}^2=\text{(6.5}\times {10}^{-7}){(1.3\times {10}^{-6})}^2=1.1\times {10}^{-18}$

   The calculations check.

Check your learning

Determine the molar solubility of MgF ₂ from its solubility product: K _sp = 6.4 $\times$ 10 ^–9 .

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