Gen Chem 2Solubility and Complex Ions

The Solubility Product

Ksp describes a slightly soluble salt in equilibrium with its ions, and gives its molar solubility.

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"Insoluble" salts from precipitation reactions do dissolve a little. A saturated solution is a heterogeneous equilibrium between the solid and its ions, and its constant is the solubility product.

Key idea

For $\mathrm{M}_a\mathrm{X}_b\mathrm{(s)} \rightleftharpoons a\mathrm{M}^{m+} + b\mathrm{X}^{x-}$, $K_{sp} = [\mathrm{M}^{m+}]^a[\mathrm{X}^{x-}]^b$ (the solid is left out). If $s$ is the molar solubility (moles of salt that dissolve per liter), the ions are $as$ and $bs$, so $K_{sp} = (as)^a(bs)^b$. Solve for $s$ with an nth root.

Why it works

Every formula unit that dissolves puts $a$ cations and $b$ anions into solution. So if $s$ mol/L dissolve, the cation concentration is $as$ and the anion's is $bs$. For $\mathrm{PbI_2}$, one $\mathrm{Pb^{2+}}$ and two $\mathrm{I^-}$ per unit: $K_{sp} = (s)(2s)^2 = 4s^3$. The 2 is applied twice: once to make $[\mathrm{I^-}] = 2s$ and once as the exponent.
Mixing silver nitrate and sodium chloride: the $\mathrm{Ag^+}$ and $\mathrm{Cl^-}$ ions come together as solid AgCl, leaving only a trace of the ions in solution.

Worked examples

Worked example: A 1:1 salt

$K_{sp}$ of AgCl is $1.8 \times 10^{-10}$. Find its molar solubility.
  1. $\mathrm{AgCl(s) \rightleftharpoons Ag^+ + Cl^-}$: $K_{sp} = s \cdot s = s^2$.
  2. $s = \sqrt{1.8 \times 10^{-10}} = 1.3 \times 10^{-5}\ \text{M}$.

Answer: $1.3 \times 10^{-5}\ \text{M}$.

Worked example: A 1:2 salt

$K_{sp}$ of $\mathrm{PbI_2}$ is $9.8 \times 10^{-9}$. Find its molar solubility and $[\mathrm{I^-}]$.
  1. $[\mathrm{Pb^{2+}}] = s$, $[\mathrm{I^-}] = 2s$. $K_{sp} = s(2s)^2 = 4s^3$.
  2. $s^3 = \dfrac{9.8 \times 10^{-9}}{4} = 2.45 \times 10^{-9}$, so $s = 1.3 \times 10^{-3}\ \text{M}$.
  3. $[\mathrm{I^-}] = 2s = 2.7 \times 10^{-3}\ \text{M}$.

Answer: $s = 1.3 \times 10^{-3}\ \text{M}$; $[\mathrm{I^-}] = 2.7 \times 10^{-3}\ \text{M}$.

Worked example: From solubility to Ksp

The molar solubility of $\mathrm{CaF_2}$ is $2.06 \times 10^{-4}\ \text{M}$. Find $K_{sp}$.
  1. $[\mathrm{Ca^{2+}}] = 2.06 \times 10^{-4}$, $[\mathrm{F^-}] = 4.12 \times 10^{-4}\ \text{M}$.
  2. $$K_{sp} = (2.06 \times 10^{-4})(4.12 \times 10^{-4})^2 = 3.5 \times 10^{-11}$$

Answer: $K_{sp} = 3.5 \times 10^{-11}$.

Common mistake

Forgetting to raise an ion's concentration to its coefficient, writing $K_{sp} = s^2$ for $\mathrm{PbI_2}$, or forgetting the 2 inside: $s \cdot s^2$. Both the count and the exponent come from the formula. Check by substituting back: $s(2s)^2$ must give the $K_{sp}$ you started with.

Solubilities calculated from $K_{sp}$ are good estimates, not exact values. In real solutions some ions pair up, some react with water (carbonate and phosphate are bases), and crowded ions behave as if they were less concentrated. These effects are small for salts like AgCl and larger for salts with highly charged or basic ions.

Molar solubility times molar mass gives grams per liter. For AgCl: $(1.3 \times 10^{-5})(143.3) = 1.9 \times 10^{-3}\ \text{g/L}$, about 2 mg in a liter.

Practice

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Sources and further reading

  • OpenStax Chemistry 2e, Section 15.1 Precipitation and Dissolution
  • OpenStax Chemistry 2e, Appendix J: Solubility Products
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