Solubility & precipitation equilibria
The idea
Sparingly soluble salts set up their own heterogeneous equilibrium: a solid in contact with its dissolved ions, governed by the solubility product Ksp. For CaF₂(s) ⇌ Ca²⁺(aq) + 2 F⁻(aq), Ksp = [Ca²⁺][F⁻]² — the solid never appears in the expression because its activity is constant. You already think in terms of dissolving and concentration; here the payoffs are computing molar solubility from Ksp, predicting precipitation by comparing the ion product Q to Ksp, and quantifying the common-ion effect.
Translate carefully between solubility s (moles of salt dissolving per liter) and ion concentrations: stoichiometry inserts coefficients and powers, so a 1:2 salt gives Ksp = s(2s)² = 4s³. The common-ion effect is Le Chatelier in action — pre-loading the solution with one product ion drives the equilibrium back toward solid and can suppress solubility by orders of magnitude, which is why selective precipitation works as a separation tool.
The trap to avoid is ranking solubilities by comparing raw Ksp values across salts of different formulas. Because the exponents differ, a 1:1 salt and a 1:2 salt with similar Ksp values can have wildly different molar solubilities — always convert to s before comparing.
Worked example
The Ksp of CaF₂ is 3.9 × 10⁻¹¹ at 25 °C. Calculate its molar solubility in pure water and in a 0.10 M NaF solution, and compare the two.
- Write the dissolution and assign concentrations: CaF₂(s) ⇌ Ca²⁺ + 2 F⁻, so if s mol/L dissolves, [Ca²⁺] = s and [F⁻] = 2s.
- Substitute into the solubility product: Ksp = s(2s)² = 4s³ = 3.9 × 10⁻¹¹, so s³ = 9.75 × 10⁻¹² and s = 2.1 × 10⁻⁴ M in pure water.
- In 0.10 M NaF the fluoride is dominated by the added salt: [F⁻] ≈ 0.10 M, since the contribution 2s will be tiny. Then s = Ksp/[F⁻]² = 3.9 × 10⁻¹¹/(0.10)² = 3.9 × 10⁻⁹ M.
- Check the approximation: 2s = 7.8 × 10⁻⁹ M is utterly negligible next to 0.10 M, so the shortcut was justified.
- Interpret: the common ion crushes solubility by a factor of 2.1 × 10⁻⁴/3.9 × 10⁻⁹ ≈ 54,000 — exactly the Le Chatelier shift toward solid you would predict when a product ion is supplied externally.
Answer. Molar solubility is about 2.1 × 10⁻⁴ M in pure water but only 3.9 × 10⁻⁹ M in 0.10 M NaF — roughly 54,000 times lower because of the common-ion effect.
Check your understanding
- Why does the exponent structure of Ksp make direct comparisons between salts of different stoichiometry meaningless?
- How does comparing Q to Ksp tell you whether mixing two solutions will produce a precipitate, and what does Q = Ksp mean physically?
- Why does the common-ion calculation square the fluoride concentration but not the calcium concentration in this salt?
- How could pH changes alter the solubility of a salt whose anion is a weak base, even though Ksp itself stays fixed?
Build the foundations first
Solubility & precipitation equilibria builds on these concepts. If any feel shaky, start there.