Short on time? This page is the recap. For explanations and worked examples, use the full study guide.
Recap
- A common ion is an ion that a dissolving salt shares with something already in the solution (for AgCl: Cl⁻ from NaCl, or Ag⁺ from AgNO₃).
- Le Châtelier: adding a product ion pushes the dissolving equilibrium back towards the solid.
- Q and K: the added ion makes Q > K_sp, so ions precipitate until Q = K_sp again.
- Result: the salt is less soluble than in pure water. This is the common-ion effect.
- K_sp does not change at constant temperature. The molar solubility falls, the other ion’s concentration falls, and the common ion’s concentration is higher.
- An ion that is not in the K_sp expression (for AgCl, Na⁺ or NO₃⁻) does not cause this effect.
- Removing one of the salt’s ions has the opposite effect: more solid dissolves.
Key relationships
| Situation | Set-up | Simplified (s ≪ c) |
|---|---|---|
| MX in c M of X⁻ | K_sp = s(c + s) | s ≈ K_sp ÷ c |
| MX₂ in c M of X⁻ | K_sp = s(c + 2s)² | s ≈ K_sp ÷ c² |
| MX₂ in c M of M²⁺ | K_sp = (c + s)(2s)² | s ≈ √(K_sp ÷ 4c) |
| K_sp from common-ion data | K_sp = [measured ion]ᵐ × [common ion]ⁿ (powers from the formula) | use the common-ion concentration, not s, for that ion |
Assumptions
- The added salt that supplies the common ion is soluble and fully dissociated.
- The salt’s own contribution to the common ion is tiny (check: xs or ys under about 5% of c).
- Temperature is constant, so K_sp is constant.
Mistakes to avoid
- Saying K_sp decreases. Solubility decreases; K_sp does not.
- Writing K_sp = s² when a common ion is present. The common ion must appear as c.
- Dropping the power. For MX₂ in X⁻ solution, use c², not c.
- Not checking the approximation after solving.
- Treating every added salt as a common ion. Only ions in the K_sp expression count.
- Using the pure-water formula on data from a common-ion solution. It gives a K_sp far too small.
Quick self-check
- A 1:1 salt has K_sp = 1.0 × 10⁻¹². What is its molar solubility in 0.010 M of a common ion? (1.0 × 10⁻¹⁰ M)
- MX₂ has K_sp = 2.0 × 10⁻¹⁰. What is its molar solubility in 0.020 M X⁻? (K_sp ÷ c² = 5.0 × 10⁻⁷ M)
- Same salt, in 0.020 M M²⁺? (√(K_sp ÷ 0.080) = 5.0 × 10⁻⁵ M)
- Adding solid NaCl to saturated AgCl: what happens to [Ag⁺] and to K_sp? ([Ag⁺] falls; K_sp unchanged)
Next: practice questions.