Solubility Equilibria (Ksp)

'Insoluble' is a white lie — even a chalky precipitate dissolves a tiny bit, and Ksp tells you exactly how much.

High schoolIntro Gen ChemUni Year 1
⏱️ About 18 min

Drop 'insoluble' silver chloride into water and it looks like nothing dissolves. But a vanishingly small amount does — and that trace sets up an equilibrium with its own constant. That constant, Ksp, is why some precipitates form and others don't.

💡
The big idea: A slightly soluble salt sits in equilibrium with its dissolved ions. The solubility product Ksp is the equilibrium constant for that dissolving — the product of the ion concentrations, each raised to its coefficient. It lets you calculate how much dissolves and predict whether a precipitate forms.
🎯 By the end, you'll be able to
  • Write the Ksp expression for a slightly soluble salt from its dissolution equation
  • Calculate molar solubility from Ksp (and Ksp from solubility)
  • Relate Ksp to solubility for 1:1, 1:2 and 2:1 salts
  • Use the reaction quotient Q vs Ksp to predict whether a precipitate forms
📎 Helpful to know first

Even 'insoluble' salts dissolve a little

When a slightly soluble ionic solid sits in water, a dynamic equilibrium forms: ions leave the solid as fast as dissolved ions rejoin it. The solution is saturated — it holds the maximum amount of dissolved salt, which for these compounds is tiny but not zero.

Because it's an equilibrium, it has an equilibrium constant. For dissolving we call it the solubility product, Ksp.

\[ \ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)} \qquad K_{sp} = [\ce{Ag+}][\ce{Cl-}] \]
The Ksp expression includes only the dissolved ions — the pure solid doesn't appear (its 'concentration' is constant).
🔑 Ions raised to their coefficients
Just like any equilibrium constant, each ion concentration is raised to its coefficient in the balanced dissolution. For CaF₂(s) ⇌ Ca²⁺ + 2 F⁻, the fluoride gets squared: Ksp = [Ca²⁺][F⁻]². Forgetting that exponent is the most common Ksp mistake.

From Ksp to molar solubility

Molar solubility (s) is how many moles of the salt dissolve per litre to saturate the solution. The link to Ksp comes from the stoichiometry: define s, express each ion in terms of s, and substitute. The exponents make the shape of the relationship depend on the salt's formula.

📝 Worked example: Write Ksp in terms of molar solubility s for CaF₂ (which dissolves to Ca²⁺ + 2 F⁻).
  1. If s mol/L of CaF₂ dissolves, it makes [Ca²⁺] = s and [F⁻] = 2s (two fluorides per formula unit).
  2. Substitute into Ksp = [Ca²⁺][F⁻]² = (s)(2s)².
  3. (2s)² = 4s², so Ksp = s × 4s² = 4s³.
✓ Ksp = 4s³ for a 1:2 salt like CaF₂.
✏️ Practice: Silver chloride has Ksp = 1.8×10⁻¹⁰. What is its molar solubility s (in mol/L)? (AgCl ⇌ Ag⁺ + Cl⁻, so Ksp = s².)
mol/L
Solution
  1. For a 1:1 salt, [Ag⁺] = [Cl⁻] = s, so Ksp = s² = 1.8×10⁻¹⁰.
  2. s = √(1.8×10⁻¹⁰).
  3. s = 1.34×10⁻⁵ mol/L — genuinely tiny, but not zero.
✏️ Practice: Magnesium hydroxide, Mg(OH)₂, has Ksp = 5.6×10⁻¹². It dissolves to Mg²⁺ + 2 OH⁻, so Ksp = 4s³. Find its molar solubility s (mol/L).
mol/L
Solution
  1. Set Ksp = 4s³ = 5.6×10⁻¹², so s³ = 5.6×10⁻¹² ÷ 4 = 1.4×10⁻¹².
  2. s = (1.4×10⁻¹²)^(1/3).
  3. s = 1.12×10⁻⁴ mol/L. (Note the cube root, because of the 4s³ form.)

Predicting a precipitate: Q vs Ksp

Mix two solutions and you might get a precipitate — or you might not. To decide, compute the reaction quotient Q (the same ion product, but with the actual mixed concentrations) and compare it with Ksp:

  • Q > Ksp: supersaturated — a precipitate forms until Q falls back to Ksp.
  • Q = Ksp: exactly saturated, at equilibrium.
  • Q < Ksp: unsaturated — everything stays dissolved.
✨ Comparing Ksp values needs care
You can only compare Ksp values directly to rank solubility when the salts share the same formula type. Two 1:1 salts? Bigger Ksp means more soluble. But a 1:1 salt versus a 1:2 salt have different Ksp expressions (s² versus 4s³), so you must convert to molar solubility before comparing.

Check your understanding

1. Why doesn't the solid appear in the Ksp expression for AgCl(s) ⇌ Ag⁺ + Cl⁻?
Pure solids (and liquids) have essentially constant activity, so they're absorbed into the constant and left out of the expression. Ksp = [Ag⁺][Cl⁻] only.
2. For CaF₂ ⇌ Ca²⁺ + 2 F⁻, the correct Ksp expression is:
Each ion is raised to its coefficient. There are two fluorides, so [F⁻] is squared: Ksp = [Ca²⁺][F⁻]². This is the most common place students slip.
3. You mix two solutions and find Q > Ksp for a salt. What happens?
Q > Ksp means the solution is supersaturated, so the excess ions precipitate out until the ion product falls back to Ksp (equilibrium).
🔎 Related lessons
Previous: Titrations & Indicators Dynamic Equilibrium & K pH, pOH & Strong Acids/Bases
✅ Key takeaways
  • A slightly soluble salt is in equilibrium with its dissolved ions; Ksp is that constant.
  • Ksp = product of ion concentrations, each raised to its coefficient; solids are omitted.
  • Molar solubility s links to Ksp via stoichiometry: s² for a 1:1 salt, 4s³ for a 1:2 salt.
  • Compare Ksp directly only for salts of the same formula type; otherwise convert to solubility.
  • Predict precipitation with Q: Q > Ksp precipitates, Q < Ksp stays dissolved.
➡️ You've now followed proton transfer from its definitions all the way to the faint equilibria of barely-soluble salts. From here, the equilibrium toolkit carries straight into electrochemistry, where electrons — not protons — do the moving.
Want to test yourself on this? Try the Chemistry practice test →