Solubility Product
Molar solubility from Ksp, the common ion effect, and whether a precipitate forms.
AgCl(s) ⇌ Ag⁺ + Cl⁻ · Ksp = [Ag⁺][Cl⁻] · M = 143.32 g/mol
in pure water
× 143.32 g/mol
[Ag⁺] = 1.3304e-5 M, [Cl⁻] = 1.3304e-5 M
| Setting up | Ksp = (1s)(1s) = 1s^2 | with s the molar solubility |
| Solving | s = (1.7700e-10 ÷ 1)^(1/2) = 1.3304e-5 M | in pure water |
| [Ag⁺] | 1.3304e-5 M | 1s |
| [Cl⁻] | 1.3304e-5 M | 1s |
| Check | 1.7700e-10 | the ion product back — must equal Ksp |
Ksp is the equilibrium constant for a salt dissolving. For AₓBᵧ it works out to Ksp = xˣyʸ·s^(x+y), where s is the molar solubility — so AgCl with Ksp 1.77 × 10⁻¹⁰ dissolves to 1.33 × 10⁻⁵ M.
Why the exponents make comparison misleading
A saturated solution is an equilibrium: solid dissolving at the same rate as ions crystallising out. The equilibrium constant omits the solid, leaving only the ion concentrations, each raised to its coefficient. For AgCl that is simply [Ag⁺][Cl⁻]; for CaF₂ it is [Ca²⁺][F⁻]², and the square is what makes the arithmetic less obvious than it looks.
The consequence trips people up constantly: a smaller Ksp does not always mean a less soluble salt. AgCl has Ksp 1.77 × 10⁻¹⁰ and CaF₂ has 3.45 × 10⁻¹¹ — smaller — yet CaF₂ is about fifteen times more soluble, because its Ksp is a product of three concentration factors rather than two. Comparing Ksp values directly is only valid between salts of the same stoichiometry.
The common ion effect
Add one of the ions from another source and Le Chatelier pushes the equilibrium back toward the solid, so less dissolves. AgCl in 0.10 M sodium chloride is about seven and a half thousand times less soluble than in pure water. The usual calculation assumes the salt’s own contribution is negligible beside the added ion — which is normally fine, and sometimes badly wrong, as the tool shows when it is.
Q has the same form from any concentrations; precipitate forms when Q > Ksp
- 1 Write the dissolution equation. CaF₂(s) ⇌ Ca²⁺ + 2F⁻. The coefficients are what the exponents come from.
- 2 Express each ion in terms of s. If s moles dissolve per litre, [Ca²⁺] = s and [F⁻] = 2s — the 2 comes from the equation, not from the charge.
- 3 Substitute into Ksp. Ksp = (s)(2s)² = 4s³.
- 4 Solve for s. s = (3.45 × 10⁻¹¹ ÷ 4)^(1/3) = 2.05 × 10⁻⁴ M.
- 5 Convert to grams if asked. 2.05 × 10⁻⁴ mol/L × 78.07 g/mol = 0.016 g/L.
Why Ksp values cannot be compared across stoichiometries
Molar solubility in pure water at 25 °C. The smaller Ksp is not always the less soluble salt.
| Salt | Type | Ksp | Ksp in terms of s | Molar solubility |
|---|---|---|---|---|
| AgCl | AB | 1.77 × 10⁻¹⁰ | s² | 1.33 × 10⁻⁵ M |
| CaF₂ | AB₂ | 3.45 × 10⁻¹¹ | 4s³ | 2.05 × 10⁻⁴ M |
| Ag₂CrO₄ | A₂B | 1.12 × 10⁻¹² | 4s³ | 6.54 × 10⁻⁵ M |
| Mg(OH)₂ | AB₂ | 5.61 × 10⁻¹² | 4s³ | 1.12 × 10⁻⁴ M |
| Ca₃(PO₄)₂ | A₃B₂ | 2.07 × 10⁻³³ | 108s⁵ | 1.14 × 10⁻⁷ M |
Where the standard shortcut fails
For a common ion problem the textbook move is to assume the salt contributes so little that the added ion concentration is effectively unchanged. For AgCl in 0.10 M chloride that assumption is correct to within a few millionths of a percent, and the algebra collapses to s = Ksp ÷ 0.10.
It stops working when the added ion is comparable to what the salt itself would supply. Magnesium hydroxide at pH 10 is the standard counterexample: the hydroxide from the water is 10⁻⁴ M while the salt alone would give about 2.2 × 10⁻⁴ M of it, so ignoring the salt’s own contribution overstates the solubility by roughly sevenfold. This page solves the full cubic numerically instead, and says by how much the shortcut would have been out.
Two further caveats worth carrying. Ksp values are quoted at 25 °C and vary with temperature — usually rising, which is why hot water dissolves more of most salts. And they assume ideal behaviour; in a concentrated solution, activity coefficients drift from one and real solubility exceeds the calculation, which is why Ksp is best treated as a good estimate rather than an exact prediction outside dilute conditions.