Solubility Product of an AB₂ Salt
Worked example: CaF2 s = 2.15e-4 M → Ksp = 4s^3 = 3.97535e-11 — press Try an example to run it live, then adjust anything.
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Grade 12Grade 12 Chemistry
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Solubility Product of an AB₂ Salt explained
When a salt dissolves into three ions instead of two the arithmetic changes shape. CaF₂(s) ⇌ Ca²⁺ + 2F⁻ releases one calcium and two fluorides per formula unit, so if s moles dissolve per litre then [Ca²⁺] = s but [F⁻] = 2s, and Ksp = s(2s)² = 4s³. Calcium fluoride's molar solubility of 2.15 × 10⁻⁴ M therefore gives Ksp = 4(2.15 × 10⁻⁴)³ = 4.0 × 10⁻¹¹, matching the handbook value.
Reversed, Mg(OH)₂ with Ksp = 5.6 × 10⁻¹² dissolves to s = ∛(5.6 × 10⁻¹² / 4) = 1.1 × 10⁻⁴ M — enough hydroxide to buffer a stomach at pH about 10 in the flask, which is the entire pharmacology of milk of magnesia. The universal trap is dropping the 4, or forgetting that the doubled ion gets both a coefficient and an exponent. Also note that Ksp alone does not rank solubility across different stoichiometries: AgCl (Ksp 1.7 × 10⁻¹⁰) is actually less soluble than Mg(OH)₂ despite the larger constant, because the exponents differ.
Solubility Product of an AB₂ Salt formula
- = Solubility product ((mol/L)³)
- = Molar solubility (M)
Missing one of these? Work it out first, then come back
- Solubility product — Solubility Product of a 1:1 Salt
- Molar solubility — Solubility Product of a 1:1 Salt, Molarity (C = n/V)