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Lessons · Equilibrium & acid–base

How much dissolves? The solubility of calcium fluoride

ICE ledger machine-checked — every dissolved row is initial + ν·s, one extent sKsp data-sourced (openstax-chemistry-2e)2 modeling assumptions (disclosed)

Calcium fluoride — the mineral fluorite — is called “insoluble,” but no salt is perfectly insoluble: a tiny amount dissolves until the solution is saturated, CaF2(s)⇌Ca2+(aq)+2 F−(aq)\mathrm{CaF_2(s)} \rightleftharpoons \mathrm{Ca^{2+}(aq)} + 2\,\mathrm{F^-(aq)}. How much? It is the ICE table again — but the dissolving species is a pure solid, whose activity is 1, so it never appears in the equilibrium expression: Ksp=[Ca2+][F−]2K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{F^-}]^2. The extent of this reaction is the molar solubility ss — solve KspK_{sp} for it.

CaF2 (s)⇌Ca2+ (aq)+2 F− (aq)\mathrm{CaF_{2}}\,\text{(s)} \rightleftharpoons \mathrm{Ca}^{2+}\,\text{(aq)} + 2\,\mathrm{F}^{-}\,\text{(aq)}
Ksp=[Ca2+][F−]2K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^{2}=4×10⁻¹¹at 25 °C
SpeciesInitial (M)Change (M)Equilibrium (M)
CaF2\mathrm{CaF_{2}} (s)pure solid · excluded from Ksp———
Ca2+\mathrm{Ca}^{2+} (aq)0+s0.000215
F−\mathrm{F}^{-} (aq)0+2s0.000431

machine-checkedThe Change row is ν·s — the very same extent ledger ci = ci,0 + νi·sas any reaction. The pure solid has activity 1, so it never enters the equilibrium expression.

Mass actions = 0.000215 M — the molar solubility, where Q = Ksp.

Ksp=[Ca2+][F−]2K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^{2}=(0.000215)(0.000431)²=4×10⁻¹¹= Ksp ✓

Put the committed equilibrium concentrations back in and the quotient reproduces Ksp to within 2.6×10⁻¹² — the solver found the extent (a cubic — solved numerically, not by a formula), and an independent check re-solves it and agrees.

molar solubility0.000215 Mof CaF₂ dissolves
solubility0.0168 g/L= s × 78.074 g/mol
saturated ions[Ca²⁺] = 0.000215 · [F⁻] = 0.000431M at equilibrium
VerificationProven at build time — not asserted.
  • ✓ Every equilibrium concentration = initial + ν·s [ICE identity]
  • ✓ The extent s re-solved independently — numerically, to high precision — reproduces Q = Ksp [mass-action root]
  • ✓ The pure solid is excluded from Q; s > 0 — no concentration goes negative [extent physical]
  • ✓ solubility (g/L) = s × molar mass [solubility consistent]
✗ Common misconception: “Ksp is small (4.0e-11) and Ksp = s^2, so the molar solubility is s = sqrt(Ksp).”

Careful with the coefficient. Each CaF₂ releases two F⁻ions, so [F⁻] = 2s, and Ksp = [Ca²⁺][F⁻]2= s·(2s)2 = 4s3 — a cubic, not s2. That gives s = (Ksp/4)1/3 = 0.000215 M, about 34× larger than √Ksp ≈ 6.3e-6. The coefficient enters twice — as a concentration factor and as an exponent.

Modeling assumptions — author-asserted, disclosed not discharged
  • model The solution is at saturation equilibrium with excess solid present, and this dissolution is the only reaction (no complex ions, no pH effect on F−\mathrm{F^-}).
  • model Activities are approximated by molar concentrations — the ideal-dilute-solution model, which is why KspK_{sp} is written with concentrations.

Concepts in this lesson

Linked into the Chemical Atlas where an entry exists; the rest fill in as the Atlas grows.

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