Lessons · Equilibrium & acid–base

The common-ion effect: calcium fluoride in a fluoride solution

ICE ledger machine-checked — every dissolved row is initial + ν·s, one extent sK_sp data-sourced (openstax-chemistry-2e)3 modeling assumptions (disclosed)

In the last lesson, $\mathrm{CaF_2}$ dissolved in pure water. Now dissolve it in water that is already **0.10 M in $\mathrm{F^-}$ ** — a solution of sodium fluoride, $\mathrm{NaF}$ , which dissociates completely. The $\mathrm{Na^+}$ is a spectator, but the $\mathrm{F^-}$ is shared with the salt: it is a common ion. The very same equilibrium, $\mathrm{CaF_2(s)} \rightleftharpoons \mathrm{Ca^{2+}(aq)} + 2\,\mathrm{F^-(aq)}$ , now starts with $\mathrm{F^-}$ already on the product side. By Le Chatelier that extra $\mathrm{F^-}$ pushes the dissolution backward — so how much less dissolves? It is the ICE table again, with one column no longer starting at zero.

\mathrm{CaF_{2}}\,\text{(s)} \rightleftharpoons \mathrm{Ca}^{2+}\,\text{(aq)} + 2\,\mathrm{F}^{-}\,\text{(aq)}

K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^{2}

=4×10⁻¹¹at 25 °C

Species	Initial (M)	Change (M)	Equilibrium (M)
$\mathrm{CaF_{2}}$ (s)pure solid · excluded from Ksp	—	—	—
$\mathrm{Ca}^{2+}$ (aq)	0	+s	0.000000004
$\mathrm{F}^{-}$ (aq)	0.10	+2s	0.1

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

Mass actions = 0.000000004 M — the molar solubility, where Q = K_sp.

$K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^{2}$ =(0.000000004)(0.1)²=4×10⁻¹¹= K_sp ✓

Put the committed equilibrium concentrations back in and the quotient reproduces K_sp to within 1.9×10⁻¹⁴ — the solver found the extent (a cubic — solved numerically, not by a formula), and an independent check re-solves it and agrees.

molar solubility0.000000004 Mof CaF₂ dissolves

solubility0.000000312 g/L= s × 78.074 g/mol

saturated ions[Ca²⁺] = 0.000000004 · [F⁻] = 0.1M at equilibrium

vs. pure water53900× less0.000215 M → 0.000000004 M

Common-ion effectIn pure water CaF₂ dissolves to 0.000215 M. Start instead with 0.10 M F⁻ already in solution and that shared ion hardly moves as the solid dissolves, so K_sp = [Ca²⁺][F⁻]²is satisfied at a far smaller [Ca²⁺]. The same cubic now givess = 0.000000004 M — about 53900× less. Le Chatelier: the pre-loaded F⁻ pushes $\mathrm{CaF_{2}}\,\text{(s)} \rightleftharpoons \mathrm{Ca}^{2+}\,\text{(aq)} + 2\,\mathrm{F}^{-}\,\text{(aq)}$ to the left.

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 = K_sp [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: “Solubility is a fixed property of the salt, so calcium fluoride dissolves just as much here as it does in pure water.”

Solubility is not a fixed property of the salt — it depends on what is already dissolved. The shared F⁻ is a product of the dissolution, so (Le Chatelier) it drives $\mathrm{CaF_{2}}\,\text{(s)} \rightleftharpoons \mathrm{Ca}^{2+}\,\text{(aq)} + 2\,\mathrm{F}^{-}\,\text{(aq)}$ back toward the solid. The ledger shows it: with 0.10 M F⁻ present, only s = 0.000000004 M of CaF₂ dissolves — about 53900× less than the 0.000215 M it reaches in pure water. K_sp is unchanged; the solubility is not. That is the common-ion effect — the same move that holds a buffer's pH near pK_a.

Modeling assumptions — author-asserted, disclosed not discharged

model The added sodium fluoride is fully dissociated, so $[\mathrm{F^-}]_0 = 0.10$ M before any $\mathrm{CaF_2}$ dissolves, and the $\mathrm{Na^+}$ is a spectator absent from the equilibrium.
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 $\mathrm{F^-}$ ).
model Activities are approximated by molar concentrations — the ideal-dilute-solution model, which is why $K_{sp}$ 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.

Practice this

The lesson goes deep on one scenario; the gym builds fluency by repetition. Drill these:

Weak-acid pH: solve the ICE table for [H⁺] →