Lesson 1 · 25 min

The solubility product

Even “insoluble” salts dissolve a little. The solubility product Ksp turns that little into a number, and the ion product Q tells you whether mixing two solutions will give a precipitate.

In this lesson

Write the dissolution equilibrium and the solubility-product expression of a sparingly soluble salt
Calculate molar solubility from Ksp for AB, AB₂, A₂B and A₂B₃ salts, and the reverse
Compare the ion product Q with Ksp to predict whether mixed solutions precipitate
Calculate how much precipitate forms and what remains in solution

Before an X-ray of the gut, patients drink a cup of white barium sulfate suspension. Soluble barium salts are highly toxic — a few grams of barium chloride can kill — yet barium sulfate is swallowed safely, because only about 2.4 mg of it dissolves per litre and hardly any $\ce{Ba^2+}$ reaches solution. “Insoluble” really means “very slightly soluble”; exactly how slightly is set by an equilibrium constant.

Predict first

$\ce{AgCl}$ has $K_\mathrm{sp} = 1.8\times10^{-10}$ and $\ce{Ag2CrO4}$ has $K_\mathrm{sp} = 1.2\times10^{-12}$ . Which has the larger molar solubility (mol/L) in water?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

AgCl, because its Ksp is 150 times larger Ag₂CrO₄, even though its Ksp is smaller They are equal Ksp cannot tell

The dissolution equilibrium and Ksp

Put a sparingly soluble salt into water and ions keep leaving the crystal surface and returning to it. When the two rates are equal the solution is saturated and a dissolution equilibrium holds:

\ce{A_xB_y(s) <=> x A^{m+}(aq) + y B^{n-}(aq)}

The solid’s “concentration” does not change and is left out of the equilibrium constant, which gives the solubility product $K_\mathrm{sp}$ : the product of the ion concentrations in a saturated solution, each raised to its coefficient. Like any equilibrium constant it depends only on temperature.

K_\mathrm{sp} = [\mathrm{A}]^x[\mathrm{B}]^y = (xs)^x(ys)^y \quad\Rightarrow\quad s = \left(\frac{K_\mathrm{sp}}{x^x y^y}\right)^{1/(x+y)}

Molar solubility s from Ksp, in pure water and with no side reactions of the ions.

$s$: molar solubility: formula units dissolved per litre of saturated solution (mol/L)
$x, y$: cations and anions per formula unit
$x^x y^y$: 1 for AB, 4 for AB₂ and A₂B, 27 for AB₃, 108 for A₂B₃ and A₃B₂

Worked example

The solubility of calcium fluoride

Fluorite, $\ce{CaF2}$ , has $K_\mathrm{sp} = 3.9\times10^{-11}$ at 25 °C. Find its molar solubility in pure water and the milligrams that dissolve per litre.

Will it precipitate? Compare Q with Ksp

When a solution containing the cation meets one containing the anion, first compute the ion product $Q$ at the moment of mixing (the same form as Ksp, but with whatever concentrations exist at that moment):

$Q < K_\mathrm{sp}$ : unsaturated — no precipitate, and more could dissolve;
$Q = K_\mathrm{sp}$ : exactly saturated;
$Q > K_\mathrm{sp}$ : supersaturated — solid forms until the remaining ions satisfy $Q = K_\mathrm{sp}$ .

Remember that mixing dilutes: combine equal volumes and every ion’s concentration halves.

Compare Q with Ksp · exact equilibrium

Mix two solutions: will it precipitate?

Reagents

\ce{Pb(NO3)2}

Cation concentration · 0.01 M Volume · 10 mL

\ce{KI}

Anion concentration · 0.01 M Volume · 10 mL

\ce{PbI2}

· bright yellow

Ksp line: saturated solutions

Q > Ksp: supersaturated — solid forms until Q = Ksp

Ion product Q at the moment of mixing: 1.25 × 10⁻⁷
Ksp: 7.94 × 10⁻⁹
After mixing (before precipitation): [Pb²⁺] 0.005 · [I⁻] 0.005 M
Left in solution at equilibrium: [Pb²⁺] 0.00328 · [I⁻] 0.00156 M
Mass of precipitate: 15.87 mg
Share of the limiting ion precipitated: 68.87 %

10 mL of 0.010 M lead nitrate plus 10 mL of 0.010 M potassium iodide: Q is about 16 times Ksp and about 16 mg of bright-yellow PbI₂ forms. Dilute both to 0.001 M and try again — Q drops below Ksp and nothing happens. On the right, the state slides along the stoichiometric path until it lands on the Ksp line.

Concept check

Equal volumes of 0.0020 M $\ce{BaCl2}$ and 0.0020 M $\ce{Na2SO4}$ are mixed ( $K_\mathrm{sp}(\ce{BaSO4}) = 1.1\times10^{-10}$ ). What happens?

No precipitate: too dilute A precipitate: Q = 1.0 × 10⁻⁶ > Ksp A precipitate: Q = 4.0 × 10⁻⁶ > Ksp Exactly saturated

Try the calculation

Find the molar solubility (mol/L) of $\ce{Ag2CrO4}$ ( $K_\mathrm{sp} = 1.2\times10^{-12}$ ) in pure water.

Worked example

How much precipitates on mixing

50.0 mL of 1.00 × 10⁻³ M $\ce{AgNO3}$ is mixed with 50.0 mL of 1.00 × 10⁻³ M $\ce{NaCl}$ ( $K_\mathrm{sp}(\ce{AgCl}) = 1.82\times10^{-10}$ , as on the mixing bench). How much AgCl precipitates?

Key takeaways

Ksp is the product of the ion concentrations in a saturated solution, each raised to its coefficient; the solid is left out.
In pure water s = (Ksp / xˣyʸ)^(1/(x+y)); only salts of the same type can be ranked by Ksp.
When Q > Ksp a precipitate forms until Q = Ksp; allow for dilution on mixing.
To find how much precipitates, remove ions in stoichiometric ratio until the remainder satisfies Ksp.

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Data and references

A. E. Martell & R. M. Smith, Critical Stability Constants, Vol. 4 (Plenum, 1976), as tabulated in D. Harvey, Analytical Chemistry 2.1, Appendix 10.
D. C. Harris, Quantitative Chemical Analysis, 9th ed. (2016), Ch. 6 (solubility product) and Appendix F.
W. M. Haynes (ed.), CRC Handbook of Chemistry and Physics, 97th ed. (2016): solubility product constants.