Coaching Centre
The solubility product, Ksp
Writing Ksp expressions, converting between solubility and Ksp, predicting whether a precipitate forms by comparing Q with Ksp, net ionic equations, the common ion effect, and the investigation of solubility rules.
Barium ions are poisonous: a few grams of barium chloride can kill. Yet every day hospital patients drink a thick white suspension containing hundreds of grams of barium sulfate, so their stomach and intestines show up on X-rays. They are safe because barium sulfate is so insoluble that the concentration of Ba²⁺ in its saturated solution is only about 10⁻⁵ mol L⁻¹. That number comes straight from its solubility product.
The solubility product
In a saturated solution of a sparingly soluble salt, the solid is in dynamic equilibrium with its ions. Because the solid is left out of the expression, the equilibrium constant is just the product of the ion concentrations, each raised to the power of its coefficient. It is called the solubility product, Ksp.
| Equilibrium | Ksp expression | Ksp at 25 °C |
|---|---|---|
| AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) | $[\text{Ag}^+][\text{Cl}^-]$ | 1.77 × 10⁻¹⁰ |
| BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq) | $[\text{Ba}^{2+}][\text{SO}_4^{2-}]$ | 1.08 × 10⁻¹⁰ |
| PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq) | $[\text{Pb}^{2+}][\text{I}^-]^2$ | 9.8 × 10⁻⁹ |
| Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq) | $[\text{Ag}^+]^2[\text{CrO}_4^{2-}]$ | 1.12 × 10⁻¹² |
| Ca₃(PO₄)₂(s) ⇌ 3Ca²⁺(aq) + 2PO₄³⁻(aq) | $[\text{Ca}^{2+}]^3[\text{PO}_4^{3-}]^2$ | 2.07 × 10⁻³³ |
Ksp values are given on the HSC data sheet. Like any K, Ksp depends on temperature.
From Ksp to solubility, and back
Let the molar solubility be s mol L⁻¹: the number of moles of solid that dissolve per litre of saturated solution. The ion concentrations follow from the formula.
Calculate the solubility of AgCl in mol L⁻¹ and in g L⁻¹ at 25 °C.
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq). If s mol L⁻¹ dissolves, [Ag⁺] = [Cl⁻] = s.
$K_{sp} = s^2 = 1.77\times10^{-10} \Rightarrow s = 1.33\times10^{-5}\text{ mol L}^{-1}$
In grams: 1.33 × 10⁻⁵ × 143.3 = 1.91 × 10⁻³ g L⁻¹, about 2 mg per litre.
Calculate the molar solubility of PbI₂ at 25 °C.
PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq). [Pb²⁺] = s; [I⁻] = 2s.
$K_{sp} = s(2s)^2 = 4s^3 = 9.8\times10^{-9}$
$s = \sqrt[3]{\dfrac{9.8\times10^{-9}}{4}} = 1.35\times10^{-3}\text{ mol L}^{-1}$
At 25 °C, 4.4 g of PbCl₂ dissolves in 1.0 L of water to form a saturated solution. Calculate Ksp. (M = 278.1 g mol⁻¹)
$s = \dfrac{4.4}{278.1} = 0.0158\text{ mol L}^{-1}$, so [Pb²⁺] = 0.0158 and [Cl⁻] = 0.0316 mol L⁻¹
$K_{sp} = (0.0158)(0.0316)^2 = 4s^3 = 1.6\times10^{-5}$, close to the data sheet value of 1.70 × 10⁻⁵
A useful summary for the common formula types:
| Formula type | Example | Ksp in terms of s |
|---|---|---|
| AB | AgCl, BaSO₄ | $s^2$ |
| AB₂ or A₂B | PbI₂, Ag₂CrO₄ | $4s^3$ |
| AB₃ or A₃B | Fe(OH)₃ | $27s^4$ |
| A₃B₂ | Ca₃(PO₄)₂ | $108s^5$ |
Don't memorise these: derive them each time from the ICE-style reasoning above.
Will a precipitate form?
When two solutions are mixed, calculate the ionic product Q, using the concentrations immediately after mixing (before any precipitate forms), in the same form as Ksp:
- Q < Ksp: the solution is unsaturated. No precipitate.
- Q = Ksp: just saturated.
- Q > Ksp: supersaturated, so a precipitate forms until the ion concentrations fall and Q = Ksp.
Remember that mixing dilutes both solutions: equal volumes halve each concentration.
50.0 mL of 0.0020 mol L⁻¹ Pb(NO₃)₂ is mixed with 50.0 mL of 0.0020 mol L⁻¹ KI. Will PbI₂ precipitate? (Ksp = 9.8 × 10⁻⁹)
Total volume is 100.0 mL, so each concentration halves: [Pb²⁺] = 0.0010 mol L⁻¹; [I⁻] = 0.0010 mol L⁻¹
$Q = [\text{Pb}^{2+}][\text{I}^-]^2 = (0.0010)(0.0010)^2 = 1.0\times10^{-9}$
Q < Ksp, so no precipitate forms. Repeat with 0.020 mol L⁻¹ solutions: Q = (0.010)(0.010)² = 1.0 × 10⁻⁶, which is greater than Ksp, so yellow PbI₂ precipitates.
The reaction that occurs is best shown by a net ionic equation, which leaves out the spectator ions (here K⁺ and NO₃⁻) that stay dissolved and take no part:
$$\text{Pb}^{2+}\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow \text{PbI}_2\text{(s)}$$
- Mix Ba²⁺ and SO₄²⁻ at 10⁻² mol L⁻¹ (log = −2). Then lower both concentrations until no precipitate forms. At about what concentration does this happen?
- Now try Ca²⁺ with SO₄²⁻ at 10⁻² mol L⁻¹. Why is there no precipitate, even though CaSO₄ is "slightly soluble"? What concentration is needed?
- Reproduce the worked example: Pb²⁺ and I⁻ at 0.0020 mol L⁻¹ (log ≈ −2.7), 50 mL each. Then raise both to 0.020 mol L⁻¹.
- Keep the concentrations fixed and change the volumes. Why does the volume ratio matter?
The common ion effect
What happens if AgCl is added to a solution that already contains Cl⁻, such as 0.10 mol L⁻¹ NaCl? By Le Chatelier's principle, the extra Cl⁻ pushes AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) to the left, so less AgCl dissolves. This is the common ion effect.
Calculate the solubility of AgCl in 0.10 mol L⁻¹ NaCl.
[Ag⁺] = s; [Cl⁻] = 0.10 + s ≈ 0.10, because s is tiny.
$K_{sp} = s(0.10) = 1.77\times10^{-10} \Rightarrow s = 1.8\times10^{-9}\text{ mol L}^{-1}$
That is about 7500 times less than in pure water (1.33 × 10⁻⁵ mol L⁻¹).
Chemists use the common ion effect to make a precipitation more complete, for example by adding a slight excess of sulfate when precipitating barium sulfate in gravimetric analysis (Module 8). The antidote for barium poisoning works the same way: a soluble sulfate, such as magnesium sulfate, is given to convert Ba²⁺ in the gut into insoluble BaSO₄.
Test yourself
What is the Ksp expression for Ag₃PO₄?
The solid is left out; the coefficient becomes a power.
The Ksp of BaSO₄ is 1.08 × 10⁻¹⁰. What is its molar solubility?
s = √(1.08 × 10⁻¹⁰).
Which pair of solutions forms a precipitate when mixed?
Cu²⁺ + 2OH⁻ → Cu(OH)₂(s), a pale blue precipitate.
The solubility of CaF₂ in 0.10 mol L⁻¹ NaF, compared with pure water, is
Ksp is unchanged; extra F⁻ shifts the equilibrium left.
Complete the precipitation flow chart.
For a salt AB₂, Ksp = . A precipitate forms when the ionic product Q is than Ksp. Ions that do not take part in a reaction are called ions. Adding NaCl to a saturated AgCl solution makes AgCl soluble; this is the effect.
25.0 mL of 1.0 × 10⁻³ mol L⁻¹ AgNO₃ is added to 75.0 mL of 2.0 × 10⁻³ mol L⁻¹ NaCl. Ksp(AgCl) = 1.77 × 10⁻¹⁰. Determine, with calculations, whether a precipitate forms, and write the net ionic equation for any reaction.
Total volume = 100.0 mL.
[Ag⁺] = 1.0 × 10⁻³ × 25.0/100.0 = 2.5 × 10⁻⁴ mol L⁻¹; [Cl⁻] = 2.0 × 10⁻³ × 75.0/100.0 = 1.5 × 10⁻³ mol L⁻¹.
Q = [Ag⁺][Cl⁻] = (2.5 × 10⁻⁴)(1.5 × 10⁻³) = 3.8 × 10⁻⁷.
Q is greater than Ksp (1.77 × 10⁻¹⁰), so a white precipitate of AgCl forms: Ag⁺(aq) + Cl⁻(aq) → AgCl(s).