Solubility and Ksp: How Much Dissolves, and When It Precipitates

Writing the solubility product expression

When a slightly soluble ionic solid is in contact with its saturated solution, dissolving and precipitating happen at equal rates. For calcium fluoride, CaF₂(s) ⇌ Ca²⁺(aq) + 2 F⁻(aq), and the equilibrium constant is Ksp = [Ca²⁺][F⁻]². The solid is left out because it is a pure phase.

Each ion concentration is raised to the power of its coefficient in the dissolution equation. For silver chromate, Ag₂CrO₄(s) ⇌ 2 Ag⁺(aq) + CrO₄²⁻(aq), Ksp = [Ag⁺]²[CrO₄²⁻]. For lead(II) iodide, Ksp = [Pb²⁺][I⁻]².

Molar solubility, s, is the number of moles of the solid that dissolve per liter of saturated solution. It is related to Ksp by the stoichiometry of the dissolution, so Ksp and solubility are not the same number and cannot be compared directly across salts of different formula types.

From Ksp to molar solubility

Worked example 1. Silver chloride has Ksp = 1.6 × 10⁻¹⁰ in Appendix J. If s mol/L dissolves, [Ag⁺] = [Cl⁻] = s, so Ksp = s². Then s = √(1.6 × 10⁻¹⁰) = 1.3 × 10⁻⁵ M. With M(AgCl) = 143.32 g/mol, that is 1.8 × 10⁻³ g per liter, a couple of milligrams.

Worked example 2. Calcium fluoride has Ksp = 4.0 × 10⁻¹¹. If s mol/L dissolves, [Ca²⁺] = s and [F⁻] = 2s, so Ksp = s(2s)² = 4s³. Then s = (4.0 × 10⁻¹¹ / 4) raised to the 1/3 power = 2.2 × 10⁻⁴ M.

Compare the two. AgCl has the larger Ksp, yet CaF₂ is about 17 times more soluble in moles per liter. The exponent changes how Ksp maps onto solubility, so you may rank solubility by Ksp only for salts with the same ion ratio.

• 1:1 salts such as AgCl or CaCO₃: Ksp = s².
• 1:2 or 2:1 salts such as CaF₂, PbI₂ or Ag₂CrO₄: Ksp = 4s³.
• 1:3 salts such as Al(OH)₃: Ksp = 27s⁴.

From a measured solubility back to Ksp

The calculation also runs the other way. Suppose a saturated solution of lead(II) iodide at 25 °C is found to contain 0.70 g of dissolved PbI₂ per liter. With M(PbI₂) = 461.0 g/mol, s = 0.70 / 461.0 = 1.52 × 10⁻³ M.

Each PbI₂ that dissolves gives one Pb²⁺ and two I⁻, so [Pb²⁺] = 1.52 × 10⁻³ M and [I⁻] = 3.04 × 10⁻³ M. Then Ksp = (1.52 × 10⁻³)(3.04 × 10⁻³)² = 1.4 × 10⁻⁸, which agrees with the Appendix J value. Two slips are common here. Forgetting the factor of 2 on [I⁻] makes Ksp four times too small. Forgetting to square [I⁻] gives the wrong power of s altogether.

Adding a common ion suppresses solubility

If the solution already contains one of the salt's ions, Le Chatelier's principle pushes the dissolution equilibrium back toward the solid. This is the common-ion effect.

Worked example. Find the solubility of AgCl in 0.10 M NaCl. Now [Cl⁻] = 0.10 + s, and s is tiny next to 0.10, so [Cl⁻] ≈ 0.10 M. Then s = Ksp / [Cl⁻] = 1.6 × 10⁻¹⁰ / 0.10 = 1.6 × 10⁻⁹ M. That is roughly 8,000 times lower than in pure water, and s is clearly negligible compared with 0.10, so the approximation holds.

For CaF₂ in 0.010 M NaF, [F⁻] ≈ 0.010 M and s = 4.0 × 10⁻¹¹ / (0.010)² = 4.0 × 10⁻⁷ M, about 540 times lower than in pure water. Remember to square the fluoride concentration here, just as in the Ksp expression.

Will it precipitate? Comparing Q with Ksp

To predict whether mixing two solutions produces a solid, compute the ion product Q from the concentrations after mixing and compare it with Ksp. If Q is greater than Ksp, the solution is supersaturated and a precipitate forms. If Q is less than or equal to Ksp, no solid forms.

Worked example. Mix 50.0 mL of 0.0010 M Pb(NO₃)₂ with 50.0 mL of 0.0020 M KI. Mixing doubles the volume, so each concentration halves: [Pb²⁺] = 5.0 × 10⁻⁴ M and [I⁻] = 1.0 × 10⁻³ M. Q = (5.0 × 10⁻⁴)(1.0 × 10⁻³)² = 5.0 × 10⁻¹⁰. That is below Ksp = 1.4 × 10⁻⁸, so no PbI₂ precipitates.

Repeat with 0.020 M KI. Now [I⁻] = 1.0 × 10⁻² M after mixing and Q = (5.0 × 10⁻⁴)(1.0 × 10⁻²)² = 5.0 × 10⁻⁸. That exceeds Ksp, so yellow PbI₂ forms. Forgetting the dilution on mixing is the most common error in these problems.

Selective precipitation: separating ions by Ksp

When two ions can each precipitate with the same reagent, the one whose salt needs the lower reagent concentration precipitates first. Slow addition can then separate them.

Worked example. A solution is 0.010 M in both Cl⁻ and CrO₄²⁻, and Ag⁺ is added slowly. AgCl starts to form when [Ag⁺] = 1.6 × 10⁻¹⁰ / 0.010 = 1.6 × 10⁻⁸ M. Ag₂CrO₄, with Ksp = 9.0 × 10⁻¹² from Appendix J, starts to form when [Ag⁺]² × 0.010 = 9.0 × 10⁻¹², so [Ag⁺] = 3.0 × 10⁻⁵ M. AgCl precipitates first.

By the time red Ag₂CrO₄ appears, [Cl⁻] has fallen to 1.6 × 10⁻¹⁰ / 3.0 × 10⁻⁵ = 5.3 × 10⁻⁶ M, so only about 0.05 percent of the chloride remains in solution. That sharp color change is the basis of the Mohr titration for chloride.

Why some salts dissolve more in acid

If a salt's anion is a base, such as OH⁻, CO₃²⁻ or F⁻, adding acid removes that anion from solution and pulls the dissolution forward. Hydroxides and carbonates dissolve far better at low pH. Salts of strong-acid anions, such as AgCl, are not affected this way.

Worked example. Magnesium hydroxide has Ksp = 8.9 × 10⁻¹². In pure water, Ksp = 4s³ gives s = 1.3 × 10⁻⁴ M and [OH⁻] = 2.6 × 10⁻⁴ M, so a saturated solution has pH 10.42. If a buffer instead holds the pH at 9.00, then [OH⁻] = 1.0 × 10⁻⁵ M and s = 8.9 × 10⁻¹² / (1.0 × 10⁻⁵)² = 0.089 M, almost 700 times more. This assumes the buffer has enough capacity to absorb the hydroxide released.

Ksp errors to check before you submit

Most errors come from the stoichiometry of the dissolution, not from the equilibrium idea itself. Writing out the dissolution equation and the s terms beneath it catches most of them.

• Setting Ksp equal to s² for a salt that releases three or more ions.
• Doubling an ion's concentration but not squaring it in the expression, or the reverse.
• Ranking solubility by Ksp across salts with different ion ratios.
• Forgetting to dilute both solutions when computing Q after mixing.
• Ignoring a common ion that is already present.
• Including the solid in the Ksp expression.