Calculating Molar Solubility from Ksp

The molar solubility s of a sparingly soluble salt is the number of moles that dissolve per liter of saturated solution, and it can be calculated from Ksp; for an AB salt s = √Ksp, and for an AB2 or A2B salt s = (Ksp/4)1/3. In brief: AgCl (Ksp = 1.8 × 10−10) has s = 1.3 × 10−5 mol/L, which is about 1.9 mg per liter.

Once Ksp is known, you can predict how much of a salt dissolves in pure water. The calculation depends on the stoichiometry of the salt, because each formula unit releases a different number of ions. This article gives a general method, a table of the formulas for the various salt types, five worked examples and a note on the effect of pH on the solubility of hydroxides.

What is the method?

  1. Write the dissolution equation and the Ksp expression.
  2. Let s be the molar solubility. Each mole of salt dissolving gives the ions in the ratio of the formula, so the ion concentrations are multiples of s.
  3. Substitute these into the Ksp expression and solve for s.
  4. If needed, convert s into g/L by multiplying by the molar mass.

This is the ICE method for a case with an initial concentration of zero for the ions: the “change” row is +ms and +ns for the ions.

Formulas for different salt types

Relation between Ksp and molar solubility for different salt types
Type Example Ksp in terms of s s in terms of Ksp
ABAgCl, BaSO4s2√Ksp
AB2 or A2BCaF2, Ag2CrO44s3(Ksp/4)1/3
AB3 or A3BAl(OH)327s4(Ksp/27)1/4
A2B3 or A3B2Ca3(PO4)2108s5(Ksp/108)1/5

The factors come from the stoichiometry: for CaF2, Ksp = s(2s)2 = 4s3, and for Ca3(PO4)2, Ksp = (3s)3(2s)2 = 27s3 × 4s2 = 108s5.

Worked examples

Example 1: silver chloride (AB)

Ksp = 1.8 × 10−10. s = √(1.8 × 10−10) = 1.34 × 10−5 mol/L. In mass units: 1.34 × 10−5 mol/L × 143.32 g/mol = 1.9 × 10−3 g/L.

Ksp = s2 = 1.8 × 10−10 s = √(1.8 × 10−10) = 1.34 × 10−5 mol/L 1.34 × 10−5 mol/L × 143.32 g/mol = 1.9 × 10−3 g/L

Example 2: calcium fluoride (AB2)

  1. CaF2(s) ⇌ Ca2+ + 2F−; Ksp = [Ca2+][F−]2 = 3.9 × 10−11.
  2. [Ca2+] = s and [F−] = 2s, so 4s3 = 3.9 × 10−11.
  3. s3 = 9.75 × 10−12, so s = 2.1 × 10−4 mol/L.
  4. [F−] = 2s = 4.3 × 10−4 M. In mass units, s × 78.07 g/mol = 0.017 g/L.
CaF2 ⇌ Ca2+ + 2F− [Ca2+] = s , [F−] = 2s Ksp = s(2s)2 = 4s3 = 3.9 × 10−11 s3 = 9.75 × 10−12 , s = 2.1 × 10−4 mol/L

Example 3: silver chromate (A2B)

Ksp = 1.1 × 10−12 = 4s3, so s3 = 2.75 × 10−13 and s = 6.5 × 10−5 mol/L. [Ag+] = 2s = 1.3 × 10−4 M.

4s3 = 1.1 × 10−12 s3 = 2.75 × 10−13 , s = 6.5 × 10−5 mol/L [Ag+] = 2s = 1.3 × 10−4 M

Example 4: magnesium hydroxide and the pH

Ksp = [Mg2+][OH−]2 = 5.6 × 10−12 = 4s3, so s3 = 1.4 × 10−12 and s = 1.1 × 10−4 mol/L. Then [OH−] = 2s = 2.2 × 10−4 M, pOH = 3.65 and pH = 10.35. A saturated solution of magnesium hydroxide, the active ingredient of milk of magnesia, is therefore mildly basic.

4s3 = 5.6 × 10−12 , s = 1.1 × 10−4 mol/L [OH−] = 2s = 2.2 × 10−4 M pOH = 3.65 , pH = 14 − 3.65 = 10.35

Example 5: calcium phosphate (A3B2)

Ksp = 2.1 × 10−33 = 108s5, so s5 = 1.94 × 10−35. Taking the logarithm: log s = −34.71 / 5 = −6.94, so s = 1.1 × 10−7 mol/L. This tiny value is a simple estimate: the phosphate ion is a base and reacts with water, so the real solubility is higher.

Comparison of the results

Molar solubilities calculated from Ksp at 25 °C
Salt Ksp Molar solubility (mol/L) Solubility (g/L)
AgCl1.8 × 10−101.3 × 10−50.0019
BaSO41.1 × 10−101.0 × 10−50.0024
CaF23.9 × 10−112.1 × 10−40.017
Mg(OH)25.6 × 10−121.1 × 10−40.0065
Ag2CrO41.1 × 10−126.5 × 10−50.022

How does pH change solubility?

If the anion of the salt is the conjugate base of a weak acid, the solubility increases in acidic solution. For example, CaCO3 dissolves in acid because carbonate ions are removed by CO32− + 2H+ → H2O + CO2, which shifts the dissolution equilibrium to the right. Hydroxides such as Mg(OH)2 dissolve in acid since H+ reacts with OH−. In contrast, salts of strong acid anions (AgCl, BaSO4) have almost no pH dependence. This is the reason why limestone statues are damaged by acid rain and why tooth enamel is attacked by acids.

What are the limits of the calculation?

The simple calculation assumes that the dissolved salt is completely dissociated into the two ions and that these do not react further. In reality, ion pairs (such as CaSO4 in solution), hydrolysis of the ions and complex-ion formation increase the measured solubility. For salts with a high solubility the activities of the ions rather than concentrations should be used.

How were solubilities first measured?

Solubilities were measured by evaporating a known mass of a saturated solution and weighing the residue, a method that is used for soluble salts. For very sparingly soluble salts, the conductivity of the saturated solution gave the concentration: Friedrich Kohlrausch and Ferdinand Rose used it in the 1890s for silver chloride and other salts. Such measurements gave the first reliable Ksp values and confirmed Nernst’s concept.

Which mistakes are common?