Molar Solubility of Silver Chloride from Ksp

Given

• Dissolution: AgCl(s) ⇌ Ag+(aq) + Cl-(aq)

• Ksp(AgCl) = 1.8 x 10^-10 at 25 C

• Molar mass of AgCl = 143.32 g/mol

• Pure water with no common ions present.

Find

The molar solubility s of AgCl in pure water (and its mass solubility in mg/L).

Principle

For a sparingly soluble salt, the solubility product Ksp equals the product of the ion concentrations at saturation, with the solid omitted. If s mol/L of AgCl dissolve, then [Ag+] = s and [Cl-] = s, so Ksp = s^2 and s = sqrt(Ksp). Multiplying by the molar mass converts molar solubility to grams per liter.

Formula

Solution

1. Let s be the molar solubility in mol/L. Each mole of AgCl that dissolves gives 1 mol Ag+ and 1 mol Cl-, so at saturation [Ag+] = s and [Cl-] = s.

2. Write the Ksp expression (solid omitted): Ksp = [Ag+][Cl-] = s x s = s^2 = 1.8 x 10^-10.

3. Take the square root: s = sqrt(1.8 x 10^-10) = sqrt(1.8) x 10^-5 = 1.3416 x 10^-5 = 1.34 x 10^-5 M.

4. Check: (1.34 x 10^-5)^2 = 1.7956 x 10^-10, which rounds to 1.8 x 10^-10, matching Ksp.

5. Convert to mass: 1.34 x 10^-5 mol/L x 143.32 g/mol = 1.92 x 10^-3 g/L = 1.92 mg/L.

Unit check

Ksp has units M^2 here (product of two concentrations), and sqrt(M^2) = M, so s comes out in mol/L as required. Multiplying mol/L by g/mol gives g/L, which converts directly to mg/L.

What this means physically

Silver chloride is classified as insoluble: only about 13 micromoles dissolve per liter of water. This tiny solubility is why adding silver nitrate to a chloride solution gives an immediate white precipitate — the ion product exceeds 1.8 x 10^-10 almost at once, and the excess AgCl leaves the solution as a solid.