← Equilibrium principles in aqueous systems

7 Solubility & solubility product (Ks)

For a sparingly soluble salt, a saturated solution is at equilibrium with the solid. The solubility product Ks (Ksp) is the product of ion concentrations, each raised to its coefficient. General \mathrm{A}_x\mathrm{B}_y(s)\rightleftharpoons x\,\mathrm{A}^{n+}+y\,\mathrm{B}^{m-}\qquad K_s=[\mathrm{A}^{n+}]^x[\mathrm{B}^{m-}]^y Pure solids are not in the expression. AgCl (1:1) K_s=[\ce{Ag+}][\ce{Cl-}]=s^2 s = molar solubility. So s = √Ks. Mg(OH)₂ (1:2) K_s=[\ce{Mg^2+}][\ce{OH-}]^2=(s)(2s)^2=4s^3 s = ∛(Ks/4). Worked example Solubility from Ks Ks(Mg(OH)₂) = 6.0×10⁻¹². Find its molar solubility s. Dissolving: Mg(OH)₂ ⇌ Mg²⁺ + 2OH⁻. If s dissolves, [Mg²⁺] = s and [OH⁻] = 2s. Ks = [Mg²⁺][OH⁻]² = s(2s)² = 4s³. s³ = Ks/4 = 6.0×10⁻¹²/4 = 1.5×10⁻¹². s = ∛(1.5×10⁻¹²) = 1.14×10⁻⁴ mol L⁻¹. ✓ s = 1.1×10⁻⁴ mol L⁻¹. ✓

Mnemonic: will it precipitate?

“Q beats K, it goes away.” If the ionic product Q > Ks, the solution is over-saturated and a solid precipitates out. Q < Ks → stays dissolved. Q = Ks → exactly saturated.

Common-ion effect: “Add a twin, less gets in.” Adding an ion already in the equilibrium pushes it left, so solubility drops.

Predicting precipitation & the common-ion effect

  • Calculate the ionic product Q with the actual concentrations. If Q > Ks → precipitate forms; Q < Ks → stays dissolved; Q = Ks → saturated.
  • Common-ion effect: adding an ion already in the equilibrium increases its concentration, so the rate of the reverse (precipitating) reaction increases and solubility decreases. e.g. AgCl is less soluble in NaCl solution.
  • Adding a ligand instead does the opposite — see Dissolving a precipitate with a complex ion ↓.

Practise just this