Solubility Equilibria Calculations

• M_pX_q(s)⇌𝑝M^m+_(aq)+𝑞Xⁿ⁻_(aq)

• These calculations are most conveniently performed using a compound’s molar solubility, measured as moles of dissolved solute per liter of saturated solution.

Fluorite, CaF₂, is a slightly soluble solid that dissolves according to the equation:

• CaF_2(s)⇌Ca²⁺_(aq)+2F⁻_(aq)

The concentration of Ca²⁺ in a saturated solution of CaF₂ is 2.15 × 10^–4 M.

• What is the solubility product of fluorite?

According to the stoichiometry of the dissolution equation, the fluoride ion molarity of a CaF₂ solution is equal to twice its calcium ion molarity:

• [F⁻]=(2 mol F⁻/1 mol Ca²⁺)=
• (2) x (2.15×10⁻⁴𝑀) =
• 4.30×10⁻⁴𝑀

• 𝐾_sp = [Ca²⁺][F⁻]² =
• (2.15×10⁻⁴) x (4.30×10⁻⁴)2=
• 3.98×10⁻¹¹

• The 𝐾_sp of copper(I) bromide, CuBr, is 6.3×10^–9. Calculate the molar solubility of copper bromide.

Substituting the equilibrium concentration terms into the solubility product expression and solving for x yields:

• 𝐾_sp=[Cu⁺][Br⁻] = 6.3×10⁻⁹
• 6.3×10⁻⁹=(𝑥)(𝑥) = 𝑥²
• 𝑥 = √(6.3×10⁻⁹) = 7.9×10⁻⁵𝑀