Q1(e). Solubility of lead phosphate

Answer in brief

The simple Ksp model gives s = 1.69 × 10⁻⁷ mol L⁻¹, or 1.37 × 10⁻⁴ g L⁻¹ of Pb₃(PO₄)₂.

Key points

• Dissolution: Pb₃(PO₄)₂(s) ⇌ 3Pb²⁺ + 2PO₄³⁻; if molar solubility is s, [Pb²⁺] = 3s and [PO₄³⁻] = 2s.
• Solubility product: Ksp = (3s)³(2s)² = 108s⁵, so s = (1.5 × 10⁻³²/108)^1/5 = 1.6925 × 10⁻⁷ mol L⁻¹.
• M = 3(207.19) + 2[30.974 + 4(15.999)] = 811.510 g mol⁻¹.
• Mass solubility = sM = (1.6925 × 10⁻⁷)(811.510) = 1.3735 × 10⁻⁴ g L⁻¹.
• This textbook calculation neglects phosphate protonation, Pb hydrolysis and activity corrections.