Determine the molar solubility for pb3 (po4) 2 in pure water. ksp for pb3 (po4) 2 is 1.0 x 10-54.

Dissociation of Pb₃ (PO₄) ₂ is;

Pb₃ (PO₄) ₂ (s) ⇆ 3Pb²⁺ (aq) + 2PO₄³⁻ (aq)

initial - -

change - X + 3X + 2X

Equilibrium 3X 2X

Ksp = [Pb²⁺ (aq) ]³ [PO₄³⁻ (aq) ]²

1.0 x 10⁻⁵⁴ = (3X) ³ (2X) ²

1.0 x 10⁻⁵⁴ = 108X⁵

X = 6.21 x 10⁻¹² M

Hence the molar solubility of Pb₃ (PO₄) ₂ is 6.21 x 10⁻¹² M.