Question

A water is in equilibrium with Ca3(PO4)2(s). The solution contains other cations and anions, but no other weak acids or bases or sources of Ca and PO4, and has a pH of 8.6. Ignore complexes.

(a) What is the solubility (S) of Ca ion in this water? (optional) Tint: Develop equations that will allow you to make log C-pH plots for phosphate species.

(b) Will solubility of Ca ion increase, decrease, or remain constant if HCl is added to this water? Why?

Answer 1

a. Ksp of Ca₃(PO₄)₂ is 1.08×10^-23.

The solubility equilibrium equation is:

Ca₃(PO₄)₂(s) ⇌ 3Ca²⁺ + 2PO₄³⁻

Ksp = [Ca²⁺]³[PO₄³⁻]²

Let S = solubility of calcium phosphate

Then [Ca²⁺] = 3S

And [PO₄³⁻] = 2S

Substituting these values back into the solubility expression:

Ksp=1.08×10⁻²³ ≡ [Ca²⁺]³[PO₄³⁻]² = (3S)³(2S)²

1.08×10⁻²³ = 108S⁵

S = 1.0×10⁻⁵ mol⋅L^-1

b. Ca₃(PO₄)₂ has very low solubility in water​. The solubility of salts containing basic anion like Ca₃(PO₄)₂ increases as [H⁺] increases or as pH decreases.

Thus, if HCl is added to the solution, the pH decreases and the solubility of Ca₃(PO₄)₂ increases.