Answer to Question #112682 in Chemistry for Jshacona Abbott

Question

1. One type of kidney stones is formed by the precipitation of calcium phosphate (Ca3(PO4)2), which has a Ksp of
1.3 x 10-32 at 25 C. A patient submitted a urine sample which contained concentrations of 1.2 x 10-4 mol dm-3 calcium
ions and 1.1 x 10-8 mol dm-3 phosphate ions.
(a) Write a balanced equation for the formation of calcium and phosphate ions from calcium phosphate.

(b) Write the expression for the solubility product constant for calcium phosphate.

(c) Calculate the ionic product of calcium phosphate in the patient's urine.

(d) State why kidney stones are likely to form in the patient's urine.

Answer 1

a) Ca₃(PO₄)₂ <=> 3Ca²⁺ + 2PO₄^3-

b) K_sp = [Ca²⁺]³ × [PO₄^3-]²

c) Q_sp = [Ca²⁺]³ × [PO₄^3-]² = (1.2 × 10^-4 mol/L)³ × (1.1 × 10^-8 mol/L)² = 2.1 × 10^-28

d) As Q_sp > K_sp (2.1 × 10^-28 > 1.3 × 10^-32), the urine is oversaturated with calcium and phosphate ions. As a result, the equilibrium will shift to the left to form precipitate. The precipitated calcium phosphate will form kidney stones in the patient's urine.