Answer:
a) S = 5.9 × 10⁻⁵ M
b) S = 9.2 × 10⁻⁸ M
Explanation:
a) The major component in bones is calcium hydroxylapatite, Ca₅(PO₄)₃OH, with a Ksp=7.3 × 10⁻³⁴. Determine the solubility of bones in pure water.
To determine the solubility (S) of hydroxylapatite in water we will use an ICE chart. We recognize 3 stages: Initial, Change and Equilibrium, and we complete each row with the concentration or change in concentration.
Ca₅(PO₄)₃OH(s) ⇄ 5 Ca²⁺(aq) + 3 PO₄³⁻(aq) + OH⁻(aq)
I 0 0 0
C +5S +3S +S
E 5S 3S S
The solubility product Kps is:
Kps = 7.3 × 10⁻³⁴ = [Ca²⁺]⁵.[PO₄³⁻]³.[OH⁻] = (5S)⁵.(3S)³.(S)=84375 . S⁹
S = 5.9 × 10⁻⁵ M
b) Calculate the molar solubility of Ca₃(PO₄)₂ ( Ksp = 7.3 × 10⁻³⁴ ) in a 0.065 M Be₃(PO₄)₂ (100% soluble) solution
First, we have to calculate the concentration of PO₄³⁻ that comes from Be₃(PO₄)₂, which is a strong electrolyte.
Be₃(PO₄)₂(aq) ⇒ 3 Be²⁺(aq) + 2 PO₄³⁻(aq)
I 0.065 M 0 0
C -0.065 M +3×0.065 M 2×0.065 M
E 0 0.195 M 0.130 M
Now, 0.130 M will be the initial concentration in the solubilization of Ca₃(PO₄)₂.
Ca₃(PO₄)₂(s) ⇄ 3 Ca²⁺(aq) + 2 PO₄³⁻(aq)
I 0 0.130
C +3S +2S
E 3S 0.130 + 2S
Kps = 7.3 × 10⁻³⁴ = [Ca²⁺]³.[PO₄³⁻]²=(3S)³.(0.130 + 2S)²
In the term (0.130 + 2S)², 2S is very small in comparison with 0.130, so we can neglect it to simplify calculations.
7.3 × 10⁻³⁴ = (3S)³.(2S)² = 108 . S⁵
S = 9.2 × 10⁻⁸ M
