Question

Please answer these question:

Answer 1

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