Question

Hydroxylapatite, Ca10(PO4)6(OH)2, has a solubility constant of Ksp = 2.34 x 10^-59. Solid hydroxylapatite is dissolved in water to form a saturated solution. What is the concentration of Ca+2 in this solution if [OH-] is somehow fixed at 5.30 x 10-6 M?

Answer 1

Answer:

The concentration of [Ca²⁺] is 8.47 x 10⁻³ M

Explanation:

We consider the solubility of hydroxyapatite,

Ca₁₀(PO₄)₆(OH)₂ ⇔ 10Ca²⁺ + 6PO₄³⁻ + 2 OH⁻

Assumed that there is a mol of hydoxyapatite disolved in water, yielding 10a mol Ca²⁺ of  and 6a mol of PO₄³⁻

We also have Ksp equation,

Ksp = [Ca²⁺]¹⁰ x [PO₄³⁻]⁶ x [OH⁻]² = 2.34 x 10⁻⁵⁹

     ⇔  10a¹⁰ x 6a⁶ x (5.30 x 10⁻⁶)² = 2.24 x 10⁻⁵⁹

     ⇔  60a¹⁶                                    = 2.24 x 10⁻⁵⁹ / 5.30 x 10⁻¹²

     ⇔  a¹⁶                                         = 0.007 x 10⁻⁴⁷ = 7 x 10⁻⁵⁰

     ⇔  a                                           = $\sqrt[16]{7 . 10^{-50} }$ =  8.47 x 10⁻⁴

Hence,

[Ca²⁺] = 10a = 8.47 x 10⁻³ M