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

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

Step-by-step 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