Question

Hydroxyapatite, Ca 10 ( PO 4 ) 6 ( OH ) 2 Ca10(PO4)6(OH)2 , has a solubility constant of Ksp = 2.34 × 10 − 59 2.34×10−59 , and dissociates according to Ca 10 ( PO 4 ) 6 ( OH ) 2 ( s ) − ⇀ ↽ − 10 Ca 2 + ( aq ) + 6 PO 3 − 4 ( aq ) + 2 OH − ( aq ) Ca10(PO4)6(OH)2(s)↽−−⇀10Ca2+(aq)+6PO43−(aq)+2OH−(aq) Solid hydroxyapatite is dissolved in water to form a saturated solution. What is the concentration of Ca 2 + Ca2+ in this solution if [ OH − ] [OH−] is fixed at 2.50 × 10 − 6 M 2.50×10−6 M ?

Answer 1

Answer:

1.315x10⁻³M = [Ca²⁺]

Explanation:

Based in the reaction:

Ca₁₀(PO₄)₆(OH)₂(s) ⇄ 10Ca²⁺(aq) + 6PO₄³⁻(aq) + 2OH⁻(aq)

Solubility product, ksp, is defined as:

ksp = [Ca²⁺]¹⁰ [PO₄³⁻]⁶ [OH⁻]²

From 1 mole of hydroxyapatite are produced  10 moles of Ca²⁺ and 6 moles of PO₄³⁻. That means moles of PO₄³⁻ are:

6/10 Ca²⁺ = PO₄³⁻

Replacing in ksp formula:

ksp = [Ca²⁺]¹⁰ [0.6Ca²⁺]⁶ [OH⁻]²

As [OH⁻] is 2.50x10⁻⁶M and ksp is 2.34x10⁻⁵⁹:

2.34x10⁻⁵⁹ =  [Ca²⁺]¹⁰ [0.6Ca²⁺]⁶ [2.50x10⁻⁶]²

3.744x10⁻⁴⁸ = 0.046656[Ca²⁺]¹⁶

1.315x10⁻³M = [Ca²⁺]

I hope it helps!