Hydroxylapatite, Ca10(PO4)6(OH)2, has a solubility constant of Ksp = 2.34 × 10^-59. Solid hydroxylapatite is dissolved in water to form a saturated solution. What is the concentration of Ca2 in this solution if [OH–] is somehow fixed at 5.30 × 10^-6 M?
To determine the concentration of Ca2+ in the solution, we need to use the solubility constant (Ksp) and the fixed concentration of OH-.
The balanced equation for the dissolution of hydroxylapatite is:
Ca10(PO4)6(OH)2 (s) ⇌ 10 Ca2+ (aq) + 6 PO43- (aq) + 2 OH- (aq)
The solubility constant expression for this reaction is:
Ksp = [Ca2+]^10 [PO43-]^6 [OH-]^2
We are given the value of Ksp (2.34 × 10^-59) and the fixed concentration of OH- (5.30 × 10^-6 M). We need to find the concentration of Ca2+.
Since the concentration of OH- is fixed, we can substitute the given value into the Ksp expression:
2.34 × 10^-59 = [Ca2+]^10 [PO43-]^6 (5.30 × 10^-6)^2
Now we need to determine the concentration of PO43-. From the balanced equation, the molar ratio of Ca2+ to PO43- is 10:6. Therefore, the concentration of PO43- can be expressed as:
[PO43-] = (6/10) [Ca2+]
Substituting this into the Ksp expression, we have:
2.34 × 10^-59 = [Ca2+]^10 ((6/10) [Ca2+])^6 (5.30 × 10^-6)^2
Simplifying the equation:
2.34 × 10^-59 = (6/10)^6 (5.30 × 10^-6)^2 [Ca2+]^16
Now we can solve for [Ca2+] by rearranging the equation and taking the 16th root of both sides:
[Ca2+] = (∛(2.34 × 10^-59 / [(6/10)^6 (5.30 × 10^-6)^2)])^(1/16)
Calculating this expression will give us the concentration of Ca2+ in the solution.