Hydroxylapatite, Ca10(PO4)6(OH)2, has a solubility constant of Ksp = 2.34 × 10-59, and dissociates according to
Ca10(PO4)6(OH)2(s) --> 10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)
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 7.30 × 10-6 M?
I am confused as to how to solve for x with such large exponents
Ca10(PO4)6(OH)2 ==> 10Ca^2+ + 6PO4^3- + 2OH^-
I would substitute
(Ca^2+) = 10x
(PO4^3-) = 6x
(OH^-) = 2x + 7.3E-6
Ksp = (Ca^2+)^10(PO4^3-)^6(OH^-)^2
Ksp = (10x)^10(6x)^6(2x + 7.3E-6)^2
Solve for x. You want x = (Ca^2+).
Thank you. I think I understand how to set it up.
To solve for the concentration of Ca2+ in the saturated solution, we can use the solubility product constant (Ksp) expression and make some assumptions.
The balanced equation for the dissociation of hydroxylapatite is:
Ca10(PO4)6(OH)2(s) → 10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)
The solubility product constant expression for this reaction is:
Ksp = [Ca2+]^10 [PO43-]^6 [OH-]^2
Since the solubility of hydroxylapatite is fixed, the concentrations of the ions can be assumed to be equal to x, except for [OH-], which is given as 7.30 × 10^-6 M.
Therefore, the solubility product expression becomes:
Ksp = (x)^10 (x)^6 (7.30 × 10^-6)^2
Simplifying the expression:
2.34 × 10^-59 = x^16 (7.30 × 10^-6)^2
Let's solve for x mathematically:
Take the square root on both sides:
√(2.34 × 10^-59) = x^8 (7.30 × 10^-6)
Simplifying:
4.83 × 10^-30 = x^8 (7.30 × 10^-6)
Divide both sides by 7.30 × 10^-6:
(4.83 × 10^-30) / (7.30 × 10^-6) = x^8
Using a calculator:
6.62 × 10^-25 = x^8
Take the 8th root of both sides:
x = (6.62 × 10^-25)^(1/8)
Using a calculator to evaluate the right side:
x ≈ 1.067 × 10^-4 M
Thus, the concentration of Ca2+ in the saturated solution is approximately 1.067 × 10^-4 M.
To solve this problem, we need to use the solubility product constant (Ksp) and the given concentration of OH- to determine the concentration of Ca2+ in the solution.
First, let's write the balanced equation again and indicate the change in molar concentrations:
Ca10(PO4)6(OH)2(s) ⇌ 10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)
Let's assume x is the concentration of Ca2+ ions in the solution. Since 2 moles of OH- ions are produced for every 10 moles of Ca2+, the concentration of OH- is 2x.
The solubility product constant (Ksp) expression for this reaction is:
Ksp = [Ca2+]^10 [PO43-]^6 [OH-]^2
Now, substitute the given values into the equation:
2.34 × 10^-59 = (x^10) [(2x)^2]^6 (7.30 × 10^-6)^2
Simplifying the equation:
2.34 × 10^-59 = (2^12) (x^10) [(7.30 × 10^-6)^2]^6
2.34 × 10^-59 = 2^12 * x^10 * (7.30 × 10^-6)^(2 * 6)
2.34 × 10^-59 = 2^12 * x^10 * (7.30 × 10^-6)^12
To solve for x, take the 10th root and divide both sides by the other factors:
x = [2.34 × 10^-59 / (2^12 * (7.30 × 10^-6)^12)]^(1/10)
Evaluating this expression with a calculator, we find:
x ≈ 2.72 × 10^-16
Therefore, the concentration of Ca2+ in the solution is approximately 2.72 × 10^-16 M.