Posted by Rudy on Sunday, January 26, 2014 at 7:24pm.

wouldn't the Ca ion cncentration be 5 times the OH conc?

First, I think you left a squared term off the OH^-.

Ca10(PO4)6(OH)2(s) --> 10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq)

(Ca^2+) = 10x
(PO4^3-) = 6x
(OH^-) = 2x

Ksp = (Ca^2+)^10(PO4^3-)^6(OH^-)^2

Substituting the x values we have
Ksp = (10x)^10(6x)^6(4.3e-6 + 2x)^2
I would make the assumption that
4.3e-6 + 2x = 4.3e-6 so we have
Ksp = (10x)^10(6x)^6(4.3e-6)^2
2.34e-59 = 10^10*x^10 (6x)^6*(1.85e-11)
2.34e-59 = 10^10*x^10*4.66e4*x^6*(1.85e-11)
2.34e-59/[(10)^10*(4.66e4)*(1.85e-11) = x^16.

To solve this problem, we need to use the solubility product expression and the given dissociation equation. Here's how you can proceed:

1. Write the solubility product expression using the molar concentrations of the ions involved in the dissociation equation:
Ksp = [Ca2+]^10 [PO43-]^6 [OH-]^2

2. In the dissociation equation, we are given that [OH-] is fixed at 4.30 × 10-6. Substitute this value into the solubility product expression:
Ksp = [Ca2+]^10 [PO43-]^6 (4.30 × 10-6)^2

3. Rearrange the equation to solve for [Ca2+]:
[Ca2+]^10 = Ksp / ([PO43-]^6 (4.30 × 10-6)^2)

4. Take the 10th root of both sides to isolate [Ca2+]:
[Ca2+] = (Ksp / ([PO43-]^6 (4.30 × 10-6)^2))^(1/10)

5. Substitute the given values to calculate [Ca2+]. You will need the value of Ksp and the concentration of the phosphate ion ([PO43-]).

Note: In order to fully solve this problem, you will need the concentration of [PO43-]. If that value is not given, you may need to use additional information or assumptions to proceed.