[mikhailconrad]

Today during our midterms, I came across this question that asked us to get the K'sp of Ca₃(PO₄)₂, given that its concentration was 2.57x10^-6.

K'sp = [Ca]³[PO₄]²
2[Ca] = 3[PO₄]
2/3[Ca] = [PO]
K'sp = [Ca]³  * 4/9 [Ca]²
K'sp = 4/9 [Ca]⁵
K'sp = 4/9 * (2.57x10-6)⁵

The answer I got wasn't among the choices, except E). None of the Above.

I feel like I did everything correctly, but I have to be sure. If I did anything wrong, please oblige. Thanks in advance.

[Borek]

2[Ca] = 3[PO₄]

Quite the opposite.

[mjc123]

No, OP is right. If you have 1 mole of calcium phosphate, you have 3 moles Ca²⁺ and 2 moles PO₄^3-. 2*3 = 3*2.
OP's mistake is in the last step. Concentration of Ca₃(PO₄)₂ is 2.57x10^-6 (M presumably, don't forget units. What are the units of K'_sp?)
What is the concentration of Ca²⁺?

[mikhailconrad]

OP's mistake is in the last step. Concentration of Ca₃(PO₄)₂ is 2.57x10^-6 (M presumably, don't forget units. What are the units of K'_sp?)
What is the concentration of Ca²⁺?

Oh. No. I get it now. My book didn't give me any problems like that where the all ions have more than 1 mol present in the equation. I usually just equate the molarity of the dissociated ion with the solid. This time the numerical coefficient changes things, doesn't it? 2.57E-6 M is the molarity of the solution but not the concentration of [Ca] ions is it? I have to multiply it by three?
So, would it be K'sp = 4/9 (7.71x10^-6 M)⁵
K'sp = 1.23E-26 M⁵

Which I remember was one of the choices! That makes so much more sense. I knew something was kinda off. Thank you. 

[AWK]

http://www.scs.illinois.edu/mgweb/Course_Notes/freshchem/Quizzes/Quiz8_10.31.08_Solutions.pdf