[Bo]

The question itself is somewhat long, so I'll just post the part that's annoying:

The Ksp of hydroxyapatite, Ca₅(PO₄)₃OH is 6.8e-37. Calculate the solubility of hydroxyapatite in pure water in moles per liter.

What I've been doing so far is doing the standard steps to a solubility problem.

6.8e-37 = [Ca]⁵[PO₄]³[OH]
Skipping a few steps, I get to:
6.8e-37=[5x⁵][3x³][1e-7] <---------- (OH = 1e-7 as water is the primary producer of OH ions, as the solubility of hydroxyapatite is extremely low.)

I ended up getting x=5.47e-5, which is slightly incorrect. The correct answer is 2.74e-5, which is exactly 1/2 of my answer.

Any help or guidance would be appreciated, as well as any tips that could help me along the lines of the conceptual and the problem-solving aspects of Aqueous Equilibria, buffering, titrations, ionic mixtures/salts, solubility, ligands/complex ions, etc. Thanks.

[Borek]

6.8e-37=[5x⁵][3x³][1e-7]

This is wrong - it should be (5x)⁵ and so on. But I guess that's what you really did, it is just a typo.

<---------- (OH = 1e-7 as water is the primary producer of OH ions, as the solubility of hydroxyapatite is extremely low.)

Good idea, but you should compare it with your final result. If the calculated solubility is 5.47e-5 M, [OH^-] is not 10^-7 M.

The correct answer is 2.74e-5, which is exactly 1/2 of my answer.

Random coincidence.

Reality is much more complicated, as PO₄^3- is a strong base that will modify pH of the solution.

[AWK]

2.74x10^-5 cannot be a correct answer since solubility product you used has 2 significant digits. 2.7... is a correct value when neglecting hydrolysis of phosphate anion.