A) Calculate the solubility in moles/liter (M) for Ba3(PO4)2 in pure water (Ksp

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Step 1. Write out the Ksp reaction and expression for Ksp, i.e.

Ba3(PO4)2 (s) --> 3Ba2+ (aq) + 2(PO4)3- (aq)
Ksp = [Ba]^3 * [PO4]^2

Remember that Ksp is just an equilibrium constant just like any other. Products over reactants, with each entity raised to the power of its coefficient. In this case the reactant is a solid, and thus entered as 1, so we ignore it.

Step 2. Let x = concentration of Ba3(PO4)2 that will dissolve in water. Now go back to your balanced equation. For every x mol of the solid that dissolve, you get 3x mol barium and 2x mol PO4. Substitute into your Ksp equation, i.e.

Ksp (given in problem) = (3x)^3 * (2x)^2

Solve this equation for x and you have your molar solubility of barium phosphate

(A) Calculate the solubility in moles/liter (M) for Ba3(PO4)2 in pure water (Ksp

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