[Rutherford]

The problem is attached.

1st part: I got [H₃PO₄]≈0.16M, [H₂PO₄^-]≈0.035M, [HPO₄^2-]≈6.2·10^-8M, [PO₄^3-]≈8.6·10^-19M. Is it okay?

2nd part:
H₃PO₄         +       2NH₃ (NH₄)₂HPO₄
0.01 mole        0.02mole        so 0.01 mole will be produced
c=0.1M and pH=(Ka(NH₄⁺)+Ka(HPO₄^2-))/2=10.78, good?

3rd part:
c=0.05M, c of magnesium ions is 0.1M. Ksp=[Mg²⁺][NH₄PO₄^2-], magnesium being in excess we can assume that all ammoniumphosphate precipitated, so n=0.05*0.2=0.01mole, m=1.37g. Is this correct? I thought that Ksp=[Mg²⁺][NH₄⁺][PO₄^3-], but I couldn't find a proof on the internet, true or not?

4th part:
Ksp=[Ca²⁺]³[PO₄^3-]²
[PO₄^3-]²=2/3[Ca²⁺]-x, x being the amount that dissociated:
Kb=x²/(2/3[Ca²⁺]-x). Now I need to express x through [Ca²⁺], but I get a square equation, which solution I need to put into the expression for Ksp and the algebra becomes pretty problematic. I very simplified the previous parts. I am sure that there is another approach for this one, but I don't know which. Help needed.

[CrazyAssasin]

1 part. I get the same answers
2 part. You have to take the value of second dissociation constant not the third one and to do it more precise you should evaluate a small excess of NH3, but the equation would become too complicated, so I think it's a good approximation. I get pH=8.22
3 part. I think that using pH from the previous part you should find concentration of PO4-3 ions, c(PO4-3)= 3.66*10^-5, so then you get Q>Ksp and 1.37g of precipitate will form.
4 part. x- amount of dissociated, then Ksp=[Ca+2]3[PO4-3]2=(3x)3(2x)2=108x^5
x= 4.6*10^-6 mol/L

[Rutherford]

2. Right. HPO₄^2- is the base so I should use Ka of the conjugated acid. Thanks.
3. Could you wrote your work in more detail (the part after obtaining the phosphate concentration)?
4. I don't think that that is valid. The amounts of calcium and phosphate are not proportionate as the phosphate will hydrolyze to a significant extent.

[CrazyAssasin]

3. Q=[Mg+2][NH4PO4-2] is greater than Ksp, therefore precipitate forms, but now I realised that we can't state that all phosphate ions would form precipitate, it should reach limit when Q<Ksp and I am confused here
4. Yes, you are right my way is not valid, however as you said equation becomes too problematic, if we knew pH of the solution, it would become easier. Maybe we can make an assumption that pH=7

[Borek]

Maybe we can make an assumption that pH=7

No. Higher than that. Hydrolysis shifts the solubility AND pH at the same time.

Edit: but to be honest, I don't see a reasonably easy method to solve the problem.

[CrazyAssasin]

I just meant that there might be buffer solution with pH=7. Anyway, this problem seems to be impossible without cubic equation

[Rutherford]

So no idea for part 4.

For part 3. if almost all ammoniumphsohpate precipitated, then the concentration of free magnesium ions is near 0.05M, and the concentration of free ammoniumphsohpate can be calculated from Ksp, and it will be very small. Why you think that my assumption is wrong?

[Borek]

So no idea for part 4.

Plenty of ideas, but It is not clear to me what tools are available and can be used, nor how much time you can use. For example, problem can be solved iteratively, my bet is that three or four repetitions should be enough to get a reasonably correct answer.

[Rutherford]

5 hours for 8 problems, calculator and pen can be used. The final equation may be solved by iteration, but obtaining the final equation, requires much algebra, too. Using wolfram I got 1.26·10^-4M.

[Rutherford]

What is the equation for the solubility constant in part 3? Ksp=[Mg²⁺][NH₄PO₄^2-] or Ksp=[Mg²⁺][NH₄⁺][PO₄^3-] How to know this?

[Borek]

The latter, that's just how K_sp is defined. No such thing as the NH₄PO₄^2- ion (or at least  - that would be the first time I hear about it).

[Rutherford]

Good. Thanks.