Concentration of solution at precipitation point.

Question

A solution has a concentration of .0035 for BOTH Ag3 and Cu3. If PbO4 is continuously added, which ion will precipitate first? What is the concentration of the first ion left in the solution after the second ion starts to precipitate?

Ag3PO4 Ksp: 1.8*10^-18

Cu3PO4 Ksp: 7.8*10^-22

Answer 1

I made an attempt at this, but I am not sure it is correct. I would appreciate another tutor's input.

Cu3PO4 has the smallest Ksp, so it will precipitate first. The Phosphate ion concentration after all the Cu3 has precipitated is obtained from Ksp = X^4, so X  (the PO4 conc.) is the fourth root of the Ksp which is 7.8 x 10^-22. X = 5.28 x 10^-6 M. The Cu ion concentration is 3 times the PO4, so [Cu] = 1.58 x 10^-5 M

For the Ag3 ion precipitation, the product [Cu]^3 x [PO4] = 0.0035^3 x 5.28 x 10^-6 =

2.26 x 10^-7 M, so Ag3PO4 is still in solution since the result is less than its Ksp.

Once the Ag ion starts to precipitate, the PO4 concentration due to the dissociation of Ag3PO4 is gotten by Ksp = [Ag]^3 [PO4) = X^4. Solving for X gives you the resulting PO4 concentration due to the dissociation of Ag3PO4. This turns out to be X = (1.8 x 10^-18)^1/4 = 3.66 x 10^-5.  The total PO4 concentration is 5.28 x 10^-6 + 3.66 x 10^-5 = 4.19 x 10^-5 M.

Putting this result into Ksp = [Cu}^3 x [PO4] or 7.8 x 10-22 = [Cu]^3 x 4.19 x 10^-5. Solving for [Cu] gives 2.7 x 10^-6 M.