What is the silver ion concentration in a solution prepared by mixing 487 mL of 0.384 M silver nitrate with 379 mL of 0.583 M sodium phosphate? The Ksp of silver phosphate is 2.8 × 10-18

Answer 1

This is a limiting reagent problem and a Ksp problem with a common ion all rolled into one.
3AgNO3 + Na3PO4 ==> Ag3PO4 + 3NaNO3
mol AgNO3 = M x L = about 0.19 but you need to do it more accurately.
mols Na3PO4 = M x L = about 0.22 but you confirm.
Convert mols AgNO3 to mols Ag3PO4. That's 1/3 x mol AgNO3 = about 0.06 (the coefficients in the equation give you the fraction.)
Convert mols Na3PO4 to mols Ag3PO4. That's 0.22.
You have two values for mols Ag3PO4 and both can't be right. The correct value in limiting reagent problems is ALWAYS the smaller of the two and the reagent responsible for that small number is the limiting reagent. Thus 0.06 mols Ag3PO4 will be formed and AgNO3 is the limiting reagent. That makes Na3PO4 the reagent in excess.
So we now have a saturated solution of Ag3PO4 with a excess of Na3PO4. How much excess? That's 0.22-(0.06/3) = about 0.2
........Ag3PO4 --> 3Ag^+ + PO4^3-
.........solid......3x......x
(Ag^+) = 3x
(PO4^3-) = x + 0.2
Ksp Ag3PO4 = (Ag^+)(PO4^3-)
Substitute the concn into Ksp expression and solve for x, then Ag^+ = 3x.
Post your work if you get stuck.

Answer 2

Ok so I get everything up until the end
Ksp = [Ag+]^3[Po43-] since Ag3PO4 <----> 3Ag+ + PO43-
so then i do ((3x)^3)(x+0.2) = 2.8e-18

is this correct?... it comes out to be a very complex mathematical problem to solve for x which is why i feel it's wrong

Answer 3

You are ok to this point. Are you sure of the 0.2? That was an estimate on my part. If that is ok, then you can solve the cubic equation or make an assumption that x + 0.2 = 0.2 and solve the resulting equation which is easy enough to do. I would make the assumption; when you finish compare 0.2 with x+0.2 and see if the assumption that x was too small to add in is ok. I think the assumption will be ok.

Answer 4

got it, thanks!