Chemistry  Please help!
posted by amanda1012 on .
What is the silver ion concentration in a solution prepared by mixing 433 mL of 0.356 M silver nitrate with 467 mL of 0.574 M sodium phosphate?
The Ksp of silver phosphate is 2.8 × 10^18

This is a limiting reagent, solubility product, and common ion effect problem all rolled into one.
3AgNO3 + Na3PO4 ==> Ag3PO4 + 3NaNO3.
mols AgNO3 = M x L = ? (apparox 0.15 but you should confirm the actual number.
mols Na3PO4 = M x L = ? (about 0.27)
Convert mols AgNO3 to mols of the product using the coefficients in the balanced equation.
0.15 x (1/3) = about 0.05
Convert mols Na3PO4 to mols Ag3PO4 = about 0.27
In limiting reagent problems the SMALLER value ALWAYS wins; therefore, we have approximately 0.05 mols Ag3PO4 formed.
How much Na3PO4 is used. That is
0.15 mol AgNO3 x 1/3 = about 0.05 mols Na3PO4.
How much Na3PO4 remains unreacted. That is 0.270.05 = about .22 mols. The concentration of the Na3PO4 is 0.22/(total volume in L) = ?M
So we have a saturated solution of Ag3PO4 with an excess of ?M Na3PO4.
........Ag3PO4 ==> 3Ag^+ + PO4^3
..........x.........3x......x
Ksp = (Ag^+)^3(PO4^3)
For Ag substitute 3x
For PO4^3 substitute x (for PO4 from Ag3PO4) + ?M from excess PO4^3 from Na3PO4. Solve for 3x = Ag^+ 
Thank you so much!!!

I am doing something wrong in my calculations.
For ?M I have 0.24
My equation is:
2.8*10^(18) = (3x)^3 * (x+0.24) 
I figured out what I was doing wrong! I have the right answer!
Yay! 
Good for you.

Hey i m not getting how you did the last part how did you solve for this equation
2.8*10^(18) = (3x)^3 * (x+0.24)
what was your answer