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?

Question

The Ksp of silver phosphate is 2.8 × 10^-18

Answer 1

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 0.27-0.05 = 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^+

Answer 2

Thank you so much!!!

Answer 3

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)

Answer 4

I figured out what I was doing wrong! I have the right answer!
Yay!

Answer 5

Good for you.

Answer 6

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