millimoles AgNO3 = mL x M = 495 x 0.369 = about 182.6
millimoles Na3PO4 = 419 x 0.471 = about 197.3
You should check these and correct to the correct number of significant figures.
............3AgNO3 + Na3PO4 ==> Ag3PO4 + 3NaNO3
I............182.6..........197.3.................0...............0
This is a limiting reagent problem as well as a solubility product problem.
How much Ag3PO4 is formed and how much of the reactants remain.
182.6 mmols AgNO3 x (1 mol Ag3PO4/3 mols AgNO3) = 60.9
197.3 mmols Na3PO4 x (1 mol Ag3PO4)/1 mol Na3PO4) = 197.4 so the LR is AgNO3 (all of it is used) with Na3PO4 excess remaining.
I............182.6..........197.3.................0...............0
C.........-182.6..........-60.9................+60.9 ........182.6
E..............0.............136.4..................60.9.......................
So we have a solution with 60.9 mmols of Ag3PO4 ppt (solid) with 0 mmols AgNO3 and 60.9 mmols Na3PO4.
The (Ag^+) will be determined by the common ion effect of the phosphate.
Ag3PO4 ==> 3Ag^+ + PO4^3-
Ksp = 2.8E-18 = (Ag^+)^3(PO4^3-)
Look above. (Ag^+) = solve for this.
(PO4^3-) = mmols Na3PO4 from above/ total volume in mL
Note that total volume is 495 mL + 419 mL = ? mL.
Check all of this to confirm each step.
What is the silver ion concentration in a solution prepared by mixing 495 mL 0.369 M silver nitrate with 419 mL 0.471 M sodium phosphate? The 𝐾sp of silver phosphate is 2.8×10−18.
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