Question

25.0 ml of a sodium phosphate solution reacts with 31.6 ml of a 0.33 mol/L lead (II) nitrate solution. What is the molar concentration of the sodium phosphate solution? Please and thank you!

Answer 1

2Na3PO4 + 3Pb(NO3)2 ==> Pb3(PO4)2 + 6NaNO3
Let me point out that the author of this problem may have overlooked the fact that Pb3(PO4)2 is a ppt; therefore, it pulls [PO4]^3- out of solution. You can calculate the beginning concentration of Na3PO4 but there is no Na3PO4 after the reaction has taken place.
millimols Pb(NO3)2 = mL x M = 31.6 mL x 0.33 = 10.428
How many millimols Na3PO4 will react exactly with that? That's
2Na3PO4 + 3Pb(NO3)2 ==> Pb3(PO4)2 + 6NaNO3
mmols Na3PO4 = 10.428 mmols Pb(NO3)2 x (2 mols Na3PO4/3 mols Pb(NO3)2 = 10.428 x 2/3 = 6.952.
Then M Na3PO4 initially = mmols/mL = 6.952 mmols/25.0 mL = ?