What volume of 0.150 M Na3PO4 is required to precipitate ...?

Question

What volume of 0.150 M Na3PO4 is required to precipitate all the lead(II) ions from 140.0 mL of 0.300 M Pb(NO3)2?

Answer 1

First write up the reaction eq'n

2Na3PO4 + 3Pb(NO3)2 = Pb3(PO4)2 + 6NaNO3
So Molar ratios are 2: 3 :: 1 : 6
Moles(Pb(NO3)2 = 0.3 x 140/1000 = 0.042
Moles(Na3PO4) = 0.042 x 2/3 = 0.028

Vol(Na3PO4) = 0.028 x 1000/0.15 = 186.666.... cm^3 THE Answer!!!!!

Answer 2

Hmmmm i got a 5 on my AP Chem test, but that was a year ago, but anyways this is what i think it is......

3Na+PO4+Pb+2(NO3) ----> 3Na+2(NO3) +(Pb.3PO4.2)
.3Molar Pb= xmols/ .14 liters .042 mols of Pb

.042 mols Pb x(times) 2mols PO4/ 3mols Pb= .028 Mols ofPO4 needed

.15Molar= .028/xLiters x=.187

Well thats that i think



wow someone answered when i was writing my answer...at least we got the same answer haha