Question
what volume 0,100M of Na2PO4 is required to precipatate all the lead(II) ions from 150 ml of 0.250M Pb(NO3)2
Answers
DrBob222
You made a typo in Na3PO4.
3Pb(NO3)2 + 2Na3PO4 ==> Pb3(PO4)2 + 6NaNO3
mols Pb(NO3)2 = M x L = ?
mols Na3PO4 = mols Pb(NO3)2 x (2 mols Na3PO4/3 mols Pb(NO3)2) = ?
Then M Na3PO4 = mols Na3PO4/L Na3PO4. YOu know mols and M, solve for L and convert to mL if you wish.
3Pb(NO3)2 + 2Na3PO4 ==> Pb3(PO4)2 + 6NaNO3
mols Pb(NO3)2 = M x L = ?
mols Na3PO4 = mols Pb(NO3)2 x (2 mols Na3PO4/3 mols Pb(NO3)2) = ?
Then M Na3PO4 = mols Na3PO4/L Na3PO4. YOu know mols and M, solve for L and convert to mL if you wish.