What volume of 0.120 M Na3PO4 is required to precipitate all the lead(II) ions from 300.0 mL of 0.200 M Pb(NO3)2?

Answer 1

you have .3*.2 moles of Pb
or Pb ion is .06 moles
in Pb3(PO4)2, with .06 moles of Pb, you need 2/3 of that moles of PO4, or
2/3*.06=.04 moles PO4 ions.

Now to get that much PO4
.04=.120*Volume in liters
Volume=.04/.120 liters