what volume of .100 M Na3PO4 is required to precipitate all the lead(II) ions from 150.0 mL of .250 M Pb(NO3)2?

Could anyone plz help me and i appreciate any help.......

2 Answers

  • First write out the chemical reaction and determine the solid,

    2 Na3PO4 + 3 Pb(NO3)2 -> 6 NaNO3 + Pb3(PO4)2 (solid)

    next find out how many moles of lead there is,

    0.250 Mols/L * 0.150 L = 0.0375 Mols Pb(NO3)2

    because there is only one atom of Pb per Pb(NO3)2, then you know there are only 0.0375 mols of Pb atoms;

    next us stoichiometry to determine the number of mols of Na3PO4 you need

    0.0375 mols of Pb* 2 mol Na3PO4/ 3 mol Pb=0.025 mols of Na3PO4

    Then using Molarity to determine the Volume

    0.025 mols/0.100 M = 0.25 L= 25 ml Na3PO4 necessary

    Answer 25 ml Na3PO4 necessary

  • Timball is correct, but .25L to ml isn't 25mL. It's .025mL

Leave a Reply

Your email address will not be published. Required fields are marked *

Related Posts