To precipitate all the lead(II) ions from 100.0 mL of 0.400M Pb(NO3)2, 400 mL of 0.100M Na3PO4 is required.
To determine the volume of 0.100M Na3PO4 required to precipitate all the lead(II) ions from 100.0 mL of 0.400M Pb(NO3)2, we need to consider the stoichiometry of the reaction between the two compounds.
From the balanced equation, 1 mole of Pb(NO3)2 reacts with 1 mole of Na3PO4 to form 1 mole of Pb3(PO4)2 precipitate.
First, calculate the number of moles of Pb(NO3)2 in 100.0 mL of the solution:
Moles of Pb(NO3)2 = volume (L) x concentration (M) = 100.0 mL x 0.400 M = 0.0400 moles
Since the stoichiometry of the reaction is 1:1, the number of moles of Na3PO4 needed to precipitate all the Pb²+ ions is also 0.0400 moles. Now, find the volume of 0.100M Na3PO4 needed:
Volume (L) = moles / concentration = 0.0400 moles / 0.100 M = 0.400 L = 400 mL
Therefore, 400 mL of 0.100M Na3PO4 is required to precipitate all the lead(II) ions.