What volume of 0.100 M Na3PO4 is required to precipitate all the lead (ii) ions from 150.0 mL of 0.250 M Pb (NO3) 2?

First, we write the reaction equation:

3Pb (NO₃) ₂ + 2Na₃PO₄ → 6NaNO₃ + 3Pb₃ (PO₄) ₂

Moles of Pb ions present:

moles = concentration x volume

= 0.15 x 0.25

= 0.0375

From the equation,

moles Pb : moles Na₃PO₄

= 3 : 2

Moles of Na₃PO₄:

2/3 x 0.0375

= 0.025

volume = moles / concentration

= 0.025 / 0.1

= 0.25 L

= 250 ml