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?
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
To determine the volume of Na3PO4 required to precipitate all the lead(II) ions from Pb(NO3)2, we need to first write out the balanced chemical equation for the reaction between Na3PO4 and Pb(NO3)2.
The balanced chemical equation for this reaction is:
3Pb(NO3)2 + 2Na3PO4 → Pb3(PO4)2 + 6NaNO3
From the balanced equation, we can see that for every 3 moles of Pb(NO3)2, we require 2 moles of Na3PO4 to precipitate all the lead(II) ions.
Step 1: Calculate the number of moles of Pb(NO3)2 in 300.0 mL of 0.200 M solution.
Number of moles = Concentration × Volume
Number of moles = 0.200 mol/L × 0.300 L
Number of moles = 0.060 mol
Step 2: Use the stoichiometry of the balanced chemical equation to determine the number of moles of Na3PO4 required.
From the balanced equation, we can see that 3 moles of Pb(NO3)2 require 2 moles of Na3PO4.
So, the number of moles of Na3PO4 required = (2/3) × 0.060 mol
Number of moles of Na3PO4 required = 0.040 mol
Step 3: Calculate the volume of 0.120 M Na3PO4 required.
Volume = Number of moles/Concentration
Volume = 0.040 mol / 0.120 mol/L
Volume = 0.333 L or 333.3 mL
Therefore, the volume of 0.120 M Na3PO4 required to precipitate all the lead(II) ions from 300.0 mL of 0.200 M Pb(NO3)2 is 333.3 mL.