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?
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? I would greatly appreciate any assistance with this calculation. Thank you!
2 Answers
Jan 08, 2025
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
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