what volume of0.100 M NaPO4 is required to precipitate all the lead(II) ions from 150ml of 0.254 M Pb(NO3)2
3Pb(NO3)2 + 2Na3PO4 --> Pb3(PO4)2 + 6NaNO3
mols Pb(NO3)2 = M x L = ?
Convert mols Pb(NO3)2 to mols Na3PO4. That is mols Pb(NO3)2 x (2 mols Na3PO4/3 mols Pb(NO3)2 = ?
Then grams Na3PO4 = mols x molar mass
To answer this question, we need to determine the stoichiometry of the reaction between sodium phosphate (Na3PO4) and lead nitrate (Pb(NO3)2), and then use that information to calculate the volume of Na3PO4 solution required.
The balanced chemical equation for the reaction between Na3PO4 and Pb(NO3)2 is as follows:
3 Na3PO4 + 2 Pb(NO3)2 -> Pb3(PO4)2 + 6 NaNO3
From the balanced equation, we can see that the ratio between Na3PO4 and Pb(NO3)2 is 3:2, meaning that for every 3 moles of Na3PO4, 2 moles of Pb(NO3)2 will react to form Pb3(PO4)2.
Now, let's calculate the number of moles of Pb(NO3)2 present in the 150 mL of 0.254 M solution:
Number of moles of Pb(NO3)2 = concentration (mol/L) × volume (L)
= 0.254 mol/L × 0.150 L
= 0.0381 moles
Using the stoichiometry, we know that 3 moles of Na3PO4 will react with 2 moles of Pb(NO3)2. Therefore, the number of moles of Na3PO4 required will be:
Number of moles of Na3PO4 = (2/3) × 0.0381 moles
= 0.0254 moles
Finally, we can calculate the volume of 0.100 M Na3PO4 solution required to obtain 0.0254 moles:
Volume of Na3PO4 = Number of moles / concentration
= 0.0254 moles / 0.100 mol/L
= 0.254 L = 254 mL
Therefore, 254 mL of 0.100 M Na3PO4 solution is required to precipitate all the lead(II) ions from 150 mL of 0.254 M Pb(NO3)2.