Question

A solution is made by dissolving 6.93 grams of lead(II) nitrate into about 50 mL of water. The volume is then precisely brought up to 100 mL and the solution is saved as stock solution. A 50.0 mL aliquot* of this stock solution is then titrated with 0.222 M sodium phosphate. What would be the minimum number of milliliters (mL) of the the phosphate solution that are needed to completely precipitate (knock out) all the lead in this aliquot? (tolerance is ±0.1 mL)

Answer 1

To find the minimum number of mL of the phosphate solution needed to completely precipitate the lead in the aliquot, calculate the moles of lead(II) nitrate in the stock solution, use the stoichiometry to determine the moles of lead(II) phosphate that can be formed, and calculate the volume of the phosphate solution needed by dividing the moles by the molarity.

Step-by-step explanation:

To find the minimum number of milliliters of the phosphate solution needed to completely precipitate the lead in the aliquot, we need to use the stoichiometry of the balanced equation and the given information. First, calculate the moles of lead(II) nitrate in the stock solution by dividing the mass by the molar mass. Then, use the stoichiometry to determine the moles of lead(II) phosphate that can be formed. Finally, calculate the volume of the phosphate solution needed by dividing the moles by the molarity.

Given: Mass of lead(II) nitrate = 6.93 g, Volume of stock solution = 50.0 mL, Molarity of sodium phosphate = 0.222 M

Using the stoichiometry of the balanced equation: Pb(NO3)2(aq) + Na3PO4(aq) → Pb3(PO4)2(s) + 6 NaNO3(aq)

Calculations: Moles of lead(II) nitrate = 6.93 g / molar mass of Pb(NO3)2, Moles of lead(II) phosphate = Moles of lead(II) nitrate / 3, Volume of phosphate solution = Moles of lead(II) phosphate / molarity of sodium phosphate

Substituting the given values into the calculations, we can find the minimum number of milliliters of the phosphate solution needed to completely precipitate the lead in the aliquot.