Final answer:
The student needs to mix slightly more than 435.65 grams of potassium sulfate with 2.00 L of 1.25 M lead(II) nitrate solution to ensure an excess of sulfate ions for Pb²+ precipitation.
Step-by-step explanation:
The question is asking how many grams of solid potassium sulfate must be added to remove Pb²+ ions from an aqueous solution by forming a precipitate. This involves a stoichiometric calculation based on the reaction between lead(II) nitrate and potassium sulfate.
We start by writing the balanced chemical equation for the reaction between lead(II) nitrate and potassium sulfate:
Pb(NO3)2 (aq) + K2SO4 (s) \(→\) PbSO4 (s) + 2 KNO3 (aq)
Using the given concentration of lead(II) nitrate (1.25 M) and the volume of the solution (2.00 L), we can calculate the moles of Pb²+:
(1.25 mol/L) \(×\) (2.00 L) = 2.50 moles of Pb²+
Since the stoichiometry of the reaction is 1 mole of Pb²+ to 1 mole of SO4²-, we also need 2.50 moles of sulfate ions. The molar mass of potassium sulfate (K2SO4) is approximately 174.26 g/mol. Thus, the mass of K2SO4 needed is:
(2.50 moles) \(×\) (174.26 g/mol) = 435.65 grams of K2SO4
To ensure there is an excess of sulfate ions, slightly more than 435.65 grams would need to be mixed with the lead(II) nitrate solution.