Final answer:
To obtain 0.750 g of PbSO₄, the student will need 19.792 mL of a 0.125 M Na₂SO₄ solution. This is determined by calculating the moles of PbSO₄ produced and using stoichiometry to find the corresponding volume of Na₂SO₄ needed.
Step-by-step explanation:
The student is asking how many milliliters of a 0.125 M Na₂SO₄ solution are needed to react with excess Pb(NO₃)₂ to obtain 0.750 g of PbSO₄.
To determine the volume of Na₂SO₄ solution needed, we first use the molar mass of PbSO₄ to find the number of moles of PbSO₄ this mass corresponds to (gravimetric analysis).
Molar mass of PbSO₄ = 303.3 g/mol
0.750 g PbSO₄ × (1 mol/303.3 g) = 0.002474 mol PbSO₄
Next, we look at the stoichiometry of the reaction:
Pb(NO₃)₂(aq) + Na₂SO₄(aq) → PbSO₄(s) + 2NaNO₃(aq)
From the reaction, 1 mole of Na₂SO₄ yields 1 mole of PbSO₄. Thus, 0.002474 mol of PbSO₄ would require 0.002474 mol of Na₂SO₄.
Now, we can calculate the volume of 0.125 M Na₂SO₄:
Volume (L) = moles / concentration
Volume (L) = 0.002474 mol / 0.125 M = 0.019792
To convert to milliliters:
0.019792 L × 1000 mL/L = 19.792 mL
So, 19.792 mL of 0.125 M Na₂SO₄ solution are needed.