Question Detail
Question
Barium ions can be precipitated from aqueous solutions by the addition of aqueous sulfate: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). Barium sulfate is virtually insoluble in water. How many grams of solid Na₂SO₄ must be added to 50.0 mL of 0.200 M Ba²⁺ solution to completely precipitate the barium? [Ba=137, S=32, O=16, Na=23]
Correct Answer
Option A is the correct answer.
Detailed Explanation
Moles Ba²⁺ = 0.0500 L × 0.200 M = 0.0100 mol. Reaction is 1:1, so moles Na₂SO₄ needed = 0.0100 mol. Molar mass Na₂SO₄ = (2×23)+32+(4×16)=142 g/mol. Mass = 0.0100 × 142 = 1.42 g.
Hint
Use the stoichiometry of the precipitation reaction. The mole ratio between Ba²⁺ and SO₄²⁻ is 1:1.