Question

A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate is filtered and dried and found to have a mass of 258 mg. What mass of barium was in the original solution assume that all of the barium was precipitated out of the solution you the reaction.

Answer 1

Answer:

• The mass of barium in the original solution was 152mg.

Explanation:

The white precipiate that is formed when sodium sulfate, Na₂SO₄, reacts with barium ions in solution, Ba²⁺, is barium sulfate, BaSO₄.

The molar mass of BaSO₄ is 233.38 g/mol. Thus, the number of moles of BaSO₄ is:

• #moles = mass in grams / molar mass
• #moles = 258mg × (1g/1,000mg) × (1mol/233.38g) = 0.00110549mol

Since 1 mole of BaSO₄ contains 1 mole of Ba, there are 0.00110549 mol of Ba²⁺ ions.

Use the atomic mass of Ba to calculate the mass in grams in the original solution:

• mass = number of moles × atomic mass
• mass = 0.00110549mol × 137.327g/mol = 0.1518g.
• mass = 151.8mg = 152mg

Then, with 3 significant figures, there were 152mg of barium in the original solution.