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.

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:

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:

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

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