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.

Question

Chemistry

Cayden

1 August, 16:26

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.

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Dario Wang · Answer 1

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.