Suppose 0.363 g of barium acetate is dissolved in 100. mL of a 25.0 m M aqueous solution of ammonium sulfate. Calculate the final molarity of barium cation in the solution. You can assume the volume of the solution doesn't change when the barium acetate is dissolved in it. Be sure your answer has the correct number of significant digits.

Molarity of Ba²⁺ is 0M

Explanation:

Some barium ion, Ba²⁺ is produced when barium acetate is dissolved (Moles of barium acetate = moles of Ba²⁺ in solution). Then, Ba²⁺ reacts with sulfate ion to produce BaSO₄, an insoluble salt. The concentration of Ba²⁺ is the initial concentration - the concentration of SO₄²⁻ in solution.

Initial moles of Ba²⁺:

Moles barium acetate (Molar mass: 255.43g/mol)

0.363g * (1mol / 255.43) = 1.421x10⁻³ moles of Ba²⁺

Moles SO₄²⁻ = Moles of ammonium sulfate:

100mL = 0.100L * (0.025mol / L) = 2.5x10⁻³ moles of SO₄²⁻.

As moles of SO₄²⁻ are higher than moles of Ba²⁺, Molarity of Ba²⁺ is 0M because all moles of Ba²⁺ reacts producing BaSO₄(s), an insoluble salt.

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