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.

Question

Suppose 

1.15g

 of barium acetate is dissolved in 350.mL

 of a 62.0mM

 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. 

Answer 1

Ba(CH₃COO)₂(aq) + (NH₄)₂SO₄(aq) ==> BaSO₄(s) + 2CH₃COONH₄(aq) ... balanced equation

moles Ba(CH₃COO)₂ = 1.15 g x 1 mol / 255 g = 0.00451 mols

moles (NH₄)₂SO₄ = 350 ml x 1 L / 1000 ml x 0.062 mol / L = 0.0217 mols

Since Ba(CH₃COO)₂ is present in limiting supply, the final [BaSO₄] will be...

0.00451 mol Ba(CH₃COO)₂ x 1 mol BaSO₄ / mol Ba(CH₃COO)₂ = 0.00451 mol/0.350L = 0.0129 mol/L

Since this far exceeds the solubility of BaSO₄, it will all precipitate and there will be no measurable Ba²⁺ in solution.