Question

Suppose 6.74g of barium acetate is dissolved in 150.mL of a 0.20 M aqueous solution of ammonium sulfate. Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the barium acetate is dissolved in it. Round your answer to 3 significant digits.

Answer 1

Barium acetate with Ammonium Sulfate will give Barium Sulfate as precipitate. But the moles of Acetate in the solution will be equal to the initial moles put in the solution. This is because neither Ammonium Acetate nor Barium Acetate is water insoluble. Barium Acetate formula is (CH₃COO)₂Ba.

Mass of Barium Acetate (CH₃COO)₂Ba = 255.43 g/mol

Moles of Barium Acetate = Mass/ Molar Mass = (6.74 g)/(255.43 g/mol) = 0.026387 mol

1 mole of Barium acetate contains 2 moles of acetate ions, which can be seen from the formula (CH₃COO)₂Ba

Moles of Acetate Ion = 2* 0.026387 mol = 0.052774 mol

Volume of Solution = 150.0 mL = 150.0 mL * (0.001 L/mL) = 0.1500 L

Molarity of acetate = Moles / Volume = 0.052774 mol / 0.150 L = 0.35182 mol/L

Thus, the answer rounded up to 3 sig fig would be 0.352 mol/L