Question

a) An impure barium chloride sample weighing about 0.38 g was dissolved in water and then treated with excess sulfuric acid. Calculate the volume of 3.0 M H₂SO₄ needed to completely precipitate the barium ions as barium sulfate, BaSO₄. (Give your answer to 2 significant figures in mL)

b)The impure barium chloride sample from Part (a) was found to have a mass of 0.3821 g. The mass of barium sulfate, BaSO₄, that precipitated from the addition of sulfuric acid was 0.1582 g. What is the percentage barium in the unknown sample? (Please give your answer to 4 significant figures)

Answer 1

a) Write the balanced chemical equation for the reaction.

BaCl₂ + H₂SO₄ --------> BaSO₄ + 2 HCl

As per the stoichiometric equation,

1 mole BaCl₂ = 1 mole H₂SO₄

Molar mass of BaCl₂ = (1*137.327 + 2*35.453) g/mol = 208.233 g/mol.

Mole(s) of BaCl₂ taken = (0.38 g)/(208.233 g/mol) = 0.0018249 mole = (0.0018249 mole)*(1000 mmole/1 mole) = 1.8249 mmole.

Mole(s) of H₂SO₄ required = 1.8429 mmole.

Volume of H₂SO₄ required = (mmole of H₂SO₄)/(molarity of H₂SO₄) = (1.8429 mmole)/(3.0 M) = 0.6083 mL ≈ 0.61 mL (ans).

b) Molar mass of BaSO₄ = (1*137.327 + 1*32.065 + 4*15.9994) g/mol = 233.3896 g/mol.

The actual mass of BaSO₄ obtained = 0.1582 g; therefore, moles of BaSO₄ obtained = (0.1582 g)/(233.3896 g/mol) = 0.0006778 mole

We know that 1 mole BaSO₄ = 1 mole Ba; therefore, moles of Ba contained in 0.1582 g BaSO₄ = 0.0006778 mole.

Molar mass of Ba = 137.327 g/mol; therefore, mass of Ba contained in 0.1582 g BaSO₄ = (0.0006778 mole)*(137.327 g/mol) = 0.09308 g.

Ba was supplied by BaCl₂; hence, the impure sample must have contained 0.09308 g Ba.

Therefore, percentage Ba in the impure sample = (0.09308 g)/(0.3821 g)*100 = 24.36% (ans).