Question

A 0.4000 g sample that contained 96.4% Na2SO4 (142.04 g/mol) required 41.25 mL of barium chloride solution to reach the endpoint in a precipitation titration. What is the molarity of the BaCl2 solution?

Answer 1

Answer:

0.0658 M

Explanation:

The reaction that takes place is:

Na₂SO₄(aq) + BaCl₂(aq) ↔ BaSO₄(s) + 2NaCl(aq)

First we calculate the number of Na₂SO₄ moles:

• 0.4000 g sample * 96.4/100 = 0.3856 g Na₂SO₄

• 0.3856 g Na₂SO₄ ÷ 142.04 g/mol = 2.715x10⁻³ mol Na₂SO₄

Now we convert to moles of BaCl₂:

• 2.715x10⁻³ mol Na₂SO₄ * 1 molBaCl₂/1 molNa₂SO₄ = 2.715x10⁻³ mol BaCl₂

Finally we divide by the volume to calculate the molarity:

• 41.25 mL ⇒ 41.25 / 1000 =  0.04125 L

• 2.715x10⁻³ mol BaCl₂ / 0.04125 L = 0.0658 M