Chemistry equilibrium

Question

20.00 mL of a 0.25 M solution of BaCl2(aq) was added to an equal volume of 0.35 M solution of Na2SO4 (aq). Show a precipitate forms then calculate the concentration of free Ba^ 2+ (aq) when the system is at equilibrium.

Answer 1

Write a correctly balanced equation:

BaCl₂(aq) + Na₂SO₄(aq) ==> BaSO₄(s) + 2NaCl(aq) ... balanced equation

mols BaCl₂ present = 20.00 ml x 1 L / 1000 ml x 0.25 mol/L = 0.00500 mols BaCl₂

mols Na₂SO₄ present = 20.00 ml x 1 L / 1000 ml x 0.35 mol/L = 0.00700 mols Na₂SO₄

Since 1 mol Ba²⁺ reacts with 1 mol SO₄^2- to form BaSO₄, there will be no free Ba²⁺ when the system is at equilibrium unless you consider that from the precipitate and for that we'd need the Ksp for BaSO₄

That being the case, we'd proceed as follows:

0.00500 mols BaCl₂ = 0.00500 mols BaSO₄

Final volume = 40.00 ml = 0.0400 L

Excess SO₄^2- = 0.00200 mols

[SO₄^2-] = 0.00200 mol / 0.040 L = 0.05 M

Ksp BaSO₄ = 1.5x10^-9

BaSO₄(s) ==> Ba²⁺ + SO₄^2-

Ksp = 1.5x10^-9 = [Ba²⁺][SO₄^2-] = (x)(0.05)

x = 1.5x10^-9 / 0.05

x = [Ba²⁺] = 3.0x10^-8 M