Precipitation General Chemistry Question

Question

Is a precipitate expected to form when 50.0 mL of 0.0010 M BaCl₂ is added to 50.00 mL of 0.00010 M Na₂SO₄? K_sp for barium sulfate is 1.1 x 10^-10.

Answer 1

When asked if a precipitate will form or not, you want to calculate a value of Q and compare it to the value for Ksp. If Q > Ksp a precipitate will form. If Q < Ksp, no precipitate will form

The reaction taking place is:

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

moles BaCl₂ = 50.0 ml x 1 L/1000 ml x 0.0010 mol/L = 5x10^-5 moles

moles Na₂SO₄ = 50.0 ml x 1 L/1000 ml x 0.00010 mol/L = 5x10-⁶ moles <--Limiting reactant

moles BaSO₄ that can form = 5x10^-6 moles BaSO₄

Final volume = 50.0 ml + 50.0 ml = 100 ml x 1 L/1000 ml = 0.1 L

Final [BaSO₄] = 5x10^-6 moles/0.1 L = 5x10^-5 M

BaSO₄(s) ==> Ba²⁺(aq) + SO₄^2-(aq)

Q = [Ba²⁺][SO₄^2-] = (5x10^-5)(5x10^-5)

Q = 2.5x10^-9 Compare this to the Ksp for BaSO₄ (1.1x10^-10)

Q is > than Ksp. A precipitate WILL form