Will a precipitate form? show calculations

Question

A 0.100 L solution of Ba(NO₃)₂ has a barium concentration of 0.037 mol/L. If you add 27.0 mL of 0.001 M sodium oxalate (Na₂C₂O₄) to the solution, will a precipitate form? Show your calculations. K_sp for BaC₂O₄ is 1.6 x 10^-6.

Answer 1

We should calculate the value of Q and then compare that value to the value of Ksp. If Q > Ksp, a precipitate will form. If Q < Ksp, a precipitate will not form.

Ba(NO₃)₃ (aq) + Na₂C₂O₄ (aq) ==> 2NaNO₃ (aq) + BaC₂O₄(s)

0.100 L x 0.037 mol/L = 0.0037 moles Ba²⁺

0.027 L x 0.001 mol/L = 2.7x10^-4 moles C₂O₄^2-

Final volume = 0.100 L + 0.027 L = 0.127 L

[Ba²⁺] = 0.0037 mol/0.127 L = 0.029 M

[C₂O₄^2-] = 2.7x10^-4 mol/0.127 L = 2.13x10^-3 M

Q = [Ba²⁺][C₂O₄^2-] = (0.029)(2.13x10^-4)

Q = 6.2x10^-6

Comparing this value to Ksp, we see that Q > Ksp and therefore a precipitate will form.