Calculate the final concentrations of K+(aq), C2O4^2-(aq), Ba^2+, and Br^-(aq)in a solution prepared by adding 0.100 L of 0.200M K2C2O4 to 0.150 L of 0.250M Babr2. Ksp value is 2.3 x 10^-8

Answer 1

Will a precipitate even form? We can use Q and compare it to Ksp to answer this question.

Q = [Ba²⁺][C₂O₄^2-]

[Ba²⁺] = 0.150 L x 0.250 mol/L = 0.0375 mol / 0.250 L = 0.15 M

[ C₂O₄^2-] = 0.100 L x 0.200 mol/L = 0.0200 mol / 0.250 L = 0.08 M

Q = (0.15)(0.18) = 0.012

Compared to Ksp of 2.3x10^-8, a precipitate will form.

K₂C₂O₄(aq) + BaBr₂(aq) ==> 2KBr(aq) + BaC₂O₄(s) .. balanced equation

Find the limiting reactant. One way to do this is to simply divide moles of each reactant by the corresponding coefficient in the balanced equation. Which ever result is less represents the limiting reactant,

For K₂C₂O₄: 0.100 L x 0.200 mol/L = 0.0200 mols (÷1->0.0200)

For BaBr₂: 0.150 L x 0.250 mol/L = 0.0375 mols (÷1->0.0375)

Since 0.02 is less than 0.0375, K₂C₂O₄ is the limiting reactant and all the C₂O₄^2- will be used up and will be in the precipitate (BaC₂O₄). Thus, the only ions actually removed from solution are Ba²⁺ and C₂O₄^2-. All others are spectators and remain constant.

K₂C₂O₄(aq) + BaBr₂(aq) ==> 2KBr(aq) + BaC₂O₄(s)

0.0200...........0.0375..................0.................0................Initial

-0.0200..........-0.0200................+0.0400...+0.0200........Change

0.....................0.0175................0.0400........0.0200........Equilibrium

This shows no K₂C₂O₄ remaining in solution, and only BaBr₂ and KBr remaining in solution. So, we now need to find concentrations of Ba²⁺, K⁺ and Br^-.

BaBr₂ => Ba²⁺ + 2Br^-

moles Ba²⁺ = 0.0175 Ba²⁺

moles Br^- = 2 x 0.0175 = 0.035 mols Br^-

KBr ==> K⁺ + Br^-

moles K⁺ = 0.0400 mols K⁺

moles Br^- = 0.0400 mols Br^-

Summarizing total moles in solution, we have...

0.0175 mols Ba²⁺

0.0400 mols K⁺

0.0750 mols Br^-

Total final volume of solution = 0.100 L + 0.150 L = 0250 L

Final concentration of all ions in solution is...

[Ba²⁺] = 0.0175 mols / 0.250 L = 0.070 M

[K⁺] = 0.0400 mols / 0.250 L = 0.160 M

[Br^-] = 0.0750 mols / 0.250 L = 0.300 M

[C₂O₄^2-] = 3.29x10^-7 M (see calculation below)

Ksp = [Ba²⁺][C₂O₄^2-]

2.3x10^-8 = [0.070][C₂O₄^2-]

[C₂O₄^2-] = 3.29x10^-7 M