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Calculate the final concentrations of Ba2+ and C2O42- in a solution prepared by mixing 0.100 L of 6.0x10-2 M K2C2O4 and 0.150 L of 1.0x10-2 BaBr2. Ksp (BaC2O4 = 2.3x10-8 M2)

Answer 1

concentration of C2O42- = 0.1 x 6.0 x 10^-2 / 0.1 + 0.150                     

                                         = 0.024 M

concentration of Ba+2 = 0.150 x 1 x 10^-2 / 0.1 + 0.150

                                    = 6.0 x 10^-3 M

C2O42-   +   Ba+2     ----------------> BaC2O4

   1                   1                                   1

0.024          0.006                

here limiting reagent is Ba+2.

concentration of [C2O42-] = 0.024 - 0.006

concentration of [C2O42-] = 0.018 M

BaC2O4   -----------------> Ba+2   + C2O42-

Ksp = [Ba+2][C2O42-]

2.3 x 10^-8 = [Ba+2] x 0.018

[Ba+2] = 1.3 x 10^-6 M