Question

A solution containing a mixture of 0.0444 M potassium chromate (K2CrO4) and 0.0664 M sodium oxalate (Na2C2O4) was titrated with a solution of barium chloride (BaCl2) for the purpose of separating CrO42– and C2O42– by precipitation with the Ba2 cation. Answer the following questions regarding this system. The solubility product constants (Ksp) for BaCrO4 and BaC2O4 are 2.10 × 10-10 and 1.30 × 10-6, respectively.

A.) Which barium salt will precipitate first?

B.) What concentration of Ba2+ must be present for BaCrO4 to begin precipiating?

Answer 1

A. BaCrO₄ will precipitate first as Ksp of BaCrO₄ is lower than Ksp of Na2C2O4.

B. Ksp = [Ba2+][CrO₄2-] = [Ba2+]*0.0444 M = 2.1*10^-10

or[Ba2+] = 4.73*10^-9 M

C. [ox2-]= 0.0664 M

10% [ox2-] = 0.00664 M

concentration of Ba2+ is required to reduce oxalate to 10% its original concentration

= Ksp / [ox2-]

= 1.30 × 10-6/0.00664 = 1.96*10^-4 M

D. When [Ba2+] = 0.006 M

[ox2-] = Ksp /[ Ba2+] = 1.30 × 10-6/0.006 = 2.17*10^-4 M

[CrO₄2-]= Ksp /[ Ba2+] = 2.1*10^­10/0.006 = 3.5*10^-8 M

[ox2-]/[CrO₄2-]= 6190