A solution containing a mixture of 0.0284 M potassium chromate (K2CrO4) and 0.07

Question

A solution containing a mixture of 0.0284 M potassium chromate (K2CrO4) and 0.0706 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 precipitating?

C) What concentration of Ba2 is required to reduce oxalate to 10% of its original concentration?

D) What is the ratio of oxalate to chromate ([C2O42–]/[CrO42–]) when the Ba2 concentration is 0.0030 M?

Explanation / Answer

A) Barium chromate will precipitate first since it solubility product is very low.

B) concentration of Cro42- ion = 0.0284

so [Ba 2+ ][ Cro42- ] should exceed 2.10 × 10-10 for precipitation to occur.

so Ba 2+ concentration must be --- 2.10 × 10-10 / 0.0284   = 73.94 × 10-10 M

C) 10% of original concetration of oxalate , 0.0706 = 0.00706 M

so oxalate to be consumed = 0.0706 - 0.00706 = 0.06354 M

now Barium ion required = Ksp / oxalate to be consumed =1.30 × 10-6 / 0.06354 = 20.54x 10-6  

D) The ratio of[C2O42–]/[CrO42–] when barium ion concentration is 0.0030 M is ratio of Ksp values, given.

   1.30 × 10-6 /   2.10 × 10-10

= 0.619 X 104

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