A solution containing a mixture of 0.0453 M potassium chromate (K_2CrO_4) and 0.

Question

A solution containing a mixture of 0.0453 M potassium chromate (K_2CrO_4) and 0.0541 M sodium oxalate (Na_2C_2O_4) was titrated with a solution of barium chloride (BaCl_2) for the purpose of separating CrO_4^2- and C_2O_4^2- by precipitation with the Ba^2+ cation. Answer the following questions regarding this system. The solubility product constants (K_sp) for BaCrO_4 and BaC_2O_4 are 2.10 times10^-10 and 1.30 times 10^-6, respectively. Which barium salt will precipitate first? BaCrO_4. BaC_2O_4. What concentration of Ba^2+ must be present for BaCrO_4 to begin precipitating? What concentration of Ba^2+ s required to reduce oxalate to 10% of its original concentration? What is the ratio of oxalate to chromate ([C_2O_4^2-]/[CrO_4^2-]) when the Ba^2+ concentration is 0.0070 M? [C_2O_4^2-]/[CrO_4^2-] =

Explanation / Answer

A) The Ksp for BaCrO4 is lower and thus, it will precipitate first

B) Ksp for BaCrO4=[Ba2+]*[CrO42-]=2.1*10-10

But [CrO42-]=0.0453

Thus [Ba2+]=2.1*10-10/0.0453=4.636*10-9 M

C)If [C2042-]=0.1*0.0541=0.00541 M

Ksp=[Ba2+]*[C2O42-]=1.3*10-6

Thus, [Ba2+]=2.403*10-4 M

D) [Ba2+]=0.07

Thus,[C2O42-]=1.3*10-6/0.07=18.57*10-6M

[CrO42-]=2.1*10-10/0.07=30*10-10

[C2O42-]/[CrO42-]=6190

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