I got a and b correct but need help solving c and d. please give detailed. A sol

Question

I got a and b correct but need help solving c and d. please give detailed.

A solution containing a mixture of 0.027 M potassium chromate (K_2CrO_4) and 0.077 M sodium oxalate MaPg8b (Na_2C_2C_4) was titrated with a solution of barium chloride (BaCl_2) for the purpose of separating CrC_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 (Ksp) for BaCrO_4 and BaC_2O_4 are 2.1 Times 10^-10 and 1.3 Tiems10^-6, respectively. Which barium salt will precipitate first? What concentration of Ba^2+ must be present for BaCrO_4 to begin precipitating? What concentration of Ba^2+ is 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 0.0010 M?

Explanation / Answer

c) we know the final required concentration of [C2O4^2-] = 10% of 0.077 M = 0.0077 M

So using the Ksp equation,

Ksp = [Ba2+][C2O4^2-]

1.3 x 10^-6 = [Ba2+][0.0077]

[Ba2+] = 0.00017 M = 1.7 x 10^-4 M

So we would need 0.017 M of Ba2+ to reduce the concentration of oxalate to 10% its original value.

d) Calculate each separately with [Ba2+] = 0.0010 M

assuming all Ba2+ is used for one precipitation reaction,

For BaCrO4

Ksp = 2.1 x 10^-10 = [0.0010][CrO4^2-]

[CrO4^2-] = 2.1 x 10^-7 M

For BaC2O4

Ksp = 1.3 x 10^-6 = [0.0010][C2O4^2-]

[C2O4^2-] = 1.3 x 10^-3 M

So the ratio of [C2O4^2-]/[CrO4^2-] = 1.3 x 10^-3/2.1 x 10^-7 = 6190.5

So, the concentration of oxalate would be much higher than chromate in solution. This is true considering the precipitate of oxalate is more soluble so more ions will be present in solution.

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