A solution containing a mixture of 0.0381 M potassium chromate (K2CrO4) and 0.0769 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 will precipitate first? BaCrO4 or BaC2O4
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 0.0050 M?
A) To determine which compound will precipitate first, we compare the solubility product constants (Ksp) of BaCrO4 and BaC2O4. The compound with the smaller Ksp will precipitate first because it has a lower solubility.
In this case, BaCrO4 has a smaller Ksp (2.10 × 10^-10) compared to BaC2O4 (1.30 × 10^-6). Therefore, BaCrO4 will precipitate first.
B) To determine the concentration of Ba2+ required for BaCrO4 to begin precipitating, we need to find the concentration at which the product of [Ba2+] and [CrO42-] reaches the Ksp of BaCrO4.
The balanced equation for the precipitation reaction is:
Ba2+ + CrO42- → BaCrO4
The stoichiometry of the balanced equation tells us that the concentration of Ba2+ is equal to the concentration of CrO42-. Let's denote this concentration as x.
The expression for the solubility product (Ksp) is:
Ksp = [Ba2+][CrO42-]
Substituting the values, we get:
2.10 × 10^-10 = x * x
Solving for x, we find:
x = √(2.10 × 10^-10) = 1.45 × 10^-5 M
Therefore, the concentration of Ba2+ required for BaCrO4 to begin precipitating is 1.45 × 10^-5 M.
C) To determine the concentration of Ba2+ required to reduce oxalate to 10% of its original concentration, we need to calculate the new concentration of oxalate ([C2O42-]).
Let's assume the original concentration of oxalate is y. After precipitation, the concentration will decrease by a factor of 0.1 (10% reduction), resulting in a new concentration of 0.1y.
Since the concentration of Ba2+ is equal to the concentration of C2O42-, we can set up the following equation:
[Ba2+][C2O42-] = Ksp (for BaC2O4)
Substituting the values, we have:
(0.1y)(y) = 1.30 × 10^-6
0.1y^2 = 1.30 × 10^-6
Solving for y, we find:
y = √(1.30 × 10^-6 / 0.1) = 1.14 × 10^-3 M
Therefore, the concentration of Ba2+ required to reduce oxalate to 10% of its original concentration is 1.14 × 10^-3 M.
D) To find the ratio of oxalate to chromate ([C2O42-]/[CrO42-]), we need to use the stoichiometry of the balanced equation between Ba2+, C2O42-, and CrO42-:
Ba2+ + C2O42- → BaC2O4
Ba2+ + CrO42- → BaCrO4
Since the concentration of Ba2+ is equal to the concentration of C2O42- and CrO42-, we can denote the concentration of Ba2+ as x.
The ratio [C2O42-]/[CrO42-] is given by:
[C2O42-]/[CrO42-] = x/x = 1/1 = 1
Therefore, when the Ba2+ concentration is 0.0050 M, the ratio of oxalate to chromate is 1.
A) To determine which compound will precipitate first, we need to compare the solubility product constants (Ksp) for BaCrO4 and BaC2O4. The compound with the smaller Ksp value will be less soluble and therefore precipitate first.
Ksp for BaCrO4 = 2.10 × 10^-10
Ksp for BaC2O4 = 1.30 × 10^-6
Since the Ksp value for BaCrO4 is smaller than that of BaC2O4, BaCrO4 will precipitate first.
B) To calculate the concentration of Ba2+ required for BaCrO4 to begin precipitating, we can use the solubility product constant expression:
Ksp = [Ba2+][CrO42-]
Since we know the value of Ksp for BaCrO4 and the concentration of CrO42- (0.0381 M), we can calculate the concentration of Ba2+ required:
2.10 × 10^-10 = [Ba2+][0.0381]
[Ba2+] = 2.10 × 10^-10 / 0.0381
[Ba2+] = 5.50 × 10^-9 M
Therefore, the concentration of Ba2+ required for BaCrO4 to begin precipitating is 5.50 × 10^-9 M.
C) To determine the concentration of Ba2+ required to reduce oxalate to 10% of its original concentration, we need to calculate the final concentration of oxalate (C2O42-) after the precipitation reaction.
Assuming the initial concentration of oxalate is 0.0769 M, to reduce it to 10% of its original concentration:
Final concentration of C2O42- = 0.10 × 0.0769
Final concentration of C2O42- = 0.00769 M
To find the concentration of Ba2+ required for this reduction, we can use the solubility product constant expression for BaC2O4:
Ksp = [Ba2+][C2O42-]
Since we know the value of Ksp for BaC2O4 and the final concentration of C2O42- (0.00769 M), we can calculate the concentration of Ba2+ required:
1.30 × 10^-6 = [Ba2+][0.00769]
[Ba2+] = 1.30 × 10^-6 / 0.00769
[Ba2+] = 1.69 × 10^-4 M
Therefore, the concentration of Ba2+ required to reduce oxalate to 10% of its original concentration is 1.69 × 10^-4 M.
D) To calculate the ratio of oxalate to chromate ([C2O42-]/[CrO42-]) when the Ba2+ concentration is 0.0050 M, we can use the solubility product constant expressions for BaCrO4 and BaC2O4.
For BaCrO4:
Ksp = [Ba2+][CrO42-]
2.10 × 10^-10 = (0.0050)[CrO42-]
[CrO42-] = 2.10 × 10^-10 / 0.0050
[CrO42-] = 4.20 × 10^-8 M
For BaC2O4:
Ksp = [Ba2+][C2O42-]
1.30 × 10^-6 = (0.0050)[C2O42-]
[C2O42-] = 1.30 × 10^-6 / 0.0050
[C2O42-] = 2.60 × 10^-4 M
Therefore, the ratio of oxalate to chromate ([C2O42-]/[CrO42-]) when the Ba2+ concentration is 0.0050 M is:
[C2O42-] / [CrO42-] = (2.60 × 10^-4) / (4.20 × 10^-8)
[C2O42-] / [CrO42-] = 6.19 × 10^3