A solution containing a mixture of 0.0492 M potassium chromate (K2CrO4) and 0.0565 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 0.0010 M?
A) Which will precipitate first? BaCrO4 or BaC2O4
Ksp BaCrO4 = (Ba^2+)(CrO4^2-) = 2.10E-10
The BaCl2 is being added drop by drop. What must (Ba^2+) = for the first ppt of BaCrO4 to appear. That will be
(Ba^2+) = Ksp/(CrO4) = 2.10E-10/0.0492 = about 4.3E-9 M. You need to redo ALL of the calculations in this problem. My answers are just close estimates.
Do the same thing for BaC2O4. I get (Ba^+) = 2.3E-5 M. You should confirm all of these numbers. Sometimes I punch in the wrong numbers on the calculator.
Obviously the BaCrO4 will ppt first because 4.3E-9 will be reached first.
D) What is the ratio of oxalate to chromate ([C2O42–]/[CrO42–]) when the Ba2 concentration is 0.0010 M?
Use Ksp for BaCrO4, plug in 0.001 for (Ba^2+) and calculate (CrO4^-).
Do the same with Ksp for BaC2O4 and calculate C2O4^2-
Then take the ratio
Remember to confirm all of this with your calculations and your thoughts.
B) What concentration of Ba2 must be present for BaCrO4 to begin precipitating?
See my work on the first response. All of the calculations are there.
C) What concentration of Ba2 is required to reduce oxalate to 10% of its original concentration?
Initial (C2O4^2-) = 0.0565 M so 0.1 of that is 0.00565 so
(Ba^2+) = 1.3E-6/0.00565 = 0.0002 but check that.
A) To determine which compound will precipitate first, we need to compare their solubility product constants (Ksp). The compound with the lower Ksp value will be less soluble and thus precipitate first.
In this case, the Ksp for BaCrO4 (2.10 × 10^(-10)) is significantly lower than the Ksp for BaC2O4 (1.30 × 10^(-6)). Therefore, BaCrO4 will precipitate first.
B) To calculate the concentration of Ba2+ required for BaCrO4 to begin precipitating, we need to set up an equilibrium expression for the solubility of BaCrO4:
BaCrO4 ⇌ Ba2+ + CrO42-
The solubility product constant expression for BaCrO4 is:
Ksp = [Ba2+][CrO42-]
Given that the Ksp for BaCrO4 is 2.10 × 10^(-10) and assuming x is the concentration of Ba2+ when precipitation begins, we can write the following equation:
2.10 × 10^(-10) = x * x
Solving for x:
x = √(2.10 × 10^(-10))
x ≈ 4.59 × 10^(-6) M
Therefore, the concentration of Ba2+ required for BaCrO4 to begin precipitating is approximately 4.59 × 10^(-6) M.
C) To determine the concentration of Ba2+ required to reduce oxalate to 10% of its original concentration, we need to consider the reaction between Ba2+ and C2O42-:
Ba2+ + C2O42- ⇌ BaC2O4
We can use the concept of common ion effect to find the concentration of Ba2+ required.
Let's assume the initial concentration of C2O42- is [C2O42-]0. The change in concentration of C2O42- due to the reaction is [C2O42-]0 - [C2O42-].
According to the reaction stoichiometry, one mole of Ba2+ reacts with one mole of C2O42-. So, the concentration of Ba2+ formed is [C2O42-]0 - [C2O42-].
Given that we want to reduce the concentration of C2O42- to 10% of its original value, the concentration of C2O42- remaining ([C2O42-]) will be 0.10 * [C2O42-]0.
Since one mole of Ba2+ reacts with one mole of C2O42-, the concentration of Ba2+ required is also 0.10 * [C2O42-]0.
D) To calculate the ratio of oxalate to chromate ([C2O42-]/[CrO42-]) when the Ba2+ concentration is 0.0010 M, we need to consider the formation of BaC2O4 and BaCrO4.
The reaction is given by:
Ba2+ + C2O42- ⇌ BaC2O4
Ba2+ + CrO42- ⇌ BaCrO4
Let's assume the initial concentration of C2O42- is [C2O42-]0 and the initial concentration of CrO42- is [CrO42-]0.
Using the concept of common ion effect, we know that the concentration of C2O42- will decrease due to the formation of BaC2O4. Let's assume the decrease in C2O42- concentration is x.
Similarly, the concentration of CrO42- will decrease due to the formation of BaCrO4. Let's assume the decrease in CrO42- concentration is y.
Since one mole of Ba2+ reacts with one mole of C2O42- to form BaC2O4, and one mole of Ba2+ reacts with one mole of CrO42- to form BaCrO4, the stoichiometry tells us that x = y.
We are given that the concentration of Ba2+ (formed due to the reaction with C2O42- and CrO42-) is 0.0010 M.
Therefore, we have the following equations:
[C2O42-]0 - x = 0.0010 M (concentration of C2O42- remaining)
[CrO42-]0 - y = 0.0010 M (concentration of CrO42- remaining)
x = y
From the above equations, we can solve for x and y:
[C2O42-]0 - x = [CrO42-]0 - y
x = y
Substituting the values, we get:
[C2O42-]0 - x = 0.0010 M
[CrO42-]0 - y = 0.0010 M
x = y
Solving the above system of equations, we find:
[C2O42-]0 = [CrO42-]0 = 0.0015 M
Therefore, the ratio of oxalate to chromate ([C2O42-]/[CrO42-]) when the Ba2+ concentration is 0.0010 M is 1:1.