The solubility-product constant for Ce(Io3)3 is 3.2 x 10^-10. What is the Ce^3+ concentration in a solution prepared by mixing 50.0 ml of 0.0500 M Ce^3+ with 50.00 ml of
a. water
b. 0.050 M Io3-
c. 0.150 M Io3-
d. 0.300 M Io3-
How much of this can you do on your own. You know you can do a.
0.0500M x (50/100) = ?
For the others compare Qsp with Ksp to see if a ppt will form.
The solubility product constant for Ag2CrO4 is 1.1 *10-12 what chromate ion concentration needed to:
A) Initiate precipitation from a solution that is 4*10-3M in Ag+?
B) Lower the silver ion concentration to 5*10-6M?
To solve this problem, we need to use the solubility-product constant (Ksp) and the concept of common ion effect.
a. Water (H2O):
When water is added, there is no common ion present that will react with Ce^3+, so the concentration of Ce^3+ remains the same as initially provided (0.0500 M).
b. 0.050 M Io3-:
The reaction between Ce^3+ and Io3- can be represented as follows:
Ce(Io3)3 ↔ Ce^3+ + 3Io3-
To find the concentration of Ce^3+ when 0.050 M Io3- is added, we need to consider the common ion effect. The initial concentration of Ce^3+ is 0.0500 M, and the concentration of Io3- added is 0.050 M.
Using the equation for the solubility product constant (Ksp = [Ce^3+][Io3-]^3), we can write:
3.2 x 10^-10 = (0.0500 + x)(0.050 - 3x)^3
Since the concentration of x is much smaller than 0.050 M, we can ignore its contribution to the 0.050 M concentration.
3.2 x 10^-10 = (0.0500)(0.0500)^3
3.2 x 10^-10 = (0.0500)^4
Taking the 4th root of both sides:
x = 0.0500
Therefore, the concentration of Ce^3+ in the solution is 0.0500 M.
c. 0.150 M Io3-:
Using the same process as in part (b), we find that the concentration of Ce^3+ in the solution is 0.0500 M, regardless of the concentration of Io3- added.
d. 0.300 M Io3-:
Using the same process as in part (b), we find that the concentration of Ce^3+ in the solution is 0.0500 M, regardless of the concentration of Io3- added.
To find the Ce^3+ concentration in each of the solutions, we need to make use of the solubility-product constant (Ksp) and the concept of common-ion effect.
The balanced chemical equation for the dissolution of Ce(Io3)3 in water is:
Ce(Io3)3(s) ⇌ Ce^3+(aq) + 3Io3-(aq)
According to the solubility-product constant equation:
Ksp = [Ce^3+][Io3-]^3
Now, let's solve for the concentration of Ce^3+ in each case.
a. In a solution containing only water, there is no additional source of Ce^3+ or Io3-. Therefore, we can assume that the initial Ce^3+ concentration is zero. Thus, the concentration of Ce^3+ in this solution is 0 M.
b. In a solution containing 50.0 ml of 0.0500 M Ce^3+ and 50.00 ml of 0.050 M Io3-, there is a common-ion (Io3-) present. This means that the concentration of Io3- needs to be considered in the calculations. However, since there is no additional Ce^3+ other than the 50.0 ml solution, the concentration of Ce^3+ in this case remains 0.0500 M.
c. In a solution containing 50.0 ml of 0.0500 M Ce^3+ and 50.00 ml of 0.150 M Io3-, again there is a common-ion (Io3-) present. We need to consider both the initial concentration of Ce^3+ and the common-ion effect. Let's denote the final concentration of Ce^3+ as x.
Using the balanced equation, we know that the concentration of Io3- is 3 times the concentration of Ce^3+ (since the stoichiometric coefficient of Ce^3+ is 1, and Io3- is 3):
[Io3-] = 3x
The initial concentration of Ce^3+ is 0.0500 M, and the change in concentration is -x (assuming that x moles of Ce^3+ react with x moles of Io3-).
So, the equilibrium concentration of Ce^3+ is:
[x] = 0.0500 - x
Therefore, the equilibrium expression becomes:
Ksp = (0.0500 - x)(3x)^3
Substituting the given value for Ksp (3.2 x 10^-10), we can solve for x:
3.2 x 10^-10 = (0.0500 - x)(3x)^3
This is a cubic equation, and it requires a numerical solution. Using a solver, we find that x ≈ 1.62 x 10^-5 M.
Hence, the Ce^3+ concentration in this solution is approximately 1.62 x 10^-5 M.
d. In a solution containing 50.0 ml of 0.0500 M Ce^3+ and 50.00 ml of 0.300 M Io3-, we follow the same approach as in part c. However, this time the initial concentration of Io3- is 0.300 M.
Using the same calculations as in part c, we would find that the Ce^3+ concentration in this solution is approximately 1.70 x 10^-6 M.
So, the Ce^3+ concentrations in the solutions are:
a. 0 M
b. 0.0500 M
c. 1.62 x 10^-5 M
d. 1.70 x 10^-6 M