Chromium (III) iodate has a Ksp at 25°C of 5.0 x 10–6.
125 mL of a solution contains 0.0050 mol L–1 Cr (III) ion.
What is the molar solubility of chromium (III) iodate at 25°C?
a. 8.4 x 10–2 mol L–1
b. 2.2 x 10–3 mol L–1
c. 6.3 x 10–2 mol L–1
d. 2.1 x 10–2 mol L–1
I know how to find molar solubility but don't know what to do with the extra given information.
125 mL of a solution contains 0.0050 mol L–1 Cr (III) ion.
The extra info could be a common ion relative to the Chromium(III)Iodate, but plugging 0.005M Cr^+3 into the equilibrium does not give a solubility that matches any of the answer choices. The only answer consistent with some of the data is solubility of Cr(III)Iodate in pure water...
-----Cr(IO3)3 <=> Cr^+3 + 3IO3^-
C(eq) ----------- x ----- 3x
Ksp = [Cr^+3][IO3^-]^3 = (x)(3x)^3
= 27x^4
x = Solubility = 4th-Root(Ksp/27)
= 4th-Root(5x10^-6/27)
= 2.1 x 10^-2M
However, if 0.005M Cr^+3 is taken (assumed) to be a common ion and inserted into the Equilibrium Concentration, the set-up gives ...
-----Cr(IO3)3 <=> Cr^+3 + 3IO3^-
C(eq)------------0.005M --- 3x^3
Ksp = [Cr^+3][IO3^-]^3 =(0.005)(3x)^3
= (0.005)(27x^3) = 0.135x^3
Solving for x = Solubility of Cr(III)Iodate in the presence of 0.005M Cr^+3
=> x = Cube Root(Ksp/0.135) = Cube Root(5x10^-6/0.135) = 0.033M which does not match any of the answer choices.
The only other thing that can be done with the given data is calculate the moles Cr^+3 from the 125-ml of 0.005M Cr(III) ...
Moles(Cr^+3) = Molarity x Volume in Liters = (0.005M)(0.125L) = 6.25 x 10^-4 moles. However, this makes no contribution to answering the question of solubility. It is only a statement of fact that 6.25 x 10^-4 moles Cr^+3 in 125-ml soln is 0.005M Cr^+3. I'd suggest going back and reviewing the source of the original problem.
I'm going to butt in here. There isn't anything wrong with the problem. Jess, I told you how to do this yesterday. I also told you that the 125 mL of 0.005M Cr^3+ was a red herring. Ignore that extra information. You work this as a saturated solution of Cr(IO3)3 and since you know how to work solubility problems (so you said in your earlier post) you will get the right answer. The correct answer is d.
Forgot to mention that if you try to put that answer back into the Ksp equation (as DrRebel points out)you can't get Ksp of 5E-6 which is the fault of the author of the problem. S/he didn't put in the right number for the red herring.
To find the molar solubility of chromium (III) iodate at 25°C, we can use the information given about the solution containing Cr (III) ion.
The molar solubility (S) of a compound is the maximum amount of the compound that can dissolve in a given volume of solvent at a specific temperature. In this case, we want to find the molar solubility of chromium (III) iodate.
The given information tells us that 125 mL of the solution contains 0.0050 mol L–1 of Cr (III) ion. This means that there are 0.0050 moles of Cr (III) ion in 1 liter (or 1000 mL) of the solution.
To find the molar solubility of chromium (III) iodate, we can set up an equation using the balanced chemical equation and the solubility product constant (Ksp) for chromium (III) iodate.
The balanced chemical equation for the dissolution of chromium (III) iodate is:
Cr(IO3)3(s) ⇌ Cr3+(aq) + 3IO3-(aq)
The Ksp expression for chromium (III) iodate is:
Ksp = [Cr3+][IO3-]^3
Since the stoichiometric ratio between Cr3+ and IO3- is 1:3, the molar solubility of chromium (III) iodate can be expressed as 3S.
Using the Ksp expression, we can substitute the molar solubility (S) and solve for the concentration of Cr3+ and IO3-.
Ksp = [Cr3+][IO3-]^3
Ksp = (S)(3S)^3
Ksp = 27S^4
We are given the value of Ksp, which is 5.0 x 10–6.
5.0 x 10–6 = 27S^4
To solve for S, we need to isolate it. Taking the fourth root on both sides:
(Solving using a calculator or software)
S ≈ 2.18 x 10–2 mol L–1
Since the question asks for the molar solubility of chromium (III) iodate, we take the molar solubility (S) calculated above. The correct answer is:
b. 2.2 x 10–3 mol L–1
So the molar solubility of chromium (III) iodate at 25°C is approximately 2.2 x 10–3 mol L–1.