when 25.0 mL of a solution containing both Fe2+ and Fe3+ ions is titrated with 23.0 mL of 0.0200 M KMnO4 (in dilute sulfuric acid). As a result, all of the Fe2+ ions are oxidized to Fe3+ ions. Next, the solution is treated with Zn metal to convert all of the Fe3+ ions to Fe2+ ions. Finally, the solution containing only the Fe2+ ions requires 40.0 mL of the same KMnO4 solution for oxidation to Fe3+. Calculate the molar concentrations of Fe2+ and Fe3+ in the original solution
I had tremendous problems with this question as well, and since I now have the answer, I'll spread the love.
First, you figure out how many moles of KMnO4 is used for both titrations using the equation
---- M=(mol)/(L) ---- (M=Molarity)
Doing this, you would get the first amount of KMnO4 to be
(.02M)(.023L)= .00046 mol Fe2+
and the second to be
(.02M)(.040L)= .00080 mol Fe2+
Then, using stoichiometry* and the provided equation
- 2+ + 2+ 3+
MnO4 +5Fe +8H -->Mn+ 5Fe + 4H20
You get .0024mol Fe2+ to begin with and .0040mol Fe2+ to end with.
To get how many moles of Fe3+, you just subtract .0040-.0023 = .0017mol Fe3+
Molar Concentration is the same as Molarity, so...
----Here's the answer!----
M Fe2+ = .0023mol/.025L = .092M
M Fe3+ = .0017mol/.025L = .068M
*I assume you know how to set up the ratios, but I'll just say that you would use 1mol KMnO4=1mol MnO4 and
1mol MnO4=5mol Fe2+
Oops the charges in the equation got messed up so instead:
MnO4- + 5Fe2+ + 8H+ -> 5Fe3+ + 4H2O
WRONG. You don't even know how to compute for the number of moles of KMnO4 solution.
It's not: (.02mol)/(.023L)
It's: (.02mol)/(1L) X (.023L)
The Molarity of KMnO4 is 0.02M, hence it's (0.02mol KMnO4/1Liter) then you multiply the given volume to get the actual number of moles. Please review your solution.
To calculate the molar concentrations of Fe2+ and Fe3+ in the original solution, we need to use the concept of oxidation-reduction (redox) reactions.
Step 1: Determine the moles of KMnO4 used in the titration:
Molarity (M) = moles/volume (L)
0.0200 M KMnO4 = moles/0.0230 L
moles = 0.0200 M x 0.0230 L = 0.00046 mol
Step 2: Fe2+ Oxidation to Fe3+:
2 Fe2+(aq) + 5 H2O(l) + 0.00046 mol KMnO4(aq) → 2 Fe3+(aq) + 5 OH^-(aq) + 0.00046 mol Mn2+(aq)
From the balanced equation, we can see that the ratio of KMnO4 used to Fe2+ oxidized is 1:2. Therefore, the moles of Fe2+ oxidized will also be 0.00046 mol.
Step 3: Conversion of Fe3+ to Fe2+ using Zn:
Fe3+(aq) + Zn(s) → Fe2+(aq) + Zn2+(aq)
Since all Fe3+ ions are converted to Fe2+ ions, the moles of Fe3+ in the solution will be equal to the moles of Fe2+ oxidized in step 2, which is 0.00046 mol.
Step 4: Calculate the molar concentration of Fe2+ in the original solution:
Volume of original Fe2+ solution = 25.0 mL = 0.0250 L
Molarity of Fe2+ = moles/volume (L)
Molarity of Fe2+ = 0.00046 mol/0.0250 L ≈ 0.0184 M
Step 5: Calculate the molar concentration of Fe3+ in the original solution:
Molarity of Fe3+ = moles/volume (L)
Molarity of Fe3+ = 0.00046 mol/0.0250 L ≈ 0.0184 M
Therefore, both the molar concentration of Fe2+ and Fe3+ in the original solution is approximately 0.0184 M.