Irvington High School  AP Chemistry
Mr. Markic
Name ____________________________________
Number ___ Date ___/___/___
4  Aqueous Reactions and Solution Stoichiometry
Redox Titrations
1. Iron (II) can be oxidized by an acidic K2Cr2O7 solution according to the net ionic equation:
Cr2O72- + 6Fe2+ + 14H+  2Cr3+ + 6Fe3+ 7H2O
If it takes 26.0 mL of 0.0250 M K2Cr2O7 to titrate 25.0 mL of a solution containing Fe2+, what is the
molar concentration of Fe2+?
The balanced equation is given in the problem. The mole ratio between Fe2 and Cr2O72 is 6:1.
First, calculate the moles of Fe2 that react with Cr2O72.
26.00 mL soln 
0.0250 mol Cr2O72
6 mol Fe2

 3.90  103 mol Fe2
2
1000 mL soln
1 mol Cr2O7
The molar concentration of Fe2 is:
M 
3.90  103 mol Fe2
25.0  103 L soln
 0.156 M
2. A sample of iron ore (containing only Fe2+ ions) weighing 0.2792 g was dissolved in dilute acid
solution, and the Fe (II) was converted to Fe (III) ions. The solution required 23.30 mL of 0.0194 M
K2Cr2O7 for titration. Calculate the percent by mass of iron in the ore. (Hint: See problem #1 for the
balanced equation)
The balanced equation is given in problem 4.91. The mole ratio between Fe2 and Cr2O72 is 6:1.
First, calculate the moles of Cr2O72 that reacted.
23.30 mL soln 
0.0194 mol Cr2 O72
 4.52  104 mol Cr2 O72
1000 mL soln
Use the mole ratio from the balanced equation to calculate the mass of iron that reacted.
(4.52  104 mol Cr2 O72 ) 
6 mol Fe2
1 mol Cr2 O72

55.85 g Fe2
1 mol Fe 2
 0.151 g Fe 2
The percent by mass of iron in the ore is:
0.151 g
 100%  54.1%
0.2792 g
3. Iodate ion, IO31- oxidizes SO32- in acidic solution. The half-reaction for the oxidation is
SO32- + H2O  SO42- + 2H+ + 2eA 100.0-mL sample of a solution containing 1.390 g of KIO3 reacts with 32.5 mL of 0.500 M Na2SO3.
What is the final oxidation state of the iodine after the reaction has occurred?
Page 1 of 3
The balanced equation shows that 2 moles of electrons are lost for each mole of SO32 that reacts. The
electrons are gained by IO3. We need to find the moles of electrons gained for each mole of IO3 that
reacts. Then, we can calculate the final oxidation state of iodine.
The number of moles of electrons lost by SO32 is:
0.500 mol SO32
2 mol e

 0.0325 mol e lost
1000 mL soln
1 mol SO32
32.5 mL 
The number of moles of iodate, IO3, that react is:
1.390 g KIO3 
1 mol KIO3
1 mol IO3

 6.495  103 mol IO3
214.0 g KIO3 1 mol KIO3
6.495  103 mole of IO3 gain 0.0325 mole of electrons. The number of moles of electrons gained per
mole of IO3 is:
0.0325 mol e 
6.495  10
3
mol
IO3
 5.00 mol e  /mol IO3
The oxidation number of iodine in IO3 is 5. Since 5 moles of electrons are gained per mole of IO3,
the final oxidation state of iodine is 5  5  0. The iodine containing product of the reaction is most
likely elemental iodine, I2.
4. A quantity of 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. The net ionic equation is
MnO41- + 5Fe2+ + 8H+  Mn2+ + 5Fe3+ + 4H2O
The first titration oxidizes Fe2 to Fe3. This titration gives the amount of Fe2 in solution. Zn metal is
added to reduce all Fe3 back to Fe2. The second titration oxidizes all the Fe2 back to Fe3. We can
find the amount of Fe3 in the original solution by difference.
Titration #1: The mole ratio between Fe2 and MnO4 is 5:1.
23.0 mL soln 
[Fe2+ ] 
0.0200 mol MnO4
5 mol Fe2

 2.30  103 mol Fe2

1000 mL soln
1 mol MnO4
mol solute
2.30  103 mol Fe 2

 0.0920 M
L of soln
25.0  103 L soln
Titration #2: The mole ratio between Fe2 and MnO4 is 5:1.
40.0 mL soln 
0.0200 mol MnO4
5 mol Fe2

 4.00  103 mol Fe2

1000 mL soln
1 mol MnO4
In this second titration, there are more moles of Fe2 in solution. This is due to Fe3 in the original
solution being reduced by Zn to Fe2. The number of moles of Fe3 in solution is:
Page 2 of 3
(4.00  103 mol)  (2.30  103 mol)  1.70  103 mol Fe3
[Fe3+ ] 
mol solute
1.70  103 mol Fe3

 0.0680 M
L of soln
25.0  103 L soln
Page 3 of 3