The Foundations of Chemistry

Solution

The reaction ratio is

MnO 4 (aq)
8H
(aq)
5Fe^2 
(aq)88n5Fe^3 
(aq)
Mn^2 
(aq)
4H 2 O(
)

rxn ratio: 1 mol 5 mol

The number of moles of Fe^2 
to be titrated is

_?_ mol Fe^2 
40.0 mL4.00 10 ^3 mol Fe^2 

We use the balanced equation to find the number of moles of MnO 4 required.

_?_ mol MnO 4 4.00 10 ^3 mol Fe^2 
8.00 10 ^4 mol MnO 4 

Each formula unit of KMnO 4 contains one MnO 4 ion, and so

1 mol KMnO 4 1 mol MnO 4 

The volume of 0.0200 MKMnO 4 solution that contains 8.00 10 ^4 mol of KMnO 4 is

_?_ mL KMnO 4 soln8.00 10 ^4 mol KMnO 4 

40.0 mL KMnO 4 soln

You should now work Exercises 66 and 68.

Potassium dichromate, K 2 Cr 2 O 7 , is another frequently used oxidizing agent. However,

an indicator must be used when reducing agents are titrated with dichromate solutions.

K 2 Cr 2 O 7 is orange, and its reduction product, Cr^3 
, is green.

Consider the oxidation of sulfite ions, SO 32 , to sulfate ions, SO 42 , by Cr 2 O 72 ions

in the presence of a strong acid such as sulfuric acid. We shall balance the equation by

the half-reaction method.

Cr 2 O 72 88nCr^3 
 (red. half-rxn)

Cr 2 O 72 88n 2Cr^3 

14H

Cr 2 O 72 88n2Cr^3 

7H 2 O

6 e
14H

Cr 2 O 72 88n2Cr^3 

7H 2 O (balanced red. half-rxn)

SO 32  88nSO 42  (ox. half-rxn)

SO 32 
H 2 O 88nSO 42 
2H


SO 32 
H 2 O88nSO 42 
2H

 2 e (balanced ox. half-rxn)

We now equalize the electron transfer, add the balanced half-reactions, and eliminate

common terms.

(6e
14H

Cr 2 O 72 88n2Cr^3 

7H 2 O) (reduction)

3(SO 32 
H 2 O88nSO 42 
2H

 2 e) (oxidation)

8H
(aq)
Cr 2 O 72 (aq)
3SO 32 (aq)88n2Cr^3 
(aq)
3SO 42 (aq)
4H 2 O(
)

The balanced equation tells us that the reaction ratio is 3 mol SO 32 /mol Cr 2 O 72 or

1 mol Cr 2 O 72 /3 mol SO 32 . Potassium dichromate is the usual source of Cr 2 O 72 ions,

1000 mL KMnO 4 soln



0.0200 mol KMnO 4

1 mol MnO 4 



5 mol Fe^2 

0.100 mol Fe^2 




1000 mL

426 CHAPTER 11: Reactions in Aqueous Solutions II: Calculations