CHM 2045C Module Eight Solutions page 5

Part F: Solution Reaction Problems    10 points

Acid-Base Neutralization:

1. Calculate the molarity of a Calcium hydroxide solution if 18.50 mL of it requires 28.27 mL of a  0.0125 M HCl to reach its neutralization point.

1st Write a Balanced Chemical Reaction:

Ca(OH)2 (aq)    +  2 HCl (aq)       à    CaCl2 (aq)   + 2 H2O

2nd Setup the Dimensional Analysis and solve

____?__M Ca(OH)2 (aq) solution =  ? moles Ca(OH)2 solute/1 L Ca(OH)2 solution

REDOX Titration:

2. A KMnO4 (aq) solution is to be standardized by titration against As2O3(s), A 0.1156 g sample of As2O3 requires 27.08 mL of KMnO4. What is the molarity of the KMnO4?

1st balance the Redox Equation

5 As2O3   +     4 MnO4 1-      +     9 H2O         +    12 H 1+    ------->   10 H3AsO4      +    4 Mn 2+

Half equation:           MnO4 1-             à       Mn 2+                                                       1. Balance Mn

MnO4 1-             à       Mn 2+        +      4H2O                 2. Balance O

8 H 1+    +     MnO4 1-             à       Mn 2+        +      4H2O                 3. Balance H

4. Charge Balance:

+8   -1    =   +2    + 0

+7    =   +2

-5    +7    =   +2

Gain 5 electrons :

5 e1-   +    8 H 1+      +     MnO4 1-               à       Mn 2+        +      4H2O

This is the Reduction Step

5. Make e1- Gain = e1- Loss:

4 Reductions = 5 Oxidations

(5 e1-   +   8 H 1+    +    MnO4 1-             à       Mn 2+        +      4H2O) x 4

20 e1-   +  32 H 1+  +   4MnO4 1-             à    4 Mn 2+        +    16H2O

Half equation:           As2O3                    à      2 H3AsO4                                  1. Balance As

5 H2O    +        As2O3                   à      2 H3AsO4                                  2. Balance O

5 H2O    +        As2O3                   à      2 H3AsO4   +    4 H 1+               3. Balance H

4. Charge Balance:

+ 0  + 0  =  +0    +4

0  =  + 4

0  =  + 4    - 4

Loss 4 electrons

5 H2O    +          As2O3                         à      2 H3AsO4   +    4 H 1+ +  4 e1-

This is the Oxidation Step

5. Make e1- Gain = e1- Loss

4 Reductions = 5 Oxidations

( 5 H2O    +  As2O3         à      2 H3AsO4   +      4 H 1+  +  4 e1- ) x 5

25 H2O    +   5As2O3  à    10 H3AsO4   +    20 H 1+  +  20 e1-

2nd Setup and Solve Dimensional Analysis Problem: