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) 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- Mn 2+ 1. Balance Mn

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

8 H 1+ + MnO4 1- 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- 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- Mn 2+ + 4H2O) x 4

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

 

Half equation: As2O3 2 H3AsO4 1. Balance As

5 H2O + As2O3 2 H3AsO4 2. Balance O

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

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

 

2nd Setup and Solve Dimensional Analysis Problem: