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:
