CHM 1025C Module Eight Solutions Name: ___Answers

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.

1^st 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: (Bonus 10 points)

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?

1^st 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

5 e^1- + 8 H ¹⁺ + MnO₄ ^1- à Mn ²⁺ + 4H₂O 4. Charge Balance:

+8 -1 = +2 + 0

(5 e^1- + 8 H ¹⁺ + MnO₄ ^1- à Mn ²⁺ + 4H₂O) x 4 Gain 5 electrons

20 e^1- + 32 H ¹⁺ + 4MnO₄ ^1- à 4 Mn ²⁺ + 16H₂O 5. Make e^1- Gain = e^1- Loss

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

5 H₂O + As₂O₃ à 2 H₃AsO₄ + 4 H ¹⁺ + 4 e^1- 4. Charge Balance:

+ 0 + 0 = +0 + +4

( 5 H₂O + As₂O₃ à 2 H₃AsO₄ + 4 H ¹⁺ + 4 e^1- ) x 5 Loss 4 electrons

25 H₂O + 5As₂O₃ à 10 H₃AsO₄ + 20 H ^{1+ +} 20 e^1- 5. Make e^1- Gain = e^1- Loss

2^nd Setup and Solve Dimensional Analysis Problem: