   # The amount of oxygen, O 2 , dissolved in a water sample at 25 °C can be determined by titration. The first step is to add solutions of MnSO 4 and NaOH to the water to convert the dissolved oxygen to MnO 2 . A solution of H 2 SO 4 and KI is then added to convert the MnO 2 to Mn 2+ , and the iodide ion is converted to I 2 . The I 2 is then titrated with standardized Na 2 S 2 O 3 . (a) Balance the equation for the reaction of Mn 2+ ions with O 2 in basic solution. (b) Balance the equation for the reaction of MnO 2 with I − in acid solution. (c) Balance the equation for the reaction of S 2 O 3 2− with I 2 . (d) Calculate the amount of O 2 in 25.0 mL of water if the titration requires 2.45 mL of 0.0112 M Na 2 S 2 O 3 solution. ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 19, Problem 104IL
Textbook Problem
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## The amount of oxygen, O2, dissolved in a water sample at 25 °C can be determined by titration. The first step is to add solutions of MnSO4 and NaOH to the water to convert the dissolved oxygen to MnO2. A solution of H2SO4 and KI is then added to convert the MnO2 to Mn2+, and the iodide ion is converted to I2. The I2 is then titrated with standardized Na2S2O3. (a) Balance the equation for the reaction of Mn2+ ions with O2 in basic solution. (b) Balance the equation for the reaction of MnO2 with I− in acid solution. (c) Balance the equation for the reaction of S2O32− with I2. (d) Calculate the amount of O2 in 25.0 mL of water if the titration requires 2.45 mL of 0.0112 M Na2S2O3 solution.

(a)

Interpretation Introduction

Interpretation:

The equation for the reaction of Mn+2 ions with O2 in basic solution has to be determined.

Concept introduction:

Steps for balancing half –reactions in ACIDIC solution:

1. 1. Balance all atoms except H and O in half reaction.
2. 2. Balance O atoms by adding water to the side missing O atoms.
3. 3. Balance the H atoms by adding H+ to the side missing H atoms.
4. 4. Balance the charge by adding electrons to side with more total positive charge.
5. 5. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

Steps for balancing half –reactions in BASIC solution:

1. 1. Balance all atoms except H and O in half reaction.
2. 2. Balance O atoms by adding water to the side missing O atoms.
3. 3. Balance the H atoms by adding H+ to the side missing H atoms.
4. 4. Balance the charge by adding electrons to side with more total positive charge.
5. 5. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.
6. 6. Add the same number of OH- groups as there are H+ present to both sides of the equation.

### Explanation of Solution

The given reaction is as follows.

Mn2+(aq) + O2(g)  MnO2(s) +H2O

The chemical reaction involved in the first step is Mn2+(aq) + O2(g)  MnO2(s) +H2O.

Steps for balancing the half reaction in BASIC solution:

1. 1. Write the oxidation and reduction reactions.

Oxidation: Mn+2(aq) MnO2(s)Reduction: O2(g) H2O (l)

2. 2.  Balance half reactions for mass

Balance all atoms except H and O in half reaction.

Oxidation: Mn+2(aq) MnO2(s)Reduction: O2(g) H2O (l)

Addition of OH- or OH- and H2O is required for mass balance in both half reactions.

Oxidation: Mn+2(aq)+2H2O(l) MnO2(s)+4H+Reduction: O2(g) +4H+(aq)2H2O (l)

3. 3. Balance the charge by adding electrons to side with more total positive charge.

Oxidation: Mn+2(aq)+4OH(aq) MnO2(s) + 2H2O(l) + 2e-Reduction: O2(g) +2H2O(l) + 4e4OH-(aq)

4. 4

(b)

Interpretation Introduction

Interpretation:

The equation for the reaction of MnO2 with I- in acid solution has to be balanced.

Concept introduction:

Steps for balancing half –reactions in ACIDIC solution:

1. 6. Balance all atoms except H and O in half reaction.
2. 7. Balance O atoms by adding water to the side missing O atoms.
3. 8. Balance the H atoms by adding H+ to the side missing H atoms.
4. 9. Balance the charge by adding electrons to side with more total positive charge.
5. 10. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

Steps for balancing half –reactions in BASIC solution:

1. 7. Balance all atoms except H and O in half reaction.
2. 8. Balance O atoms by adding water to the side missing O atoms.
3. 9. Balance the H atoms by adding H+ to the side missing H atoms.
4. 10. Balance the charge by adding electrons to side with more total positive charge.
5. 11. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.
6. 12. Add the same number of OH- groups as there are H+ present to both sides of the equation.

(c)

Interpretation Introduction

Interpretation:

The equation for the reaction of S2O32- with I2 has to be balanced.

Concept introduction:

Steps for balancing half –reactions in ACIDIC solution:

1. 11. Balance all atoms except H and O in half reaction.
2. 12. Balance O atoms by adding water to the side missing O atoms.
3. 13. Balance the H atoms by adding H+ to the side missing H atoms.
4. 14. Balance the charge by adding electrons to side with more total positive charge.
5. 15. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

Steps for balancing half –reactions in BASIC solution:

1. 13. Balance all atoms except H and O in half reaction.
2. 14. Balance O atoms by adding water to the side missing O atoms.
3. 15. Balance the H atoms by adding H+ to the side missing H atoms.
4. 16. Balance the charge by adding electrons to side with more total positive charge.
5. 17. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.
6. 18. Add the same number of OH- groups as there are H+ present to both sides of the equation.

(d)

Interpretation Introduction

The amount of O2 in 25.0 mL of water has to be calculated if the titration requires 2.45 mL of 0.0112 MNa2S2O3 solution,.

Concept introduction:

• Molarity (M): Molarity is number of moles of the solute present in the one litter of the solution.

Molarity (M) =Numberofmolesofsolute1literofsolution

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