Chapter 19, Problem 104IL

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.

1. (a) Balance the equation for the reaction of Mn²⁺ ions with O₂ in basic solution.
2. (b) Balance the equation for the reaction of MnO₂ with I⁻ in acid solution.
3. (c) Balance the equation for the reaction of S₂O₃²⁻ with I₂.
4. (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.

Interpretation Introduction

Interpretation:

The equation for the reaction of ${\text{Mn}}^{\text{+2}}$ ions with ${\text{O}}_{\text{2}}$ 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 ${\text{H}}^{\text{+}}$ 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 ${\text{H}}^{\text{+}}$ 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 ${\text{OH}}^{\text{-}}$ groups as there are ${\text{H}}^{\text{+}}$ present to both sides of the equation.

Answer to Problem 104IL

${\text{2Mn}}^{\text{2+}}{\text{(aq) + 4OH}}^{\text{-}}{\text{ (aq) + O}}_{\text{2}}\text{(g) }\to {\text{ 2MnO}}_{\text{2}}{\text{(s) + 2H}}_{\text{2}}\text{O(l)}$

Explanation of Solution

The given reaction is as follows.

${\text{Mn}}^{\text{2+}}{\text{(aq) + O}}_{\text{2}}\text{(g) }\to {\text{ MnO}}_{\text{2}}{\text{(s) +H}}_{\text{2}}\text{O}$

The chemical reaction involved in the first step is ${\text{Mn}}^{\text{2+}}{\text{(aq) + O}}_{\text{2}}\text{(g) }\to {\text{ MnO}}_{\text{2}}{\text{(s) +H}}_{\text{2}}\text{O}$.

Steps for balancing the half reaction in BASIC solution:

1. 1. Write the oxidation and reduction reactions.

   $\begin{array}{l}{\text{Oxidation: Mn}}^{\text{+2}}\text{(aq) }\to {\text{MnO}}_{\text{2}}\text{(s)}\\ {\text{Reduction: O}}_{\text{2}}\text{(g) }\to {\text{H}}_{\text{2}}\text{O (l)}\end{array}$

2. 2.  Balance half reactions for mass

   Balance all atoms except H and O in half reaction.

   $\begin{array}{l}{\text{Oxidation: Mn}}^{\text{+2}}\text{(aq) }\to {\text{MnO}}_{\text{2}}\text{(s)}\\ {\text{Reduction: O}}_{\text{2}}\text{(g) }\to {\text{H}}_{\text{2}}\text{O (l)}\end{array}$

   Addition of ${\text{OH}}^{\text{-}}$ or ${\text{OH}}^{\text{-}}{\text{ and H}}_{\text{2}}\text{O}$ is required for mass balance in both half reactions.

   $\begin{array}{l}{\text{Oxidation: Mn}}^{\text{+2}}{\text{(aq)+2H}}_2\text{O(l) }\to {\text{MnO}}_{\text{2}}{\text{(s)+4H}}^{+}\\ {\text{Reduction: O}}_{\text{2}}{\text{(g) +4H}}^{+}\text{(aq)}\to 2{\text{H}}_{\text{2}}\text{O (l)}\end{array}$

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

   $\begin{array}{l}{\text{Oxidation: Mn}}^{\text{+2}}{\text{(aq)+4OH}}^{-}\text{(aq) }\to {\text{MnO}}_{\text{2}}{\text{(s) + 2H}}_{\text{2}}{\text{O(l) + 2e}}^{\text{-}}\\ {\text{Reduction: O}}_{\text{2}}{\text{(g) +2H}}_2{\text{O(l) + 4e}}^{-}\to {\text{4OH}}^{\text{-}}\text{(aq)}\end{array}$

4. 4. Multiply the half reactions by appropriate factors.

   $\begin{array}{l}{\text{Oxidation: 2Mn}}^{\text{+2}}{\text{(aq)+8OH}}^{-}\text{(aq) }\to 2{\text{MnO}}_{\text{2}}{\text{(s) + 4H}}_{\text{2}}{\text{O(l) + 4e}}^{\text{-}}\\ {\text{Reduction: O}}_{\text{2}}{\text{(g) +2H}}_2{\text{O(l) + 4e}}^{-}\to {\text{4OH}}^{\text{-}}\text{(aq)}\end{array}$

5. 5. Add two half reactions and cancel the common atoms.

   $\begin{array}{l}{\text{Oxidation: 2Mn}}^{\text{+2}}{\text{(aq)+8OH}}^{-}\text{(aq) }\to 2{\text{MnO}}_{\text{2}}{\text{(s) + 4H}}_{\text{2}}{\text{O(l) + 4e}}^{\text{-}}\\ {\text{Reduction: O}}_{\text{2}}{\text{(g) +2H}}_2{\text{O(l) + 4e}}^{-}\to {\text{4OH}}^{\text{-}}\text{(aq)}\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ {\text{2Mn}}^{\text{2+}}{\text{(aq) + 4OH}}^{\text{-}}{\text{ (aq) + O}}_{\text{2}}\text{(g) }\to {\text{ 2MnO}}_{\text{2}}{\text{(s) + 2H}}_{\text{2}}\text{O(l)}\end{array}$

6. 6. Simplify by eliminating reactants and products that appears on both sides.

            ${\text{2Mn}}^{\text{2+}}{\text{(aq) + 4OH}}^{\text{-}}{\text{ (aq) + O}}_{\text{2}}\text{(g) }\to {\text{ 2MnO}}_{\text{2}}{\text{(s) + 2H}}_{\text{2}}\text{O(l)}$

Interpretation Introduction

Interpretation:

The equation for the reaction of ${\text{MnO}}_{\text{2}}$ with ${\text{I}}^{\text{-}}$ 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 ${\text{H}}^{\text{+}}$ 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 ${\text{H}}^{\text{+}}$ 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 ${\text{OH}}^{\text{-}}$ groups as there are ${\text{H}}^{\text{+}}$ present to both sides of the equation.

Answer to Problem 104IL

${\text{MnO}}_{\text{2}}{\text{(s) + 2I}}^{\text{-}}{\text{ + 4H}}^{\text{+}}\text{(aq) }\to {\text{ I}}_{\text{2}}{\text{(aq) + Mn}}^{\text{+2}}{\text{(aq) + 2H}}_{\text{2}}\text{O(l)}$

Explanation of Solution

The reaction ${\text{MnO}}_{\text{2}}{\text{with I}}^{\text{-}}$ are as follows.

${\text{MnO}}_{\text{2}}{\text{(s) + I}}^{\text{-}}\text{ (aq) }\to {\text{ Mn}}^{\text{2+}}{\text{(aq) + I}}_{\text{2}}\text{(aq)}$

Steps for balancing redox reactions in ACIDIC solution:

1. 1. Recognize the reaction as an oxidation and reduction reaction.

   I undergoes oxidation and ${\text{MnO}}_{\text{2}}$ undergoes reduction.

2. 2. Separate two half reactions.

   $\begin{array}{l}{\text{oxidation: I}}^{\text{-}}\text{(aq) }\to {\text{ I}}_{\text{2}}\text{(aq)}\\ {\text{Reduction: MnO}}_{\text{2}}\text{(s)}\to {\text{ Mn}}^{\text{2+}}\text{(aq)}\end{array}$

3. 3.  Balance half reactions by mass

   Balance all atoms except H and O in half reaction.

   $\begin{array}{l}{\text{oxidation: 2I}}^{\text{-}}\text{(aq) }\to {\text{ I}}_{\text{2}}\text{(aq)}\\ {\text{Reduction: MnO}}_{\text{2}}\text{(s)}\to {\text{ Mn}}^{\text{2+}}\text{(aq)}\end{array}$

   Balance O atoms by adding water to the side missing O atoms.

   $\begin{array}{l}{\text{oxidation: 2I}}^{\text{-}}\text{(aq) }\to {\text{ I}}_{\text{2}}\text{(aq)}\\ {\text{Reduction: MnO}}_{\text{2}}\text{(s)}\to {\text{ Mn}}^{\text{2+}}{\text{(aq) + 2H}}_{\text{2}}\text{O(l)}\end{array}$

   Balance the H atoms by adding ${\text{H}}^{\text{+}}$ to the side missing H atoms.

   $\begin{array}{l}{\text{oxidation: 2I}}^{\text{-}}\text{(aq) }\to {\text{ I}}_{\text{2}}\text{(aq)}\\ {\text{Reduction: MnO}}_{\text{2}}{\text{(s) + 4H}}^{\text{+}}\text{(aq)}\to {\text{ Mn}}^{\text{2+}}{\text{(aq) + 2H}}_{\text{2}}\text{O(l)}\end{array}$

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

   $\begin{array}{l}{\text{oxidation: 2I}}^{\text{-}}\text{(aq) }\to {\text{ I}}_{\text{2}}{\text{(aq) + 2e}}^{-}\\ {\text{Reduction: MnO}}_{\text{2}}{\text{(s) + 4H}}^{\text{+}}{\text{(aq) +2e}}^{-}\to {\text{ Mn}}^{\text{2+}}{\text{(aq) + 2H}}_{\text{2}}\text{O(l)}\end{array}$

5. 5. Make the number of electrons the same in both half reactions by multiplication, while avoiding fractional number of electrons.

   $\begin{array}{l}{\text{oxidation: 2I}}^{\text{-}}\text{(aq) }\to {\text{ I}}_{\text{2}}{\text{(aq) + 2e}}^{-}\\ {\text{Reduction: MnO}}_{\text{2}}{\text{(s) + 4H}}^{\text{+}}{\text{(aq) +2e}}^{-}\to {\text{ Mn}}^{\text{2+}}{\text{(aq) + 2H}}_{\text{2}}\text{O(l)}\end{array}$

6. 6.  Add the half reaction together and than simplyfy by cancelling out species that show up on both sides.

   $\begin{array}{l}{\text{oxidation: 2I}}^{\text{-}}\text{(aq) }\to {\text{ I}}_{\text{2}}{\text{(aq) + 2e}}^{-}\\ {\text{Reduction: MnO}}_{\text{2}}{\text{(s) + 4H}}^{\text{+}}{\text{(aq) +2e}}^{-}\to {\text{ Mn}}^{\text{2+}}{\text{(aq) + 2H}}_{\text{2}}\text{O(l)}\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ {\text{MnO}}_{\text{2}}{\text{(s) + 2I}}^{\text{-}}{\text{ + 4H}}^{\text{+}}\text{(aq) }\to {\text{ I}}_{\text{2}}{\text{(aq) + Mn}}^{\text{+2}}{\text{(aq) + 2H}}_{\text{2}}\text{O(l)}\end{array}$

7. 7. Confirm that the reaction is balanced in number of atoms and charge on the both sides of the arrow.

             ${\text{MnO}}_{\text{2}}{\text{(s) + 2I}}^{\text{-}}{\text{ + 4H}}^{\text{+}}\text{(aq) }\to {\text{ I}}_{\text{2}}{\text{(aq) + Mn}}^{\text{+2}}{\text{(aq) + 2H}}_{\text{2}}\text{O(l)}$

Interpretation Introduction

Interpretation:

The equation for the reaction of ${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2-}}$ with ${\text{I}}_{\text{2}}$ 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 ${\text{H}}^{\text{+}}$ 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 ${\text{H}}^{\text{+}}$ 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 ${\text{OH}}^{\text{-}}$ groups as there are ${\text{H}}^{\text{+}}$ present to both sides of the equation.

Answer to Problem 104IL

${\text{2Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{\text{(aq) + I}}_{\text{2}}\text{(aq) }\to {\text{ Na}}_{\text{2}}{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}\text{(aq) + 2NaI(aq)}$

Explanation of Solution

Balancing the ${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2-}}{\text{ with I}}_{\text{2}}$

${\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{\text{(aq) + I}}_{\text{2}}\text{(aq) }\to {\text{ Na}}_{\text{2}}{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}\text{(aq) + 2NaI(aq)}$

Let’s balance the above reaction.

${\text{2Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{\text{(aq) + I}}_{\text{2}}\text{(aq) }\to {\text{ Na}}_{\text{2}}{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}\text{(aq) + 2NaI(aq)}$

Interpretation Introduction

The amount of ${\text{O}}_{\text{2}}$ in $\text{25}\text{.0 mL}$ of water has to be calculated if the titration requires $\text{2}\text{.45 mL}$ of $\text{0}\text{.0112 M}$${\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$ solution,.

Concept introduction:

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

$\text{Molarity (M) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1}\text{liter}\text{of}\text{solution}}$

Answer to Problem 104IL

The amount of ${\text{O}}_{\text{2}}$ present in 25 ml of water is $\text{ 5}{\text{.488×10}}^{\text{-6}}\text{ g}$.

Explanation of Solution

The three equations are involved in the determination of oxygen.

$\begin{array}{l}{\text{(a) 2Mn}}^{\text{2+}}{\text{(aq) + 4OH}}^{\text{-}}{\text{(aq) + O}}_{\text{2}}\text{(g) }\to {\text{ 2MnO}}_{\text{2}}{\text{(s) + 2H}}_{\text{2}}\text{O(l)}\\ {\text{(b) MnO}}_{\text{2}}{\text{(s) + 2I}}^{\text{-}}{\text{(aq) + 4H}}^{\text{-}}\text{(aq) }\to {\text{ I}}_{\text{2}}{\text{(aq) + Mn}}^{\text{2+}}{\text{(aq) + 2H}}_{\text{2}}\text{O(l)}\\ {\text{(c) 2Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{\text{(aq) + I}}_{\text{2}}\text{(aq) }\to {\text{ Na}}_{\text{2}}{\text{S}}_{\text{4}}{\text{O}}_{\text{6}}\text{(aq) + 2NaI(aq)}\end{array}$

Then the amount of oxygen present in 25 mL of water , when the titration requires 2.45 mL of $\text{0}{\text{.0112 M Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$ is calculated as follows.

Let’s calculate number of moles of ${\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$:

                     $\begin{array}{l}\text{=Molarity × Volume}\\ \text{= 0}\text{.0112 mol/L × 0}\text{.00245 L}\\ \text{= 2}{\text{.744 × 10}}^{\text{-5}}{\text{ molofNa}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\end{array}$

The amount of ${\text{O}}_{\text{2}}$ in 1L of water is

$\begin{array}{l}\text{= 2}{\text{.744 × 10}}^{\text{-5}}{\text{ mol Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}\text{ × }\frac{{\text{1 mol I}}_{\text{2}}}{{\text{2 mol Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}}\text{ × }\frac{{\text{1 mol MnO}}_{\text{2}}}{{\text{1 mol I}}_{\text{2}}}\text{ × }\frac{{\text{1 mol O}}_{\text{2}}}{{\text{2 mol MnO}}_{\text{2}}}\text{ × }\frac{{\text{32 g of O}}_{\text{2}}}{{\text{1 mol O}}_{\text{2}}}\\ \text{= 2}{\text{.1952 × 10}}^{\text{-4}}\text{ g/1L solution}\end{array}$

Therefore,  $\text{2}{\text{.1952 × 10}}^{\text{-4}}\text{ g}$ of ${\text{O}}_{\text{2}}$ is present in 1000 mL of water.

Let’s calculate the amount of oxygen present in 25 mL of water. $\begin{array}{l}\text{=}\frac{{\text{25 mL H}}_{\text{2}}\text{O × 2}{\text{.1952 × 10}}^{\text{-4}}{\text{ g of O}}_{\text{2}}}{{\text{1000 mL of H}}_{\text{2}}\text{O}}\\ \text{= 5}{\text{.488×10}}^{\text{-6}}\text{ g}\end{array}$