Chapter 4, Problem 154CP

Triiodide ions are generated in solution by the following (unbalanced) reaction in acidic solution:

IO₃⁻(aq) + I⁻(aq) → I₃⁻(aq)

Triiodide ion concentration is determined by titration with a sodium thiosulfate (Na₂S₂O₃) solution. The products are iodide ion and tetrathionate ion (S₄O₆⁻).

a. Balance the equation for the reaction of IO₃⁻ with I⁻ ions.

b. A sample of 0.6013 g of potassium iodate was dissolved in water. Hydrochloric acid and solid potassium iodide were then added. What is the minimum mass of solid KI and the minimum volume of 3.00 M HQ required to convert all of the IO₃⁻ ions to I⁻ ions?

c. Write and balance the equation for the reaction of S₂O₃²⁻ with I₃⁻ in acidic solution.

d. A 25.00-mL sample of a 0.0100 M solution of KIO. is reacted with an excess of KI. It requires 32.04 mL of Na₂S₂O₃ solution to titrate the I₃⁻ ions present. What is the molarity of the Na₂S₂O₃ solution?

e. How would you prepare 500.0 mL of the KIO₃ solution in part d using solid KIO₃?

Interpretation Introduction

The balanced equation for the given reaction, minimum mass of $\text{KI}$ and minimum volume of $\text{3M}$ $\text{HCl}$ required for the reaction, balanced equation for the reaction between ${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}$ and ${\text{I}}_{\text{3}}{}^{-}$, molarity of ${\text{Na}}_2{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$ and preparation of ${\text{KIO}}_{\text{3}}$ in part d are needed to be determined.

Concept introduction:

Steps followed to balance a redox (reduction-oxidation) reaction occurs in acidic medium,

• Redox reactions can be separated into two half-reactions, such as oxidation reaction and reduction reaction.
• Balancing of oxygen and hydrogen are carried out by adding water and ions in the half-reactions, and balance the number of electrons then make the number of electrons in oxidation and reduction reactions equal by multiplying by integers.
• Club the both half-reactions.
• Mole ratio between the reactants of a reaction are depends upon the coefficients of reactants in a balanced chemical equation.
• Equation for number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

• Equation for finding molarity of a solution is,

$\text{Molarity}\text{(M)}\text{=}\frac{\text{number of moles of solute}\text{(mol)}}{\text{volume of solution}\text{(L)}}$

Answer to Problem 154CP

The balanced equations for the given reaction is,

${\text{IO}}_3{}^{-}\text{(aq)}+{\text{8I}}^{-}\text{(aq)}\text{+}{\text{6H}}^{+}\text{(aq)}\to {\text{3I}}_3{}^{-}\text{(aq)}\text{+}{\text{3H}}_2\text{O}\text{(aq)}$

Explanation of Solution

To determine: The balanced equations for the given reactions.

The chemical equation for the given redox reaction is,

${\text{IO}}_3{}^{-}\text{(aq)}+{\text{I}}^{-}\text{(aq)}\to {\text{I}}_3{}^{-}\text{(aq)}$

Steps (I, II and III) followed for balance the redox equation,

I- Separate the redox reaction into two, such as oxidation and reduction.

Therefore,

Oxidation reaction is,

${\text{I}}^{-}\text{(aq)}\to {\text{I}}_3{}^{-}\text{(aq)}$

Reduction reaction is,

${\text{IO}}_3{}^{-}\text{(aq)}\to {\text{I}}_3{}^{-}\text{(aq)}$

II- Balance the atoms and electrons in each half-reactions,

Balancing of oxygen and hydrogen are carried out by adding water and ions in the half-reactions

${\text{3I}}^{-}\text{(aq)}\to {\text{I}}_3{}^{-}\text{(aq)}\text{+}{\text{2e}}^{-}$

${\text{18H}}^{+}\text{(aq)}\text{+}{\text{3IO}}_3{}^{-}\text{(aq)}\text{+}{\text{16e}}^{-}\to {\text{I}}_3{}^{-}\text{(aq)}\text{+}{\text{9H}}_2\text{O}\text{(aq)}$

Equalizing the numbers of electrons in oxidation and reduction reactions,

${\text{24I}}^{-}\text{(aq)}\to {\text{8I}}_3{}^{-}\text{(aq)}\text{+}{\text{16e}}^{-}$

III- club the both half-reactions into a single equation,

${\text{24I}}^{-}\text{(aq)}+{\text{18H}}^{+}\text{(aq)}\text{+}{\text{3IO}}_3{}^{-}\text{(aq)}\text{+}{\text{16e}}^{-}\to {\text{I}}_3{}^{-}\text{(aq)}\text{+}{\text{9H}}_2\text{O}\text{(aq)}\text{+}{\text{8I}}_3{}^{-}\text{(aq)}\text{+}{\text{16e}}^{-}$

The same number of electrons and molecules in the both sides will cancel each other,

So, the balanced equation for the given equation is,

$\begin{array}{l}{\text{24I}}^{-}\text{(aq)}+{\text{18H}}^{+}\text{(aq)}\text{+}{\text{3IO}}_3{}^{-}\text{(aq)}\to {\text{9I}}_3{}^{-}\text{(aq)}\text{+}{\text{9H}}_2\text{O}\text{(aq)}\Rightarrow \\ {\text{IO}}_3{}^{-}\text{(aq)}+{\text{8I}}^{-}\text{(aq)}\text{+}{\text{6H}}^{+}\text{(aq)}\to {\text{3I}}_3{}^{-}\text{(aq)}\text{+}{\text{3H}}_2\text{O}\text{(aq)}\end{array}$

Interpretation Introduction

Interpretation:

The balanced equation for the given reaction, minimum mass of $\text{KI}$ and minimum volume of $\text{3M}$ $\text{HCl}$ required for the reaction, balanced equation for the reaction between ${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}$ and ${\text{I}}_{\text{3}}{}^{-}$, molarity of ${\text{Na}}_2{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$ and preparation of ${\text{KIO}}_{\text{3}}$ in part d are needed to be determined.

Concept introduction:

Steps followed to balance a redox (reduction-oxidation) reaction occurs in acidic medium,

• Redox reactions can be separated into two half-reactions, such as oxidation reaction and reduction reaction.
• Balancing of oxygen and hydrogen are carried out by adding water and ions in the half-reactions, and balance the number of electrons then make the number of electrons in oxidation and reduction reactions equal by multiplying by integers.
• Club the both half-reactions.
• Mole ratio between the reactants of a reaction are depends upon the coefficients of reactants in a balanced chemical equation.
• Equation for number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

• Equation for finding molarity of a solution is,

$\text{Molarity}\text{(M)}\text{=}\frac{\text{number of moles of solute}\text{(mol)}}{\text{volume of solution}\text{(L)}}$

Answer to Problem 154CP

The minimum mass of $\text{KI}$ and minimum volume of $\text{3M}$ $\text{HCl}$ required for the reaction is $\text{3}\text{.7315g}$ and $\text{0}\text{.7492}\text{×}{\text{10}}^{\text{-2}}\text{L}$.

Explanation of Solution

To determine: the minimum mass of $\text{KI}$ and minimum volume of $\text{3M}$ $\text{HCl}$ required for the reaction if $\text{0}\text{.6013g}$ of ${\text{KIO}}_3$ is used.

The mass of ${\text{KIO}}_3$ in the reaction is given as $\text{0}\text{.6013g}$.

Equation for number of moles of a substance is,

$\text{number}\text{of}\text{moles}\text{=}\frac{\text{given}\text{mass}}{\text{molar}\text{mass}}$

Therefore,

The number of moles of ${\text{KIO}}_3$ is,

$\text{number}\text{of}\text{moles}\text{=}\frac{\text{0}\text{.6013g}}{\text{214g}}=\text{0}\text{.28098}\text{×}{\text{10}}^{\text{-2}}\text{mol}$

The balanced equations for the given reaction is,

${\text{IO}}_3{}^{-}\text{(aq)}+{\text{8I}}^{-}\text{(aq)}\text{+}{\text{6H}}^{+}\text{(aq)}\to {\text{3I}}_3{}^{-}\text{(aq)}\text{+}{\text{3H}}_2\text{O}\text{(aq)}$

The mole ratio between ${\text{IO}}_3{}^{-}$ and ${\text{I}}^{-}$ is 1:8.

Therefore,

The number of moles of ${\text{I}}^{-}$ reacted with $\text{0}\text{.28098}\text{×}{\text{10}}^{\text{-2}}\text{mol}$ of ${\text{IO}}_3{}^{-}$ is,

$\text{8}\times \text{0}\text{.28098}\text{×}{\text{10}}^{\text{-2}}\text{mol}\text{=}\text{0}\text{.022478}\text{mol}$

Equation for number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

Therefore,

The minimum mass of $\text{KI}$ required in the reaction is,

$\text{0}\text{.022478}\text{mol}\text{×}\text{166}\text{.0028g}\text{=}\text{3}\text{.7315g}$

The number of moles of $\text{HCl}$ is equals to the number of mole of $\text{KI}$, that is $\text{0}\text{.022478}\text{mol}$.

The volume of $\text{3M}$ $\text{HCl}$ reacted can be find out from its molarity and number of moles,

That is,

$\text{volume of solution}\text{(L)}\text{=}\frac{\text{0}\text{.022478}\text{mol}}{\text{3mol/L}}\text{=}\text{0}\text{.7492}\text{×}{\text{10}}^{\text{-2}}\text{L}$

Interpretation Introduction

Interpretation:

The balanced equation for the given reaction, minimum mass of $\text{KI}$ and minimum volume of $\text{3M}$ $\text{HCl}$ required for the reaction, balanced equation for the reaction between ${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}$ and ${\text{I}}_{\text{3}}{}^{-}$, molarity of ${\text{Na}}_2{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$ and preparation of ${\text{KIO}}_{\text{3}}$ in part d are needed to be determined.

Concept introduction:

Steps followed to balance a redox (reduction-oxidation) reaction occurs in acidic medium,

• Redox reactions can be separated into two half-reactions, such as oxidation reaction and reduction reaction.
• Balancing of oxygen and hydrogen are carried out by adding water and ions in the half-reactions, and balance the number of electrons then make the number of electrons in oxidation and reduction reactions equal by multiplying by integers.
• Club the both half-reactions.
• Mole ratio between the reactants of a reaction are depends upon the coefficients of reactants in a balanced chemical equation.
• Equation for number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

• Equation for finding molarity of a solution is,

$\text{Molarity}\text{(M)}\text{=}\frac{\text{number of moles of solute}\text{(mol)}}{\text{volume of solution}\text{(L)}}$

Answer to Problem 154CP

The balanced equation for the reaction between ${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}$ and ${\text{I}}_{\text{3}}{}^{-}$ is,

Explanation of Solution

To determine: the balanced equations for the reaction between ${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}$ and ${\text{I}}_{\text{3}}{}^{-}$.

The chemical equation for the given redox reaction is,

${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}\text{(aq)}+{\text{I}}_3{}^{-}\text{(aq)}\to {\text{S}}_{\text{4}}{\text{O}}_{\text{6}}{}^{\text{2-}}\text{(aq)}+{\text{I}}_2{}^{-}\text{(aq)}$

Steps (I, II and III) followed for balance the redox equation,

I- Separate the redox reaction into two, such as oxidation and reduction.

Therefore,

Oxidation reaction is,

${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}\text{(aq)}\to {\text{S}}_{\text{4}}{\text{O}}_{\text{6}}{}^{\text{2-}}\text{(aq)}$

Reduction reaction is,

${\text{I}}_3{}^{-}\text{(aq)}\to {\text{I}}_2{}^{-}\text{(aq)}$

II- Balance the atoms and electrons in each half-reactions,

Balancing of oxygen and hydrogen are carried out by adding water and ions in the half-reactions

${\text{2S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}\text{(aq)}\to {\text{S}}_{\text{4}}{\text{O}}_{\text{6}}{}^{\text{2-}}\text{(aq)}\text{+}{\text{2e}}^{-}$

${\text{2I}}_3{}^{-}\text{(aq)}\text{+}{\text{1e}}^{-}\to {\text{3I}}_2{}^{-}\text{(aq)}$

Equalizing the numbers of electrons in oxidation and reduction reactions,

${\text{4I}}_3{}^{-}\text{(aq)}\text{+}{\text{2e}}^{-}\to {\text{6I}}_2{}^{-}\text{(aq)}$

III- club the both half-reactions into a single equation,

${\text{2S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}\text{(aq)}\text{+}{\text{4I}}_3{}^{-}\text{(aq)}\text{+}{\text{2e}}^{-}\to {\text{S}}_{\text{4}}{\text{O}}_{\text{6}}{}^{\text{2-}}\text{(aq)}\text{+}{\text{2e}}^{-}+{\text{6I}}_2{}^{-}\text{(aq)}$

The same number of electrons and molecules in the both sides will cancel each other,

So, the balanced equation for the given equation is,

${\text{2S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}\text{(aq)}\text{+}{\text{4I}}_3{}^{-}\text{(aq)}\to {\text{S}}_{\text{4}}{\text{O}}_{\text{6}}{}^{\text{2-}}\text{(aq)}+{\text{6I}}_2{}^{-}\text{(aq)}$

To determine: the molarity of ${\text{Na}}_2{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$ in the above reaction if $\text{25mL}$ of $\text{0}\text{.01M}$ ${\text{KIO}}_3$ is reacted with $\text{32}\text{.04mL}$ of ${\text{Na}}_2{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$.

Interpretation Introduction

Interpretation:

The balanced equation for the given reaction, minimum mass of $\text{KI}$ and minimum volume of $\text{3M}$ $\text{HCl}$ required for the reaction, balanced equation for the reaction between ${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}$ and ${\text{I}}_{\text{3}}{}^{-}$, molarity of ${\text{Na}}_2{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$ and preparation of ${\text{KIO}}_{\text{3}}$ in part d are needed to be determined.

Concept introduction:

Steps followed to balance a redox (reduction-oxidation) reaction occurs in acidic medium,

• Redox reactions can be separated into two half-reactions, such as oxidation reaction and reduction reaction.
• Balancing of oxygen and hydrogen are carried out by adding water and ions in the half-reactions, and balance the number of electrons then make the number of electrons in oxidation and reduction reactions equal by multiplying by integers.
• Club the both half-reactions.
• Mole ratio between the reactants of a reaction are depends upon the coefficients of reactants in a balanced chemical equation.
• Equation for number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

• Equation for finding molarity of a solution is,

$\text{Molarity}\text{(M)}\text{=}\frac{\text{number of moles of solute}\text{(mol)}}{\text{volume of solution}\text{(L)}}$

Answer to Problem 154CP

The molarity of ${\text{Na}}_2{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$ is $\text{0}\text{.375}\text{×}{\text{10}}^{\text{-3}}\text{mol}$.

Explanation of Solution

The molarity of ${\text{KIO}}_3$ in the reaction is given as $\text{0}\text{.01M}$.

The volume of ${\text{KIO}}_3$ in the reaction is given as $\text{25mL}$.

Equation for number of moles of a substance from its molarity and volume is,

$\text{Molarity}\text{(M)}\text{=}\frac{\text{number of moles of solute}\text{(mol)}}{\text{volume of solution}\text{(L)}}$

Therefore,

The number of moles of ${\text{KIO}}_3$ is,

$\text{number}\text{of}\text{moles}\text{=}\text{0}\text{.01mol/L}\times \text{25×}{\text{10}}^{\text{-3}}\text{L}=\text{0}\text{.25}\text{×}{\text{10}}^{\text{-3}}\text{mol}$

The balanced equations for the reaction of ${\text{IO}}_3{}^{-}$ and product ${\text{I}}_3{}^{-}$ is,

${\text{IO}}_3{}^{-}\text{(aq)}+{\text{8I}}^{-}\text{(aq)}\text{+}{\text{6H}}^{+}\text{(aq)}\to {\text{3I}}_3{}^{-}\text{(aq)}\text{+}{\text{3H}}_2\text{O}\text{(aq)}$

The mole ratio between ${\text{IO}}_3{}^{-}$ and ${\text{I}}_3{}^{-}$ is 1:3.

Therefore,

The number of moles of ${\text{I}}_3{}^{-}$ produced with $\text{0}\text{.25}\text{×}{\text{10}}^{\text{-3}}\text{mol}$ of ${\text{IO}}_3{}^{-}$ is,

$\text{3}\times \text{0}\text{.25}\text{×}{\text{10}}^{\text{-3}}\text{mol}\text{=}\text{0}\text{.75}\text{×}{\text{10}}^{\text{-3}}\text{mol}$

The balanced equations for the reaction between ${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}$ and ${\text{I}}_3{}^{-}$ is,

${\text{2S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}\text{(aq)}\text{+}{\text{4I}}_3{}^{-}\text{(aq)}\to {\text{S}}_{\text{4}}{\text{O}}_{\text{6}}{}^{\text{2-}}\text{(aq)}+{\text{6I}}_2{}^{-}\text{(aq)}$

The mole ratio between ${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}$ and ${\text{I}}_3{}^{-}$ is 1:2.

Therefore,

The number of moles of ${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}$ reacted with $\text{0}\text{.75}\text{×}{\text{10}}^{\text{-3}}\text{mol}$ of ${\text{I}}_3{}^{-}$ is,

$\frac{\text{1}}{\text{2}}\text{×}\text{0}\text{.75}\text{×}{\text{10}}^{\text{-3}}\text{mol}\text{=}\text{0}\text{.375}\text{×}{\text{10}}^{\text{-3}}\text{mol}$

To explain: the preparation of ${\text{KIO}}_{\text{3}}$ solution from solid in part (d).

Interpretation Introduction

Interpretation:

The balanced equation for the given reaction, minimum mass of $\text{KI}$ and minimum volume of $\text{3M}$ $\text{HCl}$ required for the reaction, balanced equation for the reaction between ${\text{S}}_{\text{2}}{\text{O}}_{\text{3}}{}^{\text{2-}}$ and ${\text{I}}_{\text{3}}{}^{-}$, molarity of ${\text{Na}}_2{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$ and preparation of ${\text{KIO}}_{\text{3}}$ in part d are needed to be determined.

Concept introduction:

Steps followed to balance a redox (reduction-oxidation) reaction occurs in acidic medium,

• Redox reactions can be separated into two half-reactions, such as oxidation reaction and reduction reaction.
• Balancing of oxygen and hydrogen are carried out by adding water and ions in the half-reactions, and balance the number of electrons then make the number of electrons in oxidation and reduction reactions equal by multiplying by integers.
• Club the both half-reactions.
• Mole ratio between the reactants of a reaction are depends upon the coefficients of reactants in a balanced chemical equation.
• Equation for number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

• Equation for finding molarity of a solution is,

$\text{Molarity}\text{(M)}\text{=}\frac{\text{number of moles of solute}\text{(mol)}}{\text{volume of solution}\text{(L)}}$

Answer to Problem 154CP

The preparation of ${\text{KIO}}_{\text{3}}$ in part d is by dissolving $\text{1}\text{.07g}$ of solid ${\text{KIO}}_{\text{3}}$ in water.

Explanation of Solution

The molarity of ${\text{KIO}}_3$ in the reaction is given as $\text{0}\text{.01M}$.

The volume of ${\text{KIO}}_3$ is to prepare is $\text{500mL}\text{=}\text{0}\text{.5L}$.

Equation for number of moles of a substance from its molarity and volume is,

$\text{Molarity}\text{(M)}\text{=}\frac{\text{number of moles of solute}\text{(mol)}}{\text{volume of solution}\text{(L)}}$

Therefore,

The number of moles of ${\text{KIO}}_3$ to be prepared is,

$\text{number}\text{of}\text{moles}\text{=}\text{0}\text{.01mol/L}\times \text{0}\text{.5L}=\text{0}\text{.005mol}$

Equation for number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

Therefore,

The mass of solid ${\text{KIO}}_3$ to be prepared is,

$\text{0}\text{.005}\text{×}\text{214g}\text{=}\text{1}\text{.07g}$

Dissolving $\text{1}\text{.07g}$ of ${\text{KIO}}_3$ in water of $\text{500mL}$ will give the solution of part d.