Chapter 4, Problem 4.131P

(a)

Interpretation Introduction

Interpretation:

The balanced net ionic equation for the precipitation reaction is to be determined.

Concept introduction:

There are three types of equations that are utilized to represent an ionic reaction:

1. Molecular equation

2. Total ionic equation

3. Net ionic equation

The molecular equation represents the reactants and products of the ionic reaction in undissociated form. In total ionic reaction, all the dissociated ions that are present in the reaction mixture are represented and in net ionic reaction, the useful ions that participate in the reaction are represented.

Precipitation reaction involves the reaction of two soluble ionic compounds to form an insoluble product. The insoluble product is known as a precipitate.

The reason for the precipitation reaction to occur is the formation of a product that is insoluble in nature.

Spectator ions are the ions that are not a part of the actual chemical change but are present in the reaction mixture to balance the charge on both sides of the reaction. They are represented in the total ionic reaction. These are the dissolved ions present in the reaction mixture.

Answer to Problem 4.131P

The molecular equation of the following reaction is:

${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4\left(aq\right)+{\text{CaCl}}_2\left(aq\right)\to {\text{CaC}}_{\text{2}}{\text{O}}_4\left(s\right)+2\text{NaCl}\left(aq\right)$

The total ionic equation for the given reaction is:

$2{\text{Na}}^{+}\left(aq\right)+{\text{C}}_{\text{2}}{\text{O}}_4^{2-}\left(aq\right)+{\text{Ca}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)\to {\text{CaC}}_{\text{2}}{\text{O}}_4\left(s\right)+2{\text{Na}}^{+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)$

The net ionic equation for the given reaction is:

${\text{C}}_{\text{2}}{\text{O}}_4^{2-}\left(aq\right)+{\text{Ca}}^{2+}\left(aq\right)\to {\text{CaC}}_{\text{2}}{\text{O}}_4\left(s\right)$

Explanation of Solution

Sodium oxalate $\left({\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4\right)$ reacts with calcium chloride to form calcium oxalate $\left({\text{CaC}}_{\text{2}}{\text{O}}_4\right)$ and sodium chloride. The molecular equation of the following reaction is:

${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4\left(aq\right)+{\text{CaCl}}_2\left(aq\right)\to {\text{CaC}}_{\text{2}}{\text{O}}_4\left(s\right)+2\text{NaCl}\left(aq\right)$

The total ionic equation for the given reaction is:

$2{\text{Na}}^{+}\left(aq\right)+{\text{C}}_{\text{2}}{\text{O}}_4^{2-}\left(aq\right)+{\text{Ca}}^{2+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)\to {\text{CaC}}_{\text{2}}{\text{O}}_4\left(s\right)+2{\text{Na}}^{+}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)$

${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ are the spectator ions that are present in the reaction mixture. Spectator ions are not present in the net ionic equation.

The net ionic equation for the given reaction is:

${\text{C}}_{\text{2}}{\text{O}}_4^{2-}\left(aq\right)+{\text{Ca}}^{2+}\left(aq\right)\to {\text{CaC}}_{\text{2}}{\text{O}}_4\left(s\right)$

Conclusion

${\text{CaC}}_{\text{2}}{\text{O}}_4$ is the solid insoluble substance that is formed in the reaction. ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ are the spectator ions that are present in the reaction mixture.

(b)

Interpretation Introduction

Interpretation:

The balanced net ionic equation for the titration reaction is to be determined.

Concept introduction:

There are three types of equations that are utilized to represent an ionic reaction:

1. Molecular equation

2. Total ionic equation

3. Net ionic equation

The molecular equation represents the reactants and products of the ionic reaction in undissociated form. In total ionic reaction, all the dissociated ions that are present in the reaction mixture are represented and in net ionic reaction, the useful ions that participate in the reaction are represented.

Precipitation reaction involves the reaction of two soluble ionic compounds to form an insoluble product. The insoluble product is known as a precipitate.

The reason for the precipitation reaction to occur is the formation of a product that is insoluble in nature.

Spectator ions are the ions that are not a part of the actual chemical change but are present in the reaction mixture to balance the charge on both sides of the reaction. They are represented in the total ionic reaction. These are the dissolved ions present in the reaction mixture.

Answer to Problem 4.131P

The net ionic equation of the following reaction is:

${\text{2MnO}}_4^{-}\left(aq\right)+5{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4\left(aq\right)+6{\text{H}}^{+}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+10{\text{CO}}_2\left(g\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

Potassium permanganate $\left({\text{KMnO}}_4\right)$ is a soluble salt and can be written in the dissociated form and oxalic acid is weak acid and is present in the undissociated form. ${\text{KMnO}}_4$ oxidizes the oxalic acid to ${\text{CO}}_2$ and gets reduced to form ${\text{Mn}}^{2+}$. The molecular equation of the following reaction is:

${\text{K}}^{+}\left(aq\right)+{\text{MnO}}_4^{-}\left(aq\right)+{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+{\text{CO}}_2\left(g\right)+{\text{K}}^{+}\left(aq\right)$

${\text{K}}^{+}$ is the spectator ion that is present in the reaction mixture.

The reaction takes place in an acidic medium so ${\text{H}}^{+}$ and ${\text{H}}_2\text{O}$ are used to balance the reaction. The net ionic equation of the following reaction is:

${\text{2MnO}}_4^{-}\left(aq\right)+5{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4\left(aq\right)+6{\text{H}}^{+}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+10{\text{CO}}_2\left(g\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Conclusion

Potassium permanganate $\left({\text{KMnO}}_4\right)$ is a soluble salt and written in dissociated form. Oxalic acid is a weak acid and present in dissociated form.

(c)

Interpretation Introduction

Interpretation:

The oxidizing agent in the reaction is to be identified.

Concept introduction:

A redox reaction is a type of reaction that involves the change in oxidation number of a molecule, atom or ion changes due to the transfer of an electron from one species to another.

An oxidizing agent is a substance that oxidizes another species and itself gets reduced in a chemical reaction. A reducing agent is the one that reduces another species and itself gets oxidized in a chemical reaction.

Answer to Problem 4.131P

The oxidizing agent in the reaction is ${\text{MnO}}_4^{-}$.

Explanation of Solution

The given redox reaction is:

${\text{2MnO}}_4^{-}\left(aq\right)+5{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4\left(aq\right)+6{\text{H}}^{+}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+10{\text{CO}}_2\left(g\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The oxidation number of oxygen is $-2$.

The expression to calculate the oxidation number of manganese in ${\text{MnO}}_4^{-}$ is:

$\left[4\left(\text{oxidation number of oxygen}\right)+\left(\text{oxidation number of mangenese}\right)\right]=-1$ (1)

Rearrange equation (1) for the oxidation number of manganese.

$\text{Oxidation number of mangenese}=\left[\left(-1\right)-4\left(\text{oxidation number of oxygen}\right)\right]$ (2)

Substitute $-2$ for the oxidation number of oxygen in equation (2).

$\begin{array}{c}\text{Oxidation number of mangenese}=\left[\left(-1\right)-4\left(-2\right)\right]\\ =\left[-1+8\right]\\ =+7\end{array}$

The oxidation state of ${\text{Mn}}^{2+}$ is $+2$ that is equal to the charge present on the ${\text{Mn}}^{2+}$.

The oxidation state of manganese in ${\text{MnO}}_4^{-}$ is $+7$ and it decreases to $+2$ in ${\text{Mn}}^{2+}$. ${\text{MnO}}_4^{-}$ undergoes a reduction in the reaction, therefore, it acts as the oxidizing agent.

Conclusion

An oxidizing agent is a substance that oxidizes another species and itself gets reduced in a chemical reaction. ${\text{MnO}}_4^{-}$ undergoes reduction and acts as the oxidizing agent in the reaction

(d)

Interpretation Introduction

Interpretation:

The reducing agent in the reaction is to be identified.

Concept introduction:

A redox reaction is a type of reaction that involves the change in oxidation number of a molecule, atom or ion changes due to the transfer of an electron from one species to another.

An oxidizing agent is a substance that oxidizes another species and itself gets reduced in a chemical reaction. A reducing agent is the one that reduces another species and itself gets oxidized in a chemical reaction.

Answer to Problem 4.131P

The reducing agent in the reaction is ${\text{H}}_2{\text{C}}_2{\text{O}}_4$.

Explanation of Solution

The given redox reaction is:

${\text{2MnO}}_4^{-}\left(aq\right)+5{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4\left(aq\right)+6{\text{H}}^{+}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+10{\text{CO}}_2\left(g\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The expression to calculate the oxidation number of carbon in ${\text{H}}_2{\text{C}}_2{\text{O}}_4$ is:

$\left[\begin{array}{l}2\left(\text{oxidation number of hydrogen}\right)+2\left(\text{oxidation number of carbon}\right)+\\ 4\left(\text{oxidation number of oxygen}\right)\end{array}\right]=\text{0}$ (3)

Rearrange equation (3) for the oxidation number of carbon.

$\text{Oxidation number of carbon}=\left[\frac{\begin{array}{l}-2\left(\text{oxidation number of hydrogen}\right)\\ -4\left(\text{oxidation number of oxygen}\right)\end{array}}{2}\right]$ (4)

Substitute $-2$ for oxidation number of oxygen and $+1$ for the oxidation state of hydrogen in equation (4).

$\begin{array}{c}\text{Oxidation number of carbon}=\left[\frac{-2\left(\text{+1}\right)-4\left(-2\right)}{2}\right]\\ =\left[\frac{-2+8}{2}\right]\\ =\left[\frac{+6}{2}\right]\\ =+3\end{array}$

The expression to calculate the oxidation number of carbon in ${\text{CO}}_2$ is:

$\left[2\left(\text{oxidation number of oxygen}\right)+\left(\text{oxidation number of carbon}\right)\right]=0$ (5)

Rearrange equation (5) for the oxidation number of carbon.

$\text{Oxidation number of carbon}=\left[-2\left(\text{oxidation number of oxygen}\right)\right]$ (6)

Substitute $-2$ for the oxidation number of oxygen in equation (4).

$\begin{array}{c}\text{Oxidation number of carbon}=\left[-2\left(-2\right)\right]\\ =+4\end{array}$

The oxidation state of carbon in ${\text{H}}_2{\text{C}}_2{\text{O}}_4$ is $+3$ and it increases to $+4$ in ${\text{CO}}_2$. ${\text{H}}_2{\text{C}}_2{\text{O}}_4$ undergoes oxidation in the reaction, therefore, it acts as the reducing agent.

Conclusion

A reducing agent is the one that reduces another species and itself gets oxidized in a chemical reaction. ${\text{H}}_2{\text{C}}_2{\text{O}}_4$ undergoes oxidation and acts as the reducing agent in the reaction.

(e)

Interpretation Introduction

Interpretation:

The mass percent of ${\text{CaCl}}_2$ in the original sample is to be determined.

Concept introduction:

Precipitation reaction involves the reaction of two soluble ionic compounds to form an insoluble product. The insoluble product is known as a precipitate.

The reason for the precipitation reaction to occur is the formation of a product that is insoluble in nature.

Answer to Problem 4.131P

The mass percent of ${\text{CaCl}}_2$ in the original sample is $55.06\%$.

Explanation of Solution

The given redox reaction is:

${\text{2MnO}}_4^{-}\left(aq\right)+5{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4\left(aq\right)+6{\text{H}}^{+}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+10{\text{CO}}_2\left(g\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The formula to calculate the moles of ${\text{MnO}}_4^{-}$ is,

$\text{Moles}\text{of}{\text{MnO}}_4^{-}=\left({\text{molarity of MnO}}_4^{-}\left(\text{mol/L}\right)\right)\left(\text{Volume of}{\text{MnO}}_4^{-}\left(\text{L}\right)\right)$ (7)

Substitute $0.1019\text{mol/L}$ for the molarity of ${\text{MnO}}_4^{-}$ and $37.68\text{mL}$ for the volume of ${\text{MnO}}_4^{-}$ in the equation (7).

$\begin{array}{c}\text{Moles}\text{of}{\text{MnO}}_4^{-}=\left(0.1019\text{mol/L}\right)\left(37.68\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ =0.0038396\text{mol}\end{array}$

The formula to calculate the moles of ${\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4$ is,

$\text{Moles}\text{of}{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4=\left({\text{moles of MnO}}_4^{-}\left(\text{mol}\right)\right)\left(\frac{5\text{mol}{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4}{2\text{mol}{\text{MnO}}_4^{-}}\right)$ (8)

Substitute $0.0038396\text{mol}$ for the moles of ${\text{MnO}}_4^{-}$ in the equation (8).

$\begin{array}{c}\text{Moles}\text{of}{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4=\left(0.0038396\text{mol}\right)\left(\frac{5\text{mol}{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4}{2\text{mol}{\text{MnO}}_4^{-}}\right)\\ =0.009599\text{mol}\end{array}$

The formula to calculate the mass of ${\text{CaCl}}_2$ is,

$\text{Mass}\text{of}{\text{CaCl}}_2=\left[\begin{array}{l}\left(\text{moles of}{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4\left(\text{mol}\right)\right)\left(\frac{1\text{mol}{\text{CaCl}}_2}{1\text{mol}{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4}\right)\\ \text{molecular}\text{mass}\text{of}{\text{CaCl}}_2\left(\text{g/mol}\right)\end{array}\right]$                              (9)

Substitute $0.009599\text{mol}$ for moles of ${\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4$ and $110.98\text{g/mol}$ for molecular mass of ${\text{CaCl}}_2$ in the equation (2).

$\begin{array}{c}\text{Mass}\text{of}{\text{CaCl}}_2=\left[\begin{array}{l}\left(0.009599\text{mol}\right)\left(\frac{1\text{mol}{\text{CaCl}}_2}{1\text{mol}{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_4}\right)\\ \left(110.98\text{g/mol}\right)\end{array}\right]\\ =1.06530\text{g}\end{array}$

The expression to calculate the mass percent of ${\text{CaCl}}_2$ is,

$\text{mass }\%\text{of}{\text{CaCl}}_2=\left(\frac{\text{mass}{\text{of CaCl}}_2\left(\text{g}\right)}{\text{mass of sample}\left(\text{g}\right)}\right)\left(100\right)$ (10)

Substitute $1.06530\text{g}$ for the mass of ${\text{CaCl}}_2$ and $1.9348\text{g}$ for the mass of sample in the equation (10).

$\begin{array}{c}\text{Mass }\%\text{of}{\text{CaCl}}_2=\left(\frac{1.06530\text{g}}{1.9348\text{g}}\right)\left(100\right)\\ =55.05995\%\\ \approx 55.06\%\end{array}$

Conclusion

The mass percent of ${\text{CaCl}}_2$ in the original sample is $55.06\%$.