Chapter 15, Problem 15.99AP

(a)

Interpretation Introduction

Interpretation: Whether the given reaction is a redox process is to be explained. In the given reaction acid, base, conjugate acid and the conjugate base are to be identified.

Concept introduction: An acid is defined as a compound which donates a proton and a proton acceptor is known as base according to Bronsted-Lowry theory. The reaction that shows the change of oxidation state between reactant and product is known as redox reaction.

To determine: If the given reaction is a redox process.

Answer to Problem 15.99AP

Solution

No, the given reaction is not a redox process.

Explanation of Solution

Explanation

The given reaction is,

${\text{HNO}}_{\text{3}}\left(aq\right)+{\text{2H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\stackrel{}{\to }{\text{NO}}_{\text{2}}^{\text{+}}\left(aq\right)+{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)+{\text{2HSO}}_4^{-}\left(aq\right)$

The oxidation number of nitrogen and sulfur atoms in reactant and product side is calculated to check if it is a redox reaction. When it shows the change in oxidation state between reactant and product then it is a redox reaction. Therefore, the oxidation number of nitrogen in ${\text{HNO}}_{\text{3}}$ is calculated as,

The oxidation state of hydrogen in a compound is usually $+1$ as it contains only one valence electron and of oxygen is usually $-2$ as it requires two more electrons in order to achieve stable electronic configuration. The sum of all oxidation numbers in a neutral compound is zero. The oxidation number of N is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on HNO}}_{\text{3}}=\left[\begin{array}{c}\left(\text{Number of H atoms}\times \text{oxidation number of H}\right)+\\ \left(\text{Number of N atoms}\times \text{oxidation number of N}\right)+\\ \left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on ${\text{HNO}}_{\text{3}}$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of nitrogen.

$\begin{array}{c}{\text{Charge on HNO}}_{\text{3}}=\left[\left(1\times \left(+1\right)\right)+1\times \left(x\right)+3\left(-2\right)\right]\\ 0=1+x-6\\ 0=x-5\\ x=5\end{array}$

Hence, the oxidation number of nitrogen is $+5$.

Similarly, in ${\text{NO}}_{\text{2}}^{\text{+}}$. The oxidation number of N is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on NO}}_2^{+}=\left[\begin{array}{c}\left(\text{Number of N atoms}\times \text{oxidation number of N}\right)+\\ \left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of nitrogen.

$\begin{array}{c}{\text{Charge on NO}}_2^{+}=\left[1\times \left(x\right)+2\left(-2\right)\right]\\ +1=x-4\\ x=4+1\\ x=5\end{array}$

Hence, the oxidation number of nitrogen is $+5$.

The oxidation number of sulfur in ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is calculated as,

The oxidation state of oxygen is usually $-2$ as it requires two more electrons in order to achieve stable electronic configuration and of hydrogen is $+1$ as it contains one valence electron. The oxidation number of $\text{S}$ is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on H}}_2{\text{SO}}_4=\left[\begin{array}{c}\left(\text{Number of H atoms}\times \text{oxidation number of H}\right)+\\ \left(\text{Number of S atoms}\times \text{oxidation number of S}\right)+\\ \left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$\begin{array}{c}{\text{Charge on H}}_2{\text{SO}}_4=\left[\left(2\times 1\right)+\left(1\times x\right)+\left(4\times \left(-2\right)\right)\right]\\ 0=2+x+\left(-8\right)\\ 0=x-6\\ x=6\end{array}$

Hence, the oxidation number of sulfur is $+6$.

Similarly, in ${\text{HSO}}_4^{-}$. The oxidation number of $\text{S}$ is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on HSO}}_4^{-}=\left[\begin{array}{c}\left(\text{Number of H atoms}\times \text{oxidation number of H}\right)+\\ \left(\text{Number of S atoms}\times \text{oxidation number of S}\right)+\\ \left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$\begin{array}{c}{\text{Charge on HSO}}_4^{-}=\left[\left(1\times 1\right)+\left(1\times x\right)+\left(4\times \left(-2\right)\right)\right]\\ -1=1+x+\left(-8\right)\\ -1=x-7\\ x=6\end{array}$

Hence, the oxidation number of sulfur is $+6$.

There is no change in the oxidation number of the atoms involved in the reaction. Hence, the given reaction is not a redox reaction.

(b)

Interpretation Introduction

To determine: The acid, base, conjugate acid and conjugate base in the given reaction.

Answer to Problem 15.99AP

Solution

The acid is ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, ${\text{HNO}}_{\text{3}}$ is a base, ${\text{HSO}}_4^{-}$ is a conjugate base and ${\text{NO}}_{\text{2}}^{\text{+}}$ is a conjugate acid.

Explanation of Solution

Explanation

According to Bronsted-Lowry theory, when an acid donates a proton the species formed is known as conjugate base and when the base accepts a proton the species formed is known as conjugate acid. The given reaction is,

${\text{HNO}}_{\text{3}}\left(aq\right)+{\text{2H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\stackrel{}{\to }{\text{NO}}_{\text{2}}^{\text{+}}\left(aq\right)+{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)+{\text{2HSO}}_4^{-}\left(aq\right)$

In the reaction, ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ act as an acid as it donates a proton forming ${\text{HSO}}_4^{-}$ as a conjugate base. Nitric acid $\left({\text{HNO}}_{\text{3}}\right)$ is a base and forms ${\text{NO}}_{\text{2}}^{\text{+}}$ conjugate acid.

Conclusion

1. a. The given reaction is not a redox process.
2. b. In the given reaction, ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is an acid, ${\text{HNO}}_{\text{3}}$ is a base, ${\text{HSO}}_4^{-}$ is a conjugate base and ${\text{NO}}_{\text{2}}^{\text{+}}$ is a conjugate acid.