Chapter 6, Problem 6.1P

For each of the following reactions identify the acid and the base. Also indicate which acid—base definition (Lewis, Brønsted-Lowry) applies. In some cases, more than one definition may apply.
a. $AlB{r}_3+B{r}^{-}\to AlB{r}_{4^{-}}$

b. $HCl{O}_4+C{H}_{3}CN\to C{H}_{3}CN{H}^{+}+Cl{O}_{4^{-}}$

c. $N{i}^{2+}+6N{H}_3\to {\left[Ni{\left(N{H}_3\right)}_6\right]}^{2+}$

d. $N{H}_3+CIF\to H_{3}N\cdots CIF$

e. $2Cl{O}_{3^{-}}+S{O}_2\to 2Cl{O}_2+S{O}_{4^{2-}}$

f. $C_3{H}_{7}COOH+2HF\to {\left[C_3{H}_{7}C{\left(OH\right)}_2\right]}^{+}+H{F}_{2^{-}}$

Interpretation Introduction

Interpretation: The acid and base need to be identified in the following reaction. Also, the applied acid-base definition needs to be indicated.

${\text{AlBr}}_{\text{3}}+{\text{Br}}^{-}\to {\text{AlBr}}_4^{-}$

Concept Introduction: According to Lewis concept, an acid is an electron pair acceptor. It acts as an electrophile and has vacant orbitals. On the other hand, a base is an electron-pair donor. It acts as a nucleophile and has lone pair of electrons.

According to Bronsted-Lowry acid-base theory, an acid can donate a proton $\left({\text{H}}^{+}\right)$ and a base can accept a proton. Base requires a lone pair of electrons to bond to the proton $\left({\text{H}}^{+}\right)$ .

Answer to Problem 6.1P

$\begin{array}{l}{\text{AlBr}}_{\text{3}}+{\text{Br}}^{-}\to {\text{AlBr}}_4^{-}\\ \text{Lewis Lewis}\\ \text{Acid Base}\end{array}$

Explanation of Solution

The given acid-base reaction is as follows:

${\text{AlBr}}_{\text{3}}+{\text{Br}}^{-}\to {\text{AlBr}}_4^{-}$

In the above reaction, Al in ${\text{AlBr}}_{\text{3}}$ has vacant orbitals (it can act as an electrophile) and ${\text{Br}}^{-}$ has lone pair of electrons, which can act as a nucleophile.

Here, ${\text{AlBr}}_{\text{3}}$ can accept as an electron pair from ${\text{Br}}^{-}$ to form ${\text{AlBr}}_4^{-}$ . Therefore, ${\text{AlBr}}_{\text{3}}$ is an electron pair acceptor or Lewis acid and ${\text{Br}}^{-}$ is an electron-pair donor or Lewis base.

$\begin{array}{l}{\text{AlBr}}_{\text{3}}+{\text{Br}}^{-}\to {\text{AlBr}}_4^{-}\\ \text{Lewis Lewis}\\ \text{Acid Base}\end{array}$

Interpretation Introduction

Interpretation: The acid and base need to be identified in the following reaction. Also, the applied acid-base definition needs to be indicated.

${\text{HClO}}_{\text{4}}+{\text{CH}}_{\text{3}}\text{CN}\to {\text{CH}}_{\text{3}}{\text{CNH}}^{+}+{\text{ClO}}_4^{-}$

Concept Introduction: According to Lewis concept, an acid is an electron pair acceptor. It acts as an electrophile and has vacant orbitals. On the other hand, a base is an electron-pair donor. It acts as a nucleophile and has lone pair of electrons.

According to Bronsted-Lowry acid-base theory, an acid can donate a proton $\left({\text{H}}^{+}\right)$ and a base can accept a proton. Base requires a lone pair of electrons to bond to the proton $\left({\text{H}}^{+}\right)$ .

Answer to Problem 6.1P

$\begin{array}{l}{\text{HClO}}_{\text{4}}+{\text{ CH}}_{\text{3}}\text{CN}\to {\text{CH}}_{\text{3}}{\text{CNH}}^{+}+{\text{ ClO}}_4^{-}\\ \text{Lewis \& Lewis \& Conjugate acid }\text{}\text{Conjugate base}\\ \text{Bronsted Bronsted}\\ \text{Acid Base}\end{array}$

Explanation of Solution

The given acid-base reaction is as follows:

${\text{HClO}}_{\text{4}}+{\text{CH}}_{\text{3}}\text{CN}\to {\text{CH}}_{\text{3}}{\text{CNH}}^{+}+{\text{ClO}}_4^{-}$

Here, ${\text{HClO}}_{\text{4}}$ donates a proton $\left({\text{H}}^{+}\right)$ to ${\text{CH}}_{\text{3}}\text{CN}$ to form ${\text{ClO}}_4^{-}$ and ${\text{CH}}_{\text{3}}{\text{CNH}}^{+}$ . Therefore, ${\text{HClO}}_{\text{4}}$ is a proton donor or Bronsted acid and ${\text{CH}}_{\text{3}}\text{CN}$ is a proton acceptor or Bronsted base.

In the above reaction, ${\text{HClO}}_{\text{4}}$ also accepts an electron pair (after removal of H atom)which results in a negative charge on the O atom. Similarly, ${\text{CH}}_{\text{3}}\text{CN}$ also donates an electron pair (to H atom) which results in a positive charge on the N atom. Thus, ${\text{HClO}}_{\text{4}}$ also act as a Lewis acid and ${\text{CH}}_{\text{3}}\text{CN}$ acts as a Lewis base.

Therefore, ${\text{HClO}}_{\text{4}}$ act as both Lewis and Bronsted acid and ${\text{CH}}_{\text{3}}\text{CN}$ act as both Lewis and Bronsted base.

Interpretation Introduction

Interpretation: The acid and base need to be identified in the following reaction. Also, the applied acid-base definition needs to be indicated.

${\text{Ni}}^{2+}+6{\text{NH}}_{\text{3}}\to {\left[\text{Ni}{\left({\text{NH}}_{\text{3}}\right)}_{\text{6}}\right]}^{2+}$

Concept Introduction: According to Lewis concept, an acid is an electron pair acceptor. It acts as an electrophile and has vacant orbitals. On the other hand, a base is an electron-pair donor. It acts as a nucleophile and has lone pair of electrons.

According to Bronsted-Lowry acid-base theory, an acid can donate a proton $\left({\text{H}}^{+}\right)$ and a base can accept a proton. Base requires a lone pair of electrons to bond to the proton $\left({\text{H}}^{+}\right)$ .

Answer to Problem 6.1P

$\begin{array}{l}{\text{Ni}}^{2+}+6{\text{NH}}_{\text{3}}\to {\left[\text{Ni}{\left({\text{NH}}_{\text{3}}\right)}_{\text{6}}\right]}^{2+}\\ \text{Lewis Lewis}\\ \text{Acid Base}\end{array}$

Explanation of Solution

The given acid-base reaction is as follows:

${\text{Ni}}^{2+}+6{\text{NH}}_{\text{3}}\to {\left[\text{Ni}{\left({\text{NH}}_{\text{3}}\right)}_{\text{6}}\right]}^{2+}$

In the above reaction, Ni has vacant orbitals, thus it can accept electron pairs (act as ab electrophile). Due to the lone pair of electrons on the N atom in ${\text{NH}}_{\text{3}}$ , it can donate electron pair (act as a nucleophile).

Here, ${\text{Ni}}^{2+}$ can accept an electron pair from ${\text{NH}}_{\text{3}}$ to form ${\left[\text{Ni}{\left({\text{NH}}_{\text{3}}\right)}_{\text{6}}\right]}^{2+}$ . Therefore, ${\text{Ni}}^{2+}$ is an electron pair acceptor or Lewis acid and ${\text{NH}}_{\text{3}}$ is an electron-pair donor or Lewis base.

This can be represented as follows:

$\begin{array}{l}{\text{Ni}}^{2+}+6{\text{NH}}_{\text{3}}\to {\left[\text{Ni}{\left({\text{NH}}_{\text{3}}\right)}_{\text{6}}\right]}^{2+}\\ \text{Lewis Lewis}\\ \text{Acid Base}\end{array}$

Interpretation Introduction

Interpretation: The acid and base need to be identified in the following reaction. Also, the applied acid-base definition needs to be indicated.

${\text{NH}}_{\text{3}}+\text{ClF}\to {\text{H}}_{\text{2}}\text{N}\cdots \text{ClF}$

Concept Introduction:

According to Lewis concept, an acid is an electron pair acceptor. It acts as an electrophile and has vacant orbitals. On the other hand, a base is an electron-pair donor. It acts as a nucleophile and has lone pair of electrons.

According to Bronsted-Lowry acid-base theory, an acid can donate a proton $\left({\text{H}}^{+}\right)$ and a base can accept a proton. Base requires a lone pair of electrons to bond to the proton $\left({\text{H}}^{+}\right)$ .

Answer to Problem 6.1P

$\begin{array}{l}{\text{NH}}_{\text{3}}+\text{ClF}\to {\text{H}}_{\text{2}}\text{N}\cdots \text{ClF}\\ \text{Lewis Lewis}\\ \text{Base Acid}\end{array}$

Explanation of Solution

The given acid-base reaction is as follows:

${\text{NH}}_{\text{3}}+\text{ClF}\to {\text{H}}_{\text{2}}\text{N}\cdots \text{ClF}$

In the above reaction, due to the lone pair of electrons on the N atom in ${\text{NH}}_{\text{3}}$ it can act as a nucleophile and donate an electron pair to ClF.

Here, ${\text{NH}}_{\text{3}}$ can donate an electron pair to $\text{ClF}$ to form ${\text{H}}_{\text{2}}\text{N}\cdots \text{ClF}$ type of bond. Therefore, $\text{ClF}$ is an electron pair acceptor or Lewis acid and ${\text{NH}}_{\text{3}}$ is an electron-pair donor or Lewis base.

This can be represented as follows:

$\begin{array}{l}{\text{NH}}_{\text{3}}+\text{ClF}\to {\text{H}}_{\text{2}}\text{N}\cdots \text{ClF}\\ \text{Lewis Lewis}\\ \text{Base Acid}\end{array}$

Interpretation Introduction

Interpretation: The acid and base need to be identified in the following reaction. Also, the applied acid-base definition needs to be indicated.

$2{\text{ClO}}_2^{-}+{\text{SO}}_2\to {\text{2ClO}}_{\text{2}}+{\text{SO}}_4^{2-}$

Concept Introduction: According to Lewis concept, an acid is an electron pair acceptor. It acts as an electrophile and has vacant orbitals. On the other hand, a base is an electron-pair donor. It acts as a nucleophile and has lone pair of electrons.

According to Bronsted-Lowry acid-base theory, an acid can donate a proton $\left({\text{H}}^{+}\right)$ and a base can accept a proton. Base requires a lone pair of electrons to bond to the proton $\left({\text{H}}^{+}\right)$ .

Answer to Problem 6.1P

$\begin{array}{l}2{\text{ClO}}_3^{-}+{\text{SO}}_2\to {\text{2ClO}}_{\text{2}}+{\text{SO}}_4^{2-}\\ \text{Lewis Lewis}\\ \text{Acid Base}\end{array}$

Explanation of Solution

The given acid-base reaction is as follows:

$2{\text{ClO}}_3^{-}+{\text{SO}}_2\to {\text{2ClO}}_{\text{2}}+{\text{SO}}_4^{2-}$

The above reaction is a redox reaction. Here, the oxidation state of S in ${\text{SO}}_{\text{2}}$ increases from +4 to +6 in ${\text{SO}}_4^{2-}$ , and the oxidation state of Cl in ${\text{ClO}}_3^{-}$ decreases from +5 to +4 in ${\text{ClO}}_{\text{2}}$ .

The oxidation state of S changes from +4 to +6 as S donates an electron pair. Similarly, the oxidation state of Cl changes from +5 to +4 because it accepts an electron pair (there are 2 ${\text{ClO}}_{\text{3}}{}^{-}$ on the reactant side).

Therefore, ${\text{ClO}}_{\text{3}}{}^{-}$ is an electron pair acceptor or Lewis acid and ${\text{SO}}_{\text{2}}$ is an electron-pair donor or Lewis base.

This can be represented as follows:

$\begin{array}{l}2{\text{ClO}}_3^{-}+{\text{SO}}_2\to {\text{2ClO}}_{\text{2}}+{\text{SO}}_4^{2-}\\ \text{Lewis Lewis}\\ \text{Acid Base}\end{array}$

Interpretation Introduction

Interpretation: The acid and base needto be identified in the following reaction. Also, the applied acid-base definition needs to be indicated.

${\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{COOH}+2\text{HF}\to {\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{C}{\left(\text{OH}\right)}_2\right]}^{+}+{\text{HF}}_2^{-}$

Concept Introduction: According to Lewis concept, an acid is an electron pair acceptor. It acts as an electrophile and has vacant orbitals. On the other hand, a base is an electron-pair donor. It acts as a nucleophile and has lone pair of electrons.

According to Bronsted-Lowry acid-base theory, an acid can donate a proton $\left({\text{H}}^{+}\right)$ and a base can accept a proton. Base requires a lone pair of electrons to bond to the proton $\left({\text{H}}^{+}\right)$ .

Answer to Problem 6.1P

$\begin{array}{l}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{COOH}+2\text{HF}\to {\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{C}{\left(\text{OH}\right)}_2\right]}^{+}+{\text{HF}}_2^{-}\\ \text{Lewis \& Lewis \& Conjugate acid Conjugate Base}\\ \text{Bronsted Bronsted}\\ \text{Base Acid}\end{array}$

Explanation of Solution

The given acid-base reaction is as follows:

${\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{COOH}+2\text{HF}\to {\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{C}{\left(\text{OH}\right)}_2\right]}^{+}+{\text{HF}}_2^{-}$

In the above reaction, ${\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{COOH}$ accepts a proton from $2\text{HF}$ to form ${\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{C}{\left(\text{OH}\right)}_2\right]}^{+}$ complex and ${\text{HF}}_2^{-}$ . Thus, ${\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{COOH}$ acts as a proton acceptor or Bronsted base and $2\text{HF}$ acts as a proton donor or Bronsted acid.

Here, ${\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{COOH}$ also donates electron pair from oxygen atom to form O-H bond. Similarly, in $\text{HF}$ , F also accepts electron pair which results in a negative charge on the F atom.

Therefore, HF act as both Bronsted-Lowry and Lewis acid and ${\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{COOH}$ act as both Bronsted-Lowry and Lewis base.