Chapter 16, Problem 8PS

In each of the following acid-base reactions, identify the Brønsted acid and base on the left and their conjugate partners on the right.

1. (a) C₂H₅N(aq) + CH₃CO₂H(aq) ⇄ C₅H₅NH⁺(aq) + CH₃CO₂⁻(aq)
2. (b) N₂H₄(aq) + HSO₄⁻(aq) ⇄ N₂H₅⁺(aq) + SO₄²⁻(aq)
3. (c) [Al(H₂O)₆]³⁺ (aq) + OH⁻(aq) ⇄ [Al(H₂O)₅OH]²⁺ (aq) + H₂O⁺(ℓ)

Interpretation Introduction

Interpretation:

Bronsted acid and base on the left and their conjugate partners on the right has to be identified.

${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N (aq) + CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H (l) }\rightleftharpoons {\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{\text{+}}{\text{(aq) + CH}}_{\text{3}}{\text{CO}}_{\text{2}}^{\text{-}}\text{(aq)}$

Concept Introduction:

Bronsted –Lowry definitions:

A Bronsted –Lowry acid is a proton donor, it donates a hydrogen ion, ${\text{H}}^{\text{+}}$

A Bronsted-Lowry base is a proton acceptor, it accepts a hydrogen ion ${\text{H}}^{\text{+}}$.

Bronsted –Lowry conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $\text{(HA)}$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

Answer to Problem 8PS

${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N (aq) + CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H (l) }\rightleftharpoons {\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{\text{+}}{\text{(aq) + CH}}_{\text{3}}{\text{CO}}_{\text{2}}^{\text{-}}\text{(aq)}$

From the reaction the Bronsted acid and Bronsted base is as follows.

$\begin{array}{l}{\text{CH}}_{\text{3}}\text{COOH - Bronsted acid}\\ {\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N - Bronsted base}\\ {\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{\text{+}}\text{ - Conjugate acid}\\ {\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{ - Conjugate base}\end{array}$

Explanation of Solution

${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N (aq) + CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H (l) }\rightleftharpoons {\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{\text{+}}{\text{(aq) + CH}}_{\text{3}}{\text{CO}}_{\text{2}}^{\text{-}}\text{(aq)}$

In the above reaction,

${\text{CH}}_3\text{COOH}$ is donated a proton to ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}$. Therefore, ${\text{CH}}_3\text{COOH}$ is a Bronsted acid.

${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}$ is accepted a proton from ${\text{CH}}_3\text{COOH}$. Therefore,${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{NH}$ is a Bronsted base.

The conjugate base of ${\text{CH}}_{\text{3}}\text{COOH}$ is ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}$ because it has one proton less than ${\text{CH}}_{\text{3}}\text{COOH}$.

The conjugate acid of ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}$ is ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{\text{+}}$ because it has one proton is more than ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}$.

Therefore, from the reaction the Bronsted acid and Bronsted base is as follows.

$\begin{array}{l}{\text{CH}}_{\text{3}}\text{COOH - Bronsted acid}\\ {\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N - Bronsted base}\\ {\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{\text{+}}\text{ - Conjugate acid}\\ {\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}\text{ - Conjugate base}\end{array}$

Interpretation Introduction

Interpretation:

Bronsted acid and base on the left and their conjugate partners on the right has to be identified.

${\text{ N}}_{\text{2}}{\text{H}}_{\text{4}}{\text{(aq) + HSO}}_{\text{4}}^{\text{-}}\text{ (aq) }\rightleftharpoons {\text{N}}_{\text{2}}{\text{H}}_{\text{5}}^{\text{+}}{\text{ (aq) + SO}}_{\text{4}}^{\text{2-}}\text{(aq)}$

Concept Introduction:

Bronsted –Lowry definitions:

A Bronsted –Lowry acid is a proton donor, it donates a hydrogen ion, ${\text{H}}^{\text{+}}$

A Bronsted-Lowry base is a proton acceptor, it accepts a hydrogen ion ${\text{H}}^{\text{+}}$.

Bronsted –Lowry conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $\text{(HA)}$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

Answer to Problem 8PS

${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}{\text{(aq) + HSO}}_{\text{4}}^{\text{-}}\text{ (aq) }\rightleftharpoons {\text{N}}_{\text{2}}{\text{H}}_{\text{5}}^{\text{+}}{\text{ (aq) + SO}}_{\text{4}}^{\text{2-}}\text{(aq)}$

From the reaction the Bronsted acid and Bronsted base is as follows.

$\begin{array}{l}{\text{HSO}}_4^{-}\text{ - Bronsted acid}\\ {\text{N}}_2{\text{H}}_4\text{ - Bronsted base}\\ {\text{N}}_2{\text{H}}_5^{+}\text{ - Conjugate acid}\\ {\text{SO}}_4^{2-}\text{ - Conjugate base}\end{array}$

Explanation of Solution

${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}{\text{(aq) + HSO}}_{\text{4}}^{\text{-}}\text{ (aq) }\rightleftharpoons {\text{N}}_{\text{2}}{\text{H}}_{\text{5}}^{\text{+}}{\text{ (aq) + SO}}_{\text{4}}^{\text{2-}}\text{(aq)}$

In the above reaction,

${\text{HSO}}_{\text{4}}^{\text{-}}$ is donated a proton to ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$. Therefore, ${\text{HSO}}_{\text{4}}^{\text{-}}$ is a Bronsted acid.

${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ is accepted a proton from ${\text{N}}_{\text{2}}{\text{H}}_{\text{5}}^{\text{+}}$. Therefore, ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ is a Bronsted base.

The conjugate base of ${\text{HSO}}_{\text{4}}^{\text{-}}$ is ${\text{SO}}_{\text{4}}^{\text{2-}}$ because ${\text{SO}}_{\text{4}}^{\text{2-}}$ has one proton less than ${\text{HSO}}_{\text{4}}^{\text{-}}$.

The conjugate acid of ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ is ${\text{N}}_2{\text{H}}_{\text{5}}^{\text{+}}$ because ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ has one more proton than ${\text{N}}_2{\text{H}}_{\text{5}}^{\text{+}}$.

Therefore, From the reaction the Bronsted acid and Bronsted base is as follows.

$\begin{array}{l}{\text{HSO}}_4^{-}\text{ - Bronsted acid}\\ {\text{N}}_2{\text{H}}_4\text{ - Bronsted base}\\ {\text{N}}_2{\text{H}}_5^{+}\text{ - Conjugate acid}\\ {\text{SO}}_4^{2-}\text{ - Conjugate base}\end{array}$

Interpretation Introduction

Interpretation:

Bronsted acid and base on the left and their conjugate partners on the right has to be identified.

${{\text{[Al(H}}_{\text{2}}{\text{O)}}_{\text{6}}\text{]}}^{\text{3+}}{\text{(aq) + OH}}^{\text{-}}\text{ (aq) }\rightleftharpoons {{\text{[Al(H}}_{\text{2}}{\text{O)}}_{\text{5}}\text{OH]}}^{\text{2+}}{\text{(aq) + H}}_{\text{2}}\text{O (l)}$

Concept Introduction:

Bronsted –Lowry definitions:

A Bronsted –Lowry acid is a proton donor, it donates a hydrogen ion, ${\text{H}}^{\text{+}}$

A Bronsted-Lowry base is a proton acceptor, it accepts a hydrogen ion ${\text{H}}^{\text{+}}$.

Bronsted –Lowry conjugate acid-Base pairs:

When an acid is dissolved in water, the acid $\text{(HA)}$ donates a proton to water to form a new acid (conjugate acid) and a new base (conjugate base).

Answer to Problem 8PS

${{\text{[Al(H}}_{\text{2}}{\text{O)}}_{\text{6}}\text{]}}^{\text{3+}}{\text{(aq) + OH}}^{\text{-}}\text{ (aq) }\rightleftharpoons {{\text{[Al(H}}_{\text{2}}{\text{O)}}_{\text{5}}\text{OH]}}^{\text{2+}}{\text{(aq) + H}}_{\text{2}}\text{O (l)}$

From the reaction the Bronsted acid and Bronsted base is as follows.

$\begin{array}{l}{[\text{Al}{({\text{H}}_2\text{O})}_6]}^{3+}\text{ - Bronsted acid}\\ {\text{OH}}^{\text{-}}\text{ - Bronsted base}\\ {[\text{Al}{({\text{H}}_2\text{O})}_5]}^{2+}\text{ - Conjugate acid}\\ {\text{H}}_{\text{2}}\text{O - Conjugate base}\end{array}$

Explanation of Solution

${{\text{[Al(H}}_{\text{2}}{\text{O)}}_{\text{6}}\text{]}}^{\text{3+}}{\text{(aq) + OH}}^{\text{-}}\text{ (aq) }\rightleftharpoons {{\text{[Al(H}}_{\text{2}}{\text{O)}}_{\text{5}}\text{OH]}}^{\text{2+}}{\text{(aq) + H}}_{\text{2}}\text{O (l)}$

In the above reaction,

${{\text{[Al(H}}_{\text{2}}{\text{O)}}_{\text{6}}\text{]}}^{\text{3+}}$ is donated a proton to ${\text{OH}}^{\text{-}}$. Therefore, ${{\text{[Al(H}}_{\text{2}}{\text{O)}}_{\text{6}}\text{]}}^{\text{3+}}$ is a Bronsted acid.

${\text{OH}}^{\text{-}}$ is accepted a proton from ${{\text{[Al(H}}_{\text{2}}{\text{O)}}_{\text{6}}\text{]}}^{\text{3+}}$. Therefore, ${\text{OH}}^{\text{-}}$ is a Bronsted base.

The conjugate base of ${{\text{[Al(H}}_{\text{2}}{\text{O)}}_{\text{6}}\text{]}}^{\text{3+}}$ is ${{\text{[Al(H}}_{\text{2}}{\text{O)}}_5\text{]}}^{\text{32+}}$ because ${{\text{[Al(H}}_{\text{2}}{\text{O)}}_{\text{6}}\text{]}}^{\text{3+}}$ has one proton less than ${{\text{[Al(H}}_{\text{2}}{\text{O)}}_{\text{6}}\text{]}}^{\text{3+}}$.

The conjugate acid of ${\text{OH}}^{\text{-}}$ is ${\text{H}}_{\text{2}}\text{O}$ because ${\text{H}}_{\text{2}}\text{O}$ has one more proton than ${\text{OH}}^{\text{-}}$.

Therefore, form the reaction the Bronsted acid and Bronsted base is as follows.

$\begin{array}{l}{[\text{Al}{({\text{H}}_2\text{O})}_6]}^{3+}\text{ - Bronsted acid}\\ {\text{OH}}^{\text{-}}\text{ - Bronsted base}\\ {[\text{Al}{({\text{H}}_2\text{O})}_5]}^{2+}\text{ - Conjugate acid}\\ {\text{H}}_{\text{2}}\text{O - Conjugate base}\end{array}$