Chapter 15, Problem 15.109QP

(a)

Interpretation Introduction

Interpretation:

A chemical equation for the given information has to be written and the conjugate acid, base has to be identified.

Concept introduction:

Acid:

According to Bronsted-Lowry theory, a species which donates a proton in a proton transfer-reaction is said to be an acid.

Base:

According to Bronsted-Lowry theory, a species which accepts a proton in a proton transfer-reaction is said to be base.

Conjugated acid:

In the conjugate acid-base pair, a species which can donates a proton is said to be conjugated acid.

Conjugated base:

In the conjugate acid-base pair, a species which can accepts a proton is said to be conjugated base.

Answer to Problem 15.109QP

The chemical equation and its conjugate acid, base is

$\begin{array}{l}{{\text{[Al(H}}_{\text{2}}}^{{\text{O)}}_{\text{6}}}{\text{ + NH}}_{\text{3}}\stackrel{}{\to }{{\text{[Al(H}}_{\text{2}}}^{{\text{O)}}_{\text{5}}}{\text{+ NH}}_{\text{4}}{}^{\text{+}}\\ \text{acid}\text{base}\text{base}\text{acid}\end{array}$

The conjugate acid-base pairs are $\left({{\text{[Al(H}}_{\text{2}}{\text{O)}}_{\text{6}}\text{]}}^{\text{3+}},{{\text{[Al(H}}_{\text{2}}{\text{O)}}_{\text{5}}\text{OH]}}^{\text{2+ }}\right)\text{ and }\left({\text{NH}}_{\text{4}}{}^{\text{+}}{\text{,NH}}_{\text{3}}\right)$

Explanation of Solution

The hydrated aluminum ion acts as Bronsted-Lowry acid and it can react with ammonia a Bronsted-Lowry base to produce as mentioned ions in the following reaction.

$\begin{array}{l}{{\text{[Al(H}}_{\text{2}}}^{{\text{O)}}_{\text{6}}}{\text{ + NH}}_{\text{3}}\stackrel{}{\to }{{\text{[Al(H}}_{\text{2}}}^{{\text{O)}}_{\text{5}}}{\text{+ NH}}_{\text{4}}{}^{\text{+}}\\ \text{acid}\text{base}\text{base}\text{acid}\end{array}$

The conjugate acid-base pairs are $\left({{\text{[Al(H}}_{\text{2}}{\text{O)}}_{\text{6}}\text{]}}^{\text{3+}},{{\text{[Al(H}}_{\text{2}}{\text{O)}}_{\text{5}}\text{OH]}}^{\text{2+ }}\right)\text{ and }\left({\text{NH}}_{\text{4}}{}^{\text{+}}{\text{,NH}}_{\text{3}}\right)$

(b)

Interpretation Introduction

Interpretation:

A chemical equation for the given information has to be written and the conjugate acid, base has to be identified.

Concept introduction:

Acid:

According to Bronsted-Lowry theory, a species which donates a proton in a proton transfer-reaction is said to be an acid.

Base:

According to Bronsted-Lowry theory, a species which accepts a proton in a proton transfer-reaction is said to be base.

Conjugated acid:

In the conjugate acid-base pair, a species which can donates a proton is said to be conjugated acid.

Conjugated base:

In the conjugate acid-base pair, a species which can accepts a proton is said to be conjugated base.

Answer to Problem 15.109QP

The chemical equation and its conjugate acid, base is

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{SH + H}}_{\text{2}}{\text{SO}}_{\text{4}}\stackrel{}{\to }{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{SH}}_{\text{2}}{}^{\text{+}}{\text{ + HSO}}_{\text{4}}{}^{\text{-}}\\ \text{base}\text{acid}\text{acid}\text{base}\end{array}$

The conjugate acid-base pairs are $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{SH}}_{\text{2}}{}^{\text{+}}{\text{, CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{SH}\right)\text{ and }\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}},{\text{HSO}}_{\text{4}}{}^{\text{-}}\right)$

Explanation of Solution

Ethanethiol acts as Bronsted-Lowry base and it can react with sulfuric acid a Bronsted-Lowry acid to produce as mentioned ions in the following reaction.

$\begin{array}{l}{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{SH + H}}_{\text{2}}{\text{SO}}_{\text{4}}\stackrel{}{\to }{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{SH}}_{\text{2}}{}^{\text{+}}{\text{ + HSO}}_{\text{4}}{}^{\text{-}}\\ \text{base}\text{acid}\text{acid}\text{base}\end{array}$

The conjugate acid-base pairs are $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{SH}}_{\text{2}}{}^{\text{+}}{\text{, CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{SH}\right)\text{ and }\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}},{\text{HSO}}_{\text{4}}{}^{\text{-}}\right)$