Chapter 16.1, Problem 2PPA

Identify and label the species in each reaction.

(a) ${\text{NH}}_{\text{4}}^{\text{+}}(aq)+{\text{H}}_{\text{2}}\text{O}(l)\rightleftarrows {\text{NH}}_{\text{3}}(aq)+{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}(aq)$

(b) ${\text{CN}}^{\text{-}}(aq)+{\text{H}}_{\text{2}}\text{O}(l)\rightleftarrows \text{HCN}(aq)+{\text{OH}}^{-}(aq)$

Interpretation Introduction

Interpretation:

The species in each of the given reaction has to be identified and labeled.

Concept Information:

When a Bronsted acid donates a proton, what remains of the acid is known as a conjugate base; when a Bronsted base accepts a proton, the newly formed protonated species is known as a conjugate acid.  This can be given by the below equation as shown in Figure 1.

Figure 1

The reactant that loses a proton is the acid and the reactant that gains a proton is the base.

Each product is the conjugate of one of the reactant.

Two species that differ only by a proton constitute a conjugate pair

Addition of proton to a species gives its conjugate acid whereas removal of proton from a species gives its conjugate base.

To Label: The equation ${\text{NH}}_{\text{4}}{{}^{\text{+}}}_{\text{(aq)}}{\text{ + H}}_2{\text{O}}_{\text{(l)}}\rightleftarrows {\text{NH}}_{\text{3}}{}_{\text{(aq)}}{\text{ + H}}_3{\text{O}}^{\text{+}}{}_{\text{(aq)}}$ as acid, base, conjugate acid or conjugate base.

Answer to Problem 2PPA

Answer

$\begin{array}{l}{\text{(a) NH}}_{\text{4}}{{}^{\text{+}}}_{\text{(aq)}}{\text{ + H}}_2{\text{O}}_{\text{(l)}}\rightleftarrows {\text{NH}}_{\text{3}}{}_{\text{(aq)}}\text{ + }{\text{H}}_3{\text{O}}^{\text{+}}{}_{\text{(aq)}}\\ \text{ acid base conjugate conjugate}\\ \text{ base acid}\end{array}$

Explanation of Solution

${\text{NH}}_{\text{4}}{}^{\text{+}}$ loses a proton and becomes ${\text{NH}}_3$; ${\text{H}}_2\text{O}$ gains a proton and becomes ${\text{H}}_3{\text{O}}^{\text{+}}$

Hence, ${\text{NH}}_{\text{4}}{}^{\text{+}}$ is acid and ${\text{H}}_2\text{O}$ is base.

${\text{NH}}_{\text{3}}$ loses one proton and becomes conjugate base of ${\text{NH}}_{\text{4}}{}^{\text{+}}$

${\text{H}}_3{\text{O}}^{\text{+}}$ gains one proton and become conjugate acid of ${\text{H}}_2\text{O}$

Therefore, the given equation is labeled as,

$\begin{array}{l}{\text{NH}}_{\text{4}}{{}^{\text{+}}}_{\text{(aq)}}{\text{ + H}}_2{\text{O}}_{\text{(l)}}\rightleftarrows {\text{NH}}_{\text{3}}{}_{\text{(aq)}}\text{ + }{\text{H}}_3{\text{O}}^{\text{+}}{}_{\text{(aq)}}\\ \text{acid base conjugate conjugate}\\ \text{ base acid}\end{array}$

Interpretation Introduction

Interpretation:

The species in each of the given reaction has to be identified and labeled.

Concept Information:

When a Bronsted acid donates a proton, what remains of the acid is known as a conjugate base; when a Bronsted base accepts a proton, the newly formed protonated species is known as a conjugate acid.  This can be given by the below equation as shown in Figure 1.

Figure 1

The reactant that loses a proton is the acid and the reactant that gains a proton is the base.

Each product is the conjugate of one of the reactant.

Two species that differ only by a proton constitute a conjugate pair

Addition of proton to a species gives its conjugate acid whereas removal of proton from a species gives its conjugate base.

To Label: The equation ${\text{CN}}^{\text{-}}{}_{\text{(aq)}}{\text{ + H}}_2{\text{O}}_{\text{(l)}}\rightleftarrows {\text{HCN}}_{\text{(aq)}}{\text{ + OH}}^{\text{-}}{}_{\text{(aq)}}$ as acid, base, conjugate acid or conjugate base.

Answer to Problem 2PPA

Answer

$\begin{array}{l}{\text{(b) CN}}^{\text{-}}{}_{\text{(aq)}}{\text{ + H}}_2{\text{O}}_{\text{(l)}}\rightleftarrows {\text{HCN}}_{\text{(aq)}}\text{+ }{\text{OH}}^{\text{-}}{}_{\text{(aq)}}\\ \text{ base acid conjugate conjugate}\\ \text{ acid base}\end{array}$

Explanation of Solution

${\text{H}}_2\text{O}$ loses a proton and becomes ${\text{ OH}}^{\text{-}}$; ${\text{CN}}^{\text{-}}$ gains a proton and becomes $\text{HCN}$

Hence, ${\text{H}}_2\text{O}$ is acid and ${\text{CN}}^{\text{-}}$ is base.

${\text{OH}}^{\text{-}}$ loses one proton and becomes conjugate base of ${\text{H}}_2\text{O}$

$\text{HCN}$ gains one proton and become conjugate acid of ${\text{CN}}^{\text{-}}$

Therefore, the given equation is labeled as,

$\begin{array}{l}{\text{CN}}^{\text{-}}{}_{\text{(aq)}}{\text{ + H}}_2{\text{O}}_{\text{(l)}}\rightleftarrows {\text{HCN}}_{\text{(aq)}}\text{+ }{\text{OH}}^{\text{-}}{}_{\text{(aq)}}\\ \text{base acid conjugate conjugate}\\ \text{ acid base}\end{array}$