Chapter 13, Problem 1QAP

For each of the following reactions, indicate the Brønsted-Lowry acids and bases. What are the conjugate acid/base pairs?

(a) ${\text{H}}_3{\text{O}}^{+}\left(aq\right)+{\text{CN}}^{-}\left(aq\right)\rightleftharpoons \text{HCN}\left(aq\right)+{\text{H}}_2\text{O}$

(b) ${\text{HNO}}_{\text{2}}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\rightleftharpoons {\text{NO}}_2{}^{-}\left(aq\right)+{\text{H}}_2\text{O}$

(c) ${\text{HCHO}}_{\text{2}}\left(aq\right)+{\text{H}}_2\text{O}\rightleftharpoons {\text{CHO}}_2{}^{-}\left(aq\right)+{\text{H}}_3{\text{O}}^{\text{+}}\left(aq\right)$

Interpretation Introduction

Interpretation: The Bronsted-Lowry acids, bases and conjugate acid/base pairs should be indicated for the following reaction:

  ${\text{H}}_{\text{3}}{\text{O}}^{+}\left(aq\right)+{\text{CN}}^{-}\left(aq\right)\rightleftharpoons \text{HCN}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}$

Concept Introduction:According Bronsted-Lowry acid and base theory, acids are substance which loses protons ${\text{H}}^{+}$ to form conjugate base and bases are substances which accepts protons to from conjugate acid.

For example:

  $\text{HA}\to {\text{H}}^{+}+{\text{A}}^{-}$

Here, HA is an acid as it donates a proton to form ${\text{A}}^{-}$ a conjugate base.

Similarly,

  ${\text{A}}^{-}+{\text{H}}^{+}\to \text{HA}$

Here, ${\text{A}}^{-}$ is a base as it accepts a proton to from HA which is a conjugate acid.

Answer to Problem 1QAP

  $\begin{array}{l}{\text{H}}_{\text{3}}{\text{O}}^{+}\left(aq\right)+{\text{CN}}^{-}\left(aq\right)\rightleftharpoons \text{HCN}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\\ \text{ A B CA CB}\end{array}$

Explanation of Solution

The given reaction is as follows:

  ${\text{H}}_{\text{3}}{\text{O}}^{+}\left(aq\right)+{\text{CN}}^{-}\left(aq\right)\rightleftharpoons \text{HCN}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}$

Here, ${\text{H}}_{\text{3}}{\text{O}}^{+}\left(aq\right)$ donates a proton to form ${\text{H}}_{\text{2}}\text{O}$ thus, it is an acid and ${\text{H}}_{\text{2}}\text{O}$ so formed is a conjugate base similarly, ${\text{CN}}^{-}\left(aq\right)$ accepts a proton to from $\text{HCN}\left(aq\right)$ thus, it is a base and $\text{HCN}\left(aq\right)$ so formed is a conjugate acid.

Therefore, in the given reaction, acid (A), base (B), conjugate acid (CA) and conjugate base (CB) are as follows:

  $\begin{array}{l}{\text{H}}_{\text{3}}{\text{O}}^{+}\left(aq\right)+{\text{CN}}^{-}\left(aq\right)\rightleftharpoons \text{HCN}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\\ \text{ A B CA CB}\end{array}$

Interpretation Introduction

Interpretation: The Bronsted-Lowry acids, bases and conjugate acid/base pairs should be indicated for the following reaction:

  ${\text{HNO}}_{\text{2}}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\rightleftharpoons {\text{NO}}_2^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}$

Concept Introduction: According Bronsted-Lowry acid and base theory, acids are substance which loses protons ${\text{H}}^{+}$ to form conjugate base and bases are substances which accepts protons to from conjugate acid.

For example:

  $\text{HA}\to {\text{H}}^{+}+{\text{A}}^{-}$

Here, HA is an acid as it donates a proton to form ${\text{A}}^{-}$ a conjugate base.

Similarly,

  ${\text{A}}^{-}+{\text{H}}^{+}\to \text{HA}$

Here, ${\text{A}}^{-}$ is a base as it accepts a proton to from HA which is a conjugate acid.

Answer to Problem 1QAP

  $\begin{array}{l}{\text{HNO}}_{\text{2}}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\rightleftharpoons {\text{NO}}_2^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\\ \text{ A B CB CA}\end{array}$

Explanation of Solution

The given reaction is as follows:

  ${\text{HNO}}_{\text{2}}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\rightleftharpoons {\text{NO}}_2^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}$

Here, ${\text{HNO}}_{\text{2}}\left(aq\right)$ donates a proton to form ${\text{NO}}_2^{-}\left(aq\right)$ thus, it is an acid and ${\text{NO}}_2^{-}\left(aq\right)$ so formed is a conjugate base similarly, ${\text{OH}}^{-}\left(aq\right)$ accepts a proton to from ${\text{H}}_{\text{2}}\text{O}$ thus, it is a base and ${\text{H}}_{\text{2}}\text{O}$ so formed is a conjugate acid.

Therefore, in the given reaction, acid (A), base (B), conjugate acid (CA) and conjugate base (CB) are as follows:

  $\begin{array}{l}{\text{HNO}}_{\text{2}}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\rightleftharpoons {\text{NO}}_2^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\\ \text{ A B CB CA}\end{array}$

Interpretation Introduction

Interpretation: The Bronsted-Lowry acids, bases and conjugate acid/base pairs should be indicated for the following reaction:

  ${\text{HCHO}}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\rightleftharpoons {\text{CHO}}_2^{-}\left(aq\right)+{\text{H}}_3{\text{O}}^{+}$

Concept Introduction: According Bronsted-Lowry acid and base theory, acids are substance which loses protons ${\text{H}}^{+}$ to form conjugate base and bases are substances which accepts protons to from conjugate acid.

For example:

  $\text{HA}\to {\text{H}}^{+}+{\text{A}}^{-}$

Here, HA is an acid as it donates a proton to form ${\text{A}}^{-}$ a conjugate base.

Similarly,

  ${\text{A}}^{-}+{\text{H}}^{+}\to \text{HA}$

Here, ${\text{A}}^{-}$ is a base as it accepts a proton to from HA which is a conjugate acid.

Answer to Problem 1QAP

  $\begin{array}{l}{\text{HCHO}}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\rightleftharpoons {\text{CHO}}_2^{-}\left(aq\right)+{\text{H}}_3{\text{O}}^{+}\\ \text{ A B CB CA}\end{array}$

Explanation of Solution

The given reaction is as follows:

  ${\text{HCHO}}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\rightleftharpoons {\text{CHO}}_2^{-}\left(aq\right)+{\text{H}}_3{\text{O}}^{+}$

Here, ${\text{HCHO}}_{\text{2}}\left(aq\right)$ donates a proton to form ${\text{CHO}}_2^{-}\left(aq\right)$ thus, it is an acid and ${\text{CHO}}_2^{-}\left(aq\right)$ so formed is a conjugate base similarly, ${\text{H}}_{\text{2}}\text{O}$ accepts a proton to from ${\text{H}}_3{\text{O}}^{+}$ thus, it is a base and ${\text{H}}_3{\text{O}}^{+}$ so formed is a conjugate acid.

Therefore, in the given reaction, acid (A), base (B), conjugate acid (CA) and conjugate base (CB) are as follows:

  $\begin{array}{l}{\text{HCHO}}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\rightleftharpoons {\text{CHO}}_2^{-}\left(aq\right)+{\text{H}}_3{\text{O}}^{+}\\ \text{ A B CB CA}\end{array}$