Chapter 15, Problem 15.1P

Interpretation Introduction

Interpretation: The Bronsted-Lowry acid, Bronsted-Lowry base, conjugate acid and conjugate base for the given reactions are to be identified.

Concept introduction: According to the Bronsted-Lowry, acid is the species that gives a proton to form a conjugate base and base is the species that accepts a proton to form a conjugate acid.

To determine: The Bronsted-Lowry acid, Bronsted-Lowry base, conjugate acid and conjugate base for the given reactions.

Answer to Problem 15.1P

Solution: In the given reaction, the Bronsted-Lowry acid is ${\text{H}}_{\text{2}}\text{O}$ , Bronsted-Lowry base is ${\text{C}}_5{\text{H}}_5\text{N}$ , conjugate acid is ${\text{C}}_5{\text{H}}_5{\text{NH}}^{+}$and conjugate base is ${\text{OH}}^{-}$ . In the given reaction, the Bronsted-Lowry acid is ${\text{HNO}}_3$ , Bronsted-Lowry base is ${\text{H}}_{\text{2}}\text{O}$ , conjugate acid is ${\text{H}}_3{\text{O}}^{+}$and conjugate base is ${\text{NO}}_3^{-}$ .

(a)

Explanation of Solution

The given reaction is,

  ${\text{C}}_5{\text{H}}_5\text{N}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

In the above reaction, ${\text{H}}_{\text{2}}\text{O}$donates its ${\text{H}}^{+}$ion to ${\text{C}}_5{\text{H}}_5\text{N}$so, it is the Bronsted-Lowry acid and ${\text{C}}_5{\text{H}}_5\text{N}$which accepts ${\text{H}}^{+}$ion is the Bronsted-Lowry base.

The ion formed after ${\text{C}}_5{\text{H}}_5\text{N}$accepts ${\text{H}}^{+}$ion is ${\text{C}}_5{\text{H}}_5{\text{NH}}^{+}$so, it is the conjugate acid and ${\text{OH}}^{-}$formed after ${\text{H}}_{\text{2}}\text{O}$donates its ${\text{H}}^{+}$ion to ${\text{C}}_5{\text{H}}_5\text{N}$is the conjugate base.

(b)

The given reaction is,

${\text{HNO}}_3\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{+}\left(aq\right)+{\text{NO}}_3^{-}\left(aq\right)$

In the above reaction, ${\text{HNO}}_3$donates its ${\text{H}}^{+}$ion to ${\text{H}}_{\text{2}}\text{O}$so, it is the Bronsted-Lowry acid and ${\text{H}}_{\text{2}}\text{O}$which accepts ${\text{H}}^{+}$ion is the Bronsted-Lowry base.

Hydronium ion ${\text{H}}_3{\text{O}}^{+}$is formed after ${\text{H}}_{\text{2}}\text{O}$accepts ${\text{H}}^{+}$ion from ${\text{HNO}}_3$so, it is the conjugate acid and ${\text{NO}}_3^{-}$formed after ${\text{HNO}}_3$donates its ${\text{H}}^{+}$ion to ${\text{H}}_{\text{2}}\text{O}$is the conjugate base.