Chapter 15, Problem 15.1P

Interpretation Introduction

Interpretation:

The balanced equation for the dissociation of Bronsted- Lowry acids in water should be determined.

Concept introduction:A Bronsted-Lowry acid is a species that reacts with water to donate a proton, H⁺:

$HA+H_{2}O\rightleftharpoons A^{-}+H_3{O}^{+}$

A Bronsted-Lowry base is a species that reacts with water to accept a proton:

$B+H_{2}O\rightleftharpoons B{H}^{+}+O{H}^{-}$

Answer to Problem 15.1P

Solution:

  $HS{O}_4{}^{-}+H_{2}O\rightleftharpoons S{O}_4{}^{2-}+H_3{O}^{+};conjugatebase=S{O}_4{}^{2-}$

Explanation of Solution

The given species is $HS{O}_4{}^{-}$it can donate its proton to water to from $S{O}_4{}^{2-}$and $H_3{O}^{+}$ions. Here, water accepts a proton thus, it will act as a Bronsted base and $HS{O}_4{}^{-}$donates a proton thus, it will act as Bronsted acid. The reaction will be as follows:

  $HS{O}_4{}^{-}+H_{2}O\rightleftharpoons S{O}_4{}^{2-}+H_3{O}^{+}$

Here, the conjugate base is SO₄^2-.

(c)

Interpretation Introduction

Interpretation:

The balanced equation for the dissociation of Bronsted- Lowry acids in water should be determined.

Concept introduction:A Bronsted-Lowry acid is a species that reacts with water to donate a proton, H⁺:

$HA+H_{2}O\rightleftharpoons A^{-}+H_3{O}^{+}$

A Bronsted-Lowry base is a species that reacts with water to accept a proton:

$B+H_{2}O\rightleftharpoons B{H}^{+}+O{H}^{-}$

Answer to Problem 15.1P

Solution:

  $H_{2}O+H_3{O}^{+}\rightleftharpoons H_3{O}^{+}+H_{2}O;conjugatebase=H_{2}O$

Explanation of Solution

The given species is $H_3{O}^{+}$, it can donate its proton to water to form $H_{2}O$and $H_3{O}^{+}$. Here, water accepts a proton thus, it acts as a base and $H_3{O}^{+}$donates a proton thus, it acts as an acid.

The reaction is as follows:

  $H_3{O}^{+}+H_{2}O\rightleftharpoons H_{2}O+H_3{O}^{+}$

Here, the conjugate base is H₂O.

(d)

Interpretation Introduction

Interpretation:

The balanced equation for the dissociation of Bronsted- Lowry acids in water should be determined.

Concept introduction:A Bronsted-Lowry acid is a species that reacts with water to donate a proton, H⁺:

$HA+H_{2}O\rightleftharpoons A^{-}+H_3{O}^{+}$

A Bronsted-Lowry base is a species that reacts with water to accept a proton:

$B+H_{2}O\rightleftharpoons B{H}^{+}+O{H}^{-}$

Answer to Problem 15.1P

Solution:

  $N{H}_4{}^{+}+H_{2}O\rightleftharpoons N{H}_3+H_3{O}^{+}conjugatebase=N{H}_3$

Explanation of Solution

The given species is $N{H}_4{}^{+}$it can donate its proton to water to form $N{H}_3$and $H_3{O}^{+}$. Here, water accepts proton thus, it acts as a base and $N{H}_4{}^{+}$act as an acid.

The reaction is as follows:

  $N{H}_4{}^{+}+H_{2}O\rightleftharpoons N{H}_3+H_3{O}^{+}$

Here, the conjugate base is NH₃.