Chapter 16, Problem 16.1P

Write a balanced equation for the dissociation of the Bronsted-Lowry acid, ${\text{HSO4}}^{\text{-}}$ , in water.

Interpretation Introduction

Interpretation:

The balanced equation for the dissociation of the Bronsted-Lowry acid, ${\text{HSO}}_{\text{4}}{}^{-}$ in water should be written.

Concept introduction:

Bronsted-Lowry acid is defined as a species that has the capability to donate proton ( $H^{+}$ ). Bronsted-Lowry base is defined as a species that has capability to accept protons ( $H^{+}$ ).

The general reaction is given as:

$\begin{array}{l}\text{HA + B}\rightleftharpoons {\text{ A}}^{-}+{\text{ BH}}^{\text{+}}\\ \text{Acid Base Conjugate Conjugate}\\ \text{ base acid }\\ \end{array}$

Answer to Problem 16.1P

The balanced equation for the dissociation of the Bronsted-Lowry acid, ${\text{HSO}}_{\text{4}}{}^{-}$ in water is:

${\text{HSO}}_{\text{4}}{}^{-}(aq)+{\text{H}}_{\text{2}}\text{O}(l)\rightleftharpoons {\text{SO}}_{\text{4}}{}^{2-}(aq)+{\text{H}}_{\text{3}}{\text{O}}^{+}(aq)$

Explanation of Solution

The given species is ${\text{HSO}}_{\text{4}}{}^{-}$.

${\text{HSO}}_{\text{4}}{}^{-}$ has capability to donate one proton. Thus, when ${\text{HSO}}_{\text{4}}{}^{-}$ reacts with water, ${\text{HSO}}_{\text{4}}{}^{-}$ donate one proton and water accepts that proton results in the formation of hydronium ion and sulfate ion.

Here, ${\text{HSO}}_{\text{4}}{}^{-}$ is Bronsted-Lowry acid as it has the capability to donate one proton and water acts as a Bronsted-Lowry base as it has the capability to accept proton.

The dissociation reaction is written as:

${\text{HSO}}_{\text{4}}{}^{-}(aq)+{\text{H}}_{\text{2}}\text{O}(l)\rightleftharpoons {\text{SO}}_{\text{4}}{}^{2-}(aq)+{\text{H}}_{\text{3}}{\text{O}}^{+}(aq)$