Chapter 16, Problem 79QP

Interpretation Introduction

Interpretation:

All the species present in the phosphoric acid solution are to be listed, and categorized into acids or bases, or both.

Concept introduction:

According to the Bronsted-Lowry theory, a substance that has the capability to donate a proton, acts as an acid. The acid is known as a Bronsted Acid.

A Bronsted base is a substance that accepts a proton and forms its conjugate acid.

The first ionization of the triprotic acid takes place as:

${\text{H}}_3{\text{A}}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{A}}^{-}{}_{\left(\text{aq}\right)}$

The second ionization of the triprotic acid takes place as:

${\text{H}}_2{\text{A}}^{-}{}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}+{\text{HA}}^{-}{}_{\left(\text{aq}\right)}$

The third ionization of the triprotic acid takes place as:

${\text{HA}}^{-}{}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}+{\text{A}}^{-}{}_{\left(\text{aq}\right)}$

The species ${\text{HA}}^{-}$ and ${\text{H}}_2{\text{A}}^{-}$ can also act as a base according to the reactions:

$\begin{array}{c}{\text{HA}}^{-}{}_{\left(\text{aq}\right)}+{\text{H}}_2\text{O}\rightleftharpoons {\text{H}}_{\text{2}}\text{A}+{\text{OH}}^{-}\\ {\text{H}}_2{\text{A}}^{-}{}_{\left(\text{aq}\right)}+{\text{H}}_2\text{O}\rightleftharpoons {\text{H}}_{\text{3}}\text{A}+{\text{OH}}^{-}\end{array}$

Answer to Problem 79QP

Solution:

In the solution of phosphoric acid, ${\text{H}}_3{\text{PO}}_4,{\text{ H}}_2{\text{PO}}_4{}^{-},{\text{ HPO}}_4{}^{2-}{\text{, PO}}_4{}^{3-}{\text{, and H}}_3{\text{O}}^{+}$ are present. The species ${\text{H}}_3{\text{O}}^{+}$ and ${\text{H}}_3{\text{PO}}_{\text{4}}$ act only as acids, ${\text{PO}}_4{}^{3-}$ acts only as a base, and ${\text{H}}_2{\text{PO}}_4{}^{-}{\text{ and HPO}}_{\text{4}}{}^{2-}$ act as both, an acid and a base.

Explanation of Solution

In the solution of phosphoric acid the ${\text{H}}_3{\text{PO}}_4,{\text{ H}}_2{\text{PO}}_4{}^{-},{\text{ HPO}}_4{}^{2-}{\text{, PO}}_4{}^{3-}{\text{, and H}}_3{\text{O}}^{+}$ species are present.

The species, ${\text{H}}_3{\text{PO}}_{\text{4}}$ can act as an acid as a proton can be removed from it to produce hydronium ions and its conjugate base according to the reaction:

$\begin{array}{l}{\text{H}}_3{\text{PO}}_{\text{4}}{}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{PO}}_{\text{4}}{{}^{-}}_{\left(\text{aq}\right)}\\ \left(\text{Acid}\right)\end{array}$

${\text{H}}_3{\text{PO}}_{\text{4}}$ cannot act as a base as ${\text{H}}_3{\text{PO}}_{\text{4}}$ cannot accept a proton.

The species, ${\text{H}}_2{\text{PO}}_{\text{4}}{}^{-}$ can act as both acid and a base as it can either remove a proton or can accept a proton, according to the reactions:

$\begin{array}{l}{\text{H}}_2{\text{PO}}_{\text{4}}{{}^{-}}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}+{\text{HPO}}_{\text{4}}{{}^{2-}}_{\left(\text{aq}\right)}\\ \left(\text{Acid}\right)\\ {\text{H}}_2{\text{PO}}_{\text{4}}{{}^{-}}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{OH}}^{-}{}_{\left(\text{aq}\right)}+{\text{H}}_3{\text{PO}}_{\text{4}}{}_{\left(\text{aq}\right)}\\ \left(\text{Base}\right)\end{array}$

The species, ${\text{HPO}}_{\text{4}}{}^{2-}$ can act as both acid and a base as it can either remove a proton or can accept a proton, according to the reactions:

$\begin{array}{l}{\text{HPO}}_{\text{4}}{{}^{2-}}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}+{\text{PO}}_{\text{4}}{{}^{2-}}_{\left(\text{aq}\right)}\\ \left(\text{Acid}\right)\\ {\text{HPO}}_{\text{4}}{{}^{2-}}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{OH}}^{-}{}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{PO}}_{\text{4}}{{}^{-}}_{\left(\text{aq}\right)}\\ \left(\text{Base}\right)\end{array}$

The species ${\text{PO}}_4{}^{3-}$ has no proton that can be removed from it. Thus, it acts only as a base and can accept a proton according to the reaction:

$\begin{array}{l}{\text{PO}}_4{{}^{3-}}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{OH}}^{-}{}_{\left(\text{aq}\right)}+{\text{HPO}}_{\text{4}}{{}^{2-}}_{\left(\text{aq}\right)}\\ \left(\text{Base}\right)\end{array}$

The species ${\text{H}}_3{\text{O}}^{+}$ is proton rich and so it cannot accept more protons but has an extra proton to be donated. Thus, it acts only as an acid according to the reaction:

$\begin{array}{l}{\text{H}}_3{\text{O}}^{+}{}_{\left(\text{aq}\right)}\rightleftharpoons {\text{H}}^{\text{+}}{}_{\left(\text{aq}\right)}+{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{aq}\right)}\\ \left(\text{Acid}\right)\end{array}$

Therefore, ${\text{H}}_3{\text{O}}^{+}$ and ${\text{H}}_3{\text{PO}}_{\text{4}}$ act only as an acid, ${\text{PO}}_4{}^{3-}$ acts only as a base, and ${\text{H}}_2{\text{PO}}_4{}^{-}{\text{ and HPO}}_{\text{4}}{}^{2-}$ act as both, an acid and a base.