Chapter 17, Problem 25E

Interpretation Introduction

Interpretation:

The proton transfer reaction, ${\text{HC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}\left(\text{aq}\right)+{\text{PO}}_4{{}^3}^{-}\left(\text{aq}\right)\rightleftharpoons$ is to be completed. Also, the direction in which the resulting equilibrium will be favored is to be stated.

Concept introduction:

An acid-base reaction occurs through a transfer of proton from an acid to a base. The reaction results in the formation of a conjugate acid and a conjugate base as the products. The type of products formed affects the direction of reaction at equilibrium.

Answer to Problem 25E

The complete proton transfer reaction is given below.

$\underset{\begin{array}{l}\text{Stong}\\ \text{acid}\end{array}}{{\text{HC}}_3{\text{H}}_5{\text{O}}_2}\left(\text{aq}\right)+\underset{\begin{array}{l}\text{Strong}\\ \text{base}\end{array}}{{\text{PO}}_4{}^{3-}\left(\text{aq}\right)}\rightleftharpoons \underset{\begin{array}{l}\text{Weak}\\ \text{conjugate base}\end{array}}{{\text{C}}_3{\text{H}}_5{\text{O}}_2{}^{-}\left(\text{aq}\right)}+\underset{\begin{array}{l}\text{Weak}\\ \text{conjugate acid}\end{array}}{{\text{HPO}}_{\text{4}}{{}^{\text{2}}}^{\text{-}}\left(\text{aq}\right)}$

The resulting equilibrium is favored in the forward direction.

Explanation of Solution

Consider the acid-base reaction given below.

${\text{HC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}\left(\text{aq}\right)+{\text{PO}}_4{{}^3}^{-}\left(\text{aq}\right)\rightleftharpoons$

In the given reaction, ${\text{HC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}$ is a strong acid; that is, it has a strong tendency to donate a proton and ${\text{PO}}_4{{}^3}^{-}$ is a strong base; that is, it has a strong tendency to accept a proton. ${\text{HC}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}$ loses a proton to form a weak conjugate base ${\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{O}}_{\text{2}}{}^{-}$. ${\text{PO}}_4{{}^3}^{-}$ accepts a proton to form a weak conjugate acid, ${\text{HPO}}_{\text{4}}{{}^{\text{2}}}^{\text{-}}$ respectively.

The complete acid-base reaction is shown below.

$\underset{\begin{array}{l}\text{Stong}\\ \text{acid}\end{array}}{{\text{HC}}_3{\text{H}}_5{\text{O}}_2}\left(\text{aq}\right)+\underset{\begin{array}{l}\text{Strong}\\ \text{base}\end{array}}{{\text{PO}}_4{}^{3-}\left(\text{aq}\right)}\rightleftharpoons \underset{\begin{array}{l}\text{Weak}\\ \text{conjugate base}\end{array}}{{\text{C}}_3{\text{H}}_5{\text{O}}_2{}^{-}\left(\text{aq}\right)}+\underset{\begin{array}{l}\text{Weak}\\ \text{conjugate acid}\end{array}}{{\text{HPO}}_{\text{4}}{{}^{\text{2}}}^{\text{-}}\left(\text{aq}\right)}$

Since, both the products are weak conjugate acid and weak conjugate base, they have less tendency to dissociate, therefore, they are stable, so the equilibrium will be favored in the forward direction.

Conclusion

The transfer of a proton from an acid to base in the reaction will shift the equilibrium in the forward direction.

$\underset{\begin{array}{l}\text{Stong}\\ \text{acid}\end{array}}{{\text{HC}}_3{\text{H}}_5{\text{O}}_2}\left(\text{aq}\right)+\underset{\begin{array}{l}\text{Strong}\\ \text{base}\end{array}}{{\text{PO}}_4{}^{3-}\left(\text{aq}\right)}\rightleftharpoons \underset{\begin{array}{l}\text{Weak}\\ \text{conjugate base}\end{array}}{{\text{C}}_3{\text{H}}_5{\text{O}}_2{}^{-}\left(\text{aq}\right)}+\underset{\begin{array}{l}\text{Weak}\\ \text{conjugate acid}\end{array}}{{\text{HPO}}_{\text{4}}{{}^{\text{2}}}^{\text{-}}\left(\text{aq}\right)}$