Chapter 15, Problem 15.59QE

Interpretation Introduction

Interpretation:

$K_{\text{a}}$ of propanoic acid has to be determined.

Concept Introduction:

A weak acid in water produces a hydrogen ion and conjugate base. When weak acid dissolves in water, some acid molecules transfer proton to water.

In solution of weak acid, the actual concentration of the acid molecules becomes less because partial dissociation of acid has occurred and lost protons to form hydrogen ions.

The reaction is as follows:

  $\text{HA}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(\text{aq}\right)+{\text{A}}^{-}\left(\text{aq}\right)$

The reaction is as follows:

  $\text{HA}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(\text{aq}\right)+{\text{A}}^{-}\left(\text{aq}\right)$

The expression for $K_{\text{a}}$ is as follows:

  $K_{\text{a}}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$

Negative logarithm of molar concentration of hydronium ion is called $\text{pH}$. The expression for pH is as follows:

    $\text{pH}=-{\text{log}}_{\text{10}}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$        (1)