Chapter 15, Problem 15.55QE

Interpretation Introduction

Interpretation:

$K_{\text{a}}$ and $\text{p}{K}_{\text{a}}$ for butyric acid has to be determined.

Concept Introduction:

Mole is the S.I. unit. The expression for the number of moles is as follows:

  $\text{Number of moles}=\frac{\text{mass of the compound}}{\text{molar mass of the compound}}$        (1)

The molarity and is equal to the number of moles of the solute dissolved in one liter of the solution.

The expression of molarity $\left(M\right)$ is as follows:

  $\text{Molarity}\left(M\right)=\frac{\text{number of moles}\left(\text{mol}\right)}{\text{Volume}\left(\text{L}\right)}$        (2)

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

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

Also, the relation between pH and concentration is as follows:

    $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]={\text{10}}^{-\text{pH}}$        (4)

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 of $K_{\text{a}}$ is as follows:

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