Chapter 19, Problem 102A

Interpretation Introduction

Interpretation: The acid dissociation constant of saccharin is to be calculated.

Concept introduction: The ratio of the concentration of the ionized form of an acid to the concentration of the unionized form of the acid is known as the dissociation constant or ionization constant. It is denoted as ${\text{K}}_{\text{a}}$. In general, the dissociation of acid HA is written as follows:

  $HA\rightleftharpoons H^{+}+A^{-}$

The dissociation constant ${\text{K}}_{\text{a}}$ of the acid HA is written as,

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

Answer to Problem 102A

The dissociation constant ${\text{K}}_{\text{a}}$ of saccharin is $3.68\times {10}^{-4}$ .

Explanation of Solution

Given:

The concentration of saccharin is $1.00M$ .

pH of the aqueous solution of saccharin is $\text{1}\text{.71}$ .

The chemical equation for the dissociation of the saccharin is written as follows:

  $HN{C}_7{H}_4{\text{SO}}_3(aq)\rightleftharpoons N{C}_7{H}_4{\text{SO}}_3^{-}(aq)+H^{+}(aq)$

The concentration of hydrogen ions is calculated by the use of pH value.

  $\begin{array}{c}\left[{\text{H}}^{+}\right]=\text{antilog}\left(-\text{pH}\right)\\ =\text{antilog}\left(-1.71\right)\\ =0.019\end{array}$

The ICE table is constructed below:

Concentration	$\left[HN{C}_7{H}_4{\text{SO}}_3\right]$	$\left[H^{+}\right]$	$\left[N{C}_7{H}_4{\text{SO}}_3^{-}\right]$
Initial	$1.0$	0	0
Change	$-0.019$	$0.019$	$0.019$
Equilibrium	$0.981$	$0.019$	$0.019$

The expression for the equilibrium constant ${\text{K}}_{\text{a}}$ is written below:

  $\begin{array}{l}{\text{K}}_{\text{a}}=\frac{\left[H^{+}\right]\left[N{C}_7{H}_4{\text{SO}}_3^{-}\right]}{\left[HN{C}_7{H}_4{\text{SO}}_3\right]}\\ =\frac{0.019\times 0.019}{0.981}\\ =3.68\times {10}^{-4}\end{array}$