Chapter 18.2, Problem 21SSC

Interpretation Introduction

Interpretation:

Using the data, seven acids are to be arranged in increasing order of electrical conductivity.

Concept introduction:

The value of the acid ionization constant tells about the state of whether reactants or products are more favored in the reaction.

For a reaction,

$\text{H}\text{A}+{\text{H}}_2\text{O}\rightleftharpoons {\text{A}}^{-}+{\text{H}}_3{\text{O}}^{+}$

The acid ionization constant is given by:

$\begin{array}{l}{\text{k}}_{\text{a}}=\frac{\left[{\text{A}}^{-}\right]\times \left[{\text{H}}_3{\text{O}}^{+}\right]}{\left[\text{H}\text{A}\right]}\end{array}$

  $\begin{array}{c}\text{w}\text{h}\text{e}\text{r}\text{e},\end{array}$

  $\begin{array}{c}\left[{\text{A}}^{-}\right]\text{i}\text{s}\text{c}\text{o}\text{n}\text{c}\text{e}\text{n}\text{t}\text{r}\text{a}\text{t}\text{i}\text{o}\text{n}\text{o}\text{f}{\text{A}}^{-}\end{array}$

  $\begin{array}{c}\left[{\text{H}}_3{\text{O}}^{+}\right]\text{i}\text{s}\text{c}\text{o}\text{n}\text{c}\text{e}\text{n}\text{t}\text{r}\text{a}\text{t}\text{i}\text{o}\text{n}\text{o}\text{f}{\text{H}}_3{\text{O}}^{+}\end{array}$

  $\begin{array}{c}\left[\text{H}\text{A}\right]\text{i}\text{s}\text{c}\text{o}\text{n}\text{c}\text{e}\text{n}\text{t}\text{r}\text{a}\text{t}\text{i}\text{o}\text{n}\text{o}\text{f}\text{H}\text{A}\end{array}$

The more the value of k_a , the stronger is the acid and more is the electrical conductivity.

Answer to Problem 21SSC

The increasing order of electrical conductivity is given by:

  $\text{H}{\text{S}}^{-}<\text{H}\text{C}{\text{O}}_3{}^{-}<\text{H}\text{C}\text{N}<{\text{H}}_2\text{S}<{\text{H}}_2\text{C}{\text{O}}_3<\text{C}{\text{H}}_3\text{C}\text{O}\text{O}\text{H}<\text{H}\text{F}$

Explanation of Solution

The more the value of k_a , the stronger is the acid and more is the electrical conductivity.

Acid	ka
${\text{H}}_2\text{S}$	$8.9\times {10}^{-8}$
$\text{H}{\text{S}}^{-}$	$1\times {10}^{-19}$
$\text{H}\text{F}$	$6.3\times {10}^{-4}$
$\text{H}\text{C}\text{N}$	$6.2\times {10}^{-10}$
$\text{C}{\text{H}}_3\text{C}\text{O}\text{O}\text{H}$	$1.8\times {10}^{-5}$
${\text{H}}_2\text{C}{\text{O}}_3$	$4.5\times {10}^{-7}$
$\text{H}\text{C}{\text{O}}_3{}^{-}$	$4.7\times {10}^{-11}$

Therefore, the increasing order of electrical conductivity is given by:

  $\text{H}{\text{S}}^{-}<\text{H}\text{C}{\text{O}}_3{}^{-}<\text{H}\text{C}\text{N}<{\text{H}}_2\text{S}<{\text{H}}_2\text{C}{\text{O}}_3<\text{C}{\text{H}}_3\text{C}\text{O}\text{O}\text{H}<\text{H}\text{F}$

Conclusion

Thus, the more the value of acid ionization constant, the more is the electrical conductivity of the solution.