Chapter U6, Problem 16STP

Interpretation Introduction

Interpretation:

The correct statement about dissociation of benzoic acid at equation on the basis of information needs to be determined.

Concept introduction:

An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. A weak acid ionized partially and reaches to equilibrium.

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

The acid dissociation constant is the ratio of concentration of product and reactant for the equilibrium reaction of weak acid in solution. For the given weak acid HA, it can be written as:

$\begin{array}{l}{\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] [A}}^{\text{-}}\text{]}}{\text{[HA]}}\end{array}$

Answer to Problem 16STP

Option (d) is correct.

Explanation of Solution

Reason for correct option:

Benzoic acid are weak acid therefore it partially dissociates to its respective ions as given below;

${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COOH}}_{}{}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+ C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}$

The Ka expression for benzoic acid can be written as:

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COOH}}_{}{}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+ C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] [C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}\text{]}}{{\text{[C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COOH}}_{}{}_{\text{(aq)}}\text{]}}\end{array}$

Since the value of Ka is very less, it stands that there are more ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{COOH}$ molecules than ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COO}}^{\text{-}}$ and ${\text{H}}^{\text{+}}$ ions.

Given:

[ ${\text{H}}^{\text{+}}$ ] = 0.0079 M

Calculate pH:

$\begin{array}{l}{\text{pH = - log [H}}^{\text{+}}\text{] }\\ \text{pH = - log [0}\text{.0079]}\\ \text{pH = 2}\text{.1}\end{array}$

Thus the pH of solution lies between 2 and 3.

Conclusion

Reasons for incorrect options: Benzoic acid is a weak acid therefore partial dissociation will form less ions compare to benzoic acid molecule. The pH value lies between 2 to 3 as concentration of H+ ions is 0.0079 M.