Chapter 7, Problem 64E

(a)

Interpretation Introduction

Interpretation:

The percent dissociation of 0.50 M acetic acid should be calculated.

Concept Introduction :

The formula for the percent dissociation is represented as follows:

  $\text{Percent dissociation = }\frac{\text{amount dissociated (mol/L)}}{\text{initial concentration (mol/L)}}\times 100\%$

Answer to Problem 64E

0.60 %

Explanation of Solution

The ICE table for the equilibrium reaction is represented as follows:

  $\begin{array}{l}{\text{ CH}}_3{\text{COOH(aq) + H}}_2\text{O(l) }\rightleftharpoons {\text{ H}}_3{\text{O}}^{+}{\text{(aq) + CH}}_3{\text{COO}}^{-}\text{(aq)}\\ \text{initial 0}\text{.50 M - -}\\ \text{change }-x\text{ M +}x\text{ M +}x\text{ M}\\ \text{at eqm }\left(0.50-x\right)\text{ M }x\text{ M }x\text{ M}\end{array}$

The acid dissociation constant can be represented as follows:

  $\begin{array}{l}{K}_a=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{CH}}_3{\text{COO}}^{-}\right]}{\left[{\text{CH}}_3\text{COOH}\right]}\\ 1.8\times {10}^{-5}=\frac{x\times x}{\left(0.50-x\right)}\\ 9.0\times {10}^{-6}-1.8\times {10}^{-5}x=x^2\\ x^2+1.8\times {10}^{-5}x-9.0\times {10}^{-6}=0\\ x=0.003\text{ M}\end{array}$

The percent dissociation can be calculated as follows:

  $\begin{array}{l}\text{Percent dissociation = }\frac{0.003\text{ M}}{0.50\text{ M}}\times 100\%\\ \text{ = 0}\text{.60 \%}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The percent dissociation of 0.050 M acetic acid should be calculated.

Concept Introduction :

The formula for the percent dissociation is represented as follows:

  $\text{Percent dissociation = }\frac{\text{amount dissociated (mol/L)}}{\text{initial concentration (mol/L)}}\times 100\%$

Answer to Problem 64E

1.8 %

Explanation of Solution

The acid dissociation reaction can be represented as follows:

  $\begin{array}{l}{\text{ CH}}_3{\text{COOH(aq) + H}}_2\text{O(l) }\rightleftharpoons {\text{ H}}_3{\text{O}}^{+}{\text{(aq) + CH}}_3{\text{COO}}^{-}\text{(aq)}\\ \text{initial 0}\text{.050 M - -}\\ \text{change }-x\text{ M +}x\text{ M +}x\text{ M}\\ \text{at eqm }\left(0.050-x\right)\text{ M }x\text{ M }x\text{ M}\end{array}$

The expression for the acid dissociation reaction is as follows:

  $\begin{array}{l}{K}_a=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{CH}}_3{\text{COO}}^{-}\right]}{\left[{\text{CH}}_3\text{COOH}\right]}\\ 1.8\times {10}^{-5}=\frac{x\times x}{\left(0.050-x\right)}\\ 9.0\times {10}^{-7}-1.8\times {10}^{-5}x=x^2\\ x^2+1.8\times {10}^{-5}x-9.0\times {10}^{-7}=0\\ x=0.0009\text{ M}\end{array}$

Thus,

  $\begin{array}{l}\text{Percent dissociation = }\frac{0.0009\text{ M}}{0.050\text{ M}}\times 100\%\\ \text{ = 1}\text{.8 \%}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The percent dissociation of 0.0050 M acetic acid should be calculated.

Concept Introduction :

The percent dissociation is calculated as follows:

  $\text{Percent dissociation = }\frac{\text{amount dissociated (mol/L)}}{\text{initial concentration (mol/L)}}\times 100\%$

Answer to Problem 64E

6.0 %

Explanation of Solution

The equilibrium reaction is represented as follows:

  $\begin{array}{l}{\text{ CH}}_3{\text{COOH(aq) + H}}_2\text{O(l) }\rightleftharpoons {\text{ H}}_3{\text{O}}^{+}{\text{(aq) + CH}}_3{\text{COO}}^{-}\text{(aq)}\\ \text{initial 0}\text{.0050 M - -}\\ \text{change }-x\text{ M +}x\text{ M +}x\text{ M}\\ \text{at eqm }\left(0.0050-x\right)\text{ M }x\text{ M }x\text{ M}\end{array}$

  $\begin{array}{l}{K}_a=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{CH}}_3{\text{COO}}^{-}\right]}{\left[{\text{CH}}_3\text{COOH}\right]}\\ 1.8\times {10}^{-5}=\frac{x\times x}{\left(0.0050-x\right)}\\ 9.0\times {10}^{-8}-1.8\times {10}^{-5}x=x^2\\ x^2+1.8\times {10}^{-5}x-9.0\times {10}^{-8}=0\\ x=0.0003\text{ M}\end{array}$

  $\begin{array}{l}\text{Percent dissociation = }\frac{0.0003\text{ M}}{0.0050\text{ M}}\times 100\%\\ \text{ = 6}\text{.0 \%}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The reason for increase in the percent dissociation due to decrease in the concentration of a weak acid should be explained.

Concept Introduction:

The Le Chatelier’s principle states that if a change occurs in concentration, pressure or temperature in a system at equilibrium, the equilibrium will shift in such a way that it counteracts that change.

Answer to Problem 64E

Equilibrium shifts to right side when the concentration of weak acid decreases. Hence, the percent dissociation of a weak acid increases when the concentration of weak acid decreases.

Explanation of Solution

The dissociation reaction is represented as follows:

  ${\text{CH}}_3{\text{COOH(aq) + H}}_2\text{O(l) }\rightleftharpoons {\text{ H}}_3{\text{O}}^{+}{\text{(aq) + CH}}_3{\text{COO}}^{-}\text{(aq)}$

According to the Le Chatelier’s principle, when concentration of a species in an equilibrium reaction changes, the equilibrium will shift in such a way to counteract that change. So, when the concentration of the solution decreases, all the concentrations of the species in the system decrease. So, equilibrium shifts to the side with greater number of particles. That means above equilibrium shifts to right side when the concentration of weak acid decreases. Hence, the percent dissociation of a weak acid increases when the concentration of weak acid decreases.

(e)

Interpretation Introduction

Interpretation:

The reason for decrease in [H⁺] from a to c should be explained.

Concept Introduction :

The percent dissociation is represented as follows:

  $\text{Percent dissociation = }\frac{\text{amount dissociated (mol/L)}}{\text{initial concentration (mol/L)}}\times 100\%$

Explanation of Solution

The concentration of hydrogen ion or [H⁺] depends on the initial concentration of a weak acid and the dissociation constant. Here, initial concentration of the acetic acid decreases rapidly more than the increase in percent dissociation. So, H+ concentration decreases.