Problem 54 - Chemistry: An Atoms-Focused Approach

Question

Rancid Butter The odor of spoiled butter is due in part to butanoic acid, which results from the chemical breakdown of butterfat. A \(0.100 M\) solution of butanoic acid is \(1.23 \%\) ionized. Calculate the value of \(K_{\mathrm{a}}\) for butanoic acid.

Step-by-Step Solution

Answer

The Ka for butanoic acid is 1.53 x 10^-5.

Step 1: Calculate the concentration of ionized butanoic acid

Since we know the percent ionization of butanoic acid, we can use it to determine the concentration of ionized butanoic acid. The percent ionization is \(1.23 \%\), and we have \(0.100 M\) solution of butanoic acid: $$ \text{Ionized concentration} = \frac{\text{Percent ionization}}{100} \times \text{Initial concentration} $$ $$ \text{Ionized concentration} = \frac{1.23}{100} \times 0.100 M = 0.00123 M $$

Step 2: Write the equation for the ionization of butanoic acid

Next, we need to write the ionization reaction for butanoic acid. Butanoic acid (C4H8O2) ionizes into butanoate ion (C4H7O2-) and a hydrogen ion (H+) in water: $$ \text{C}_4\text{H}_8\text{O}_2 ⇌ \text{C}_4\text{H}_7\text{O}_2^- + \text{H}^+ $$

Step 3: Calculate the concentrations at equilibrium

Since the ionization is \(1.23 \%\), both the butanoate ion concentration and the hydrogen ion concentration are \(0.00123 M\). This means that butanoic acid concentration at equilibrium will be the initial concentration minus the ionized concentration: $$ [\text{C}_4\text{H}_8\text{O}_2]_{\mathrm{equilibrium}} = [\text{C}_4\text{H}_8\text{O}_2]_{\mathrm{initial}} - [\text{C}_4\text{H}_7\text{O}_2^-]_{\mathrm{equilibrium}} $$ $$ [\text{C}_4\text{H}_8\text{O}_2]_{\mathrm{equilibrium}} = 0.100 M - 0.00123 M = 0.09877 M $$

Step 4: Calculate the value of Ka

Now that we have the equilibrium concentrations, we can calculate the value of Ka using the formula: $$ K_a = \frac{[\text{C}_4\text{H}_7\text{O}_2^-][\text{H}^+]}{[\text{C}_4\text{H}_8\text{O}_2]} $$ Plugging in the equilibrium concentrations: $$ K_a = \frac{(0.00123)(0.00123)}{0.09877} = 1.53 \times 10^{-5} $$ The value of Ka for butanoic acid is \(1.53 \times 10^{-5}\).