Question

Caproic acid \(\left(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{COOH}\right)\) is found in small amounts in coconut and palm oils and is used in making artificial flavors. A saturated solution of the acid contains \(11 \mathrm{~g} / \mathrm{L}\) and has a pH of 2.94. Calculate \(K_{a}\) for the acid.

Short answer

The \(K_a\) value for caproic acid is \(1.40 \times 10^{-5}\).

Step-by-step solution

Convert the pH to \(\mathrm{H}^+\) concentration

Given the pH value is 2.94, we can find the concentration of \(\mathrm{H}^+\) ions using the definition of pH: \[\mathrm{pH} = -\mathrm{log}_{10}{[\mathrm{H}^+]}\] Rearranging to find \([\mathrm{H}^+]\): \[[\mathrm{H}^+] = 10^{-\mathrm{pH}}\] \[[\mathrm{H}^+] = 10^{-2.94} = 1.15 \times 10^{-3} \mathrm{M}\]

Write the dissociation equation for caproic acid

Caproic acid, having the formula \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{COOH}\), dissociates in water as follows: \[\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{COOH} \rightleftharpoons \mathrm{C}_{5} \mathrm{H}_{11} \mathrm{COO}^- + \mathrm{H}^+\]

Set up an ICE table

An ICE (Initial-Change-Equilibrium) table will help us understand the changes in concentrations of the species involved in the reaction. In this case, the initial concentration of caproic acid is 11 g/L. Let's convert that to molarity: Molar mass of caproic acid = \((5\times 12.01) + (11\times 1.01) + 12.01 + 16 + 16.00 + 1.01 = 116.16\) g/mol. Initial concentration of caproic acid: \(\frac{11\,\mathrm{g/L}}{116.16\,\mathrm{g/mol}} = 9.47 \times 10^{-2}\,\mathrm{M}\) Now let's set up the ICE table: | |\(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{COOH}\) | \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{COO}^-\) | \(\mathrm{H}^+\) | |----------|-----------------------------|---------------------------|---------| | Initial |\(9.47 \times 10^{-2}\,\mathrm{M}\) |0 |0 | | Change |\(-x\) |\(+x\) |\(+x\) | | Equilibrium |\(9.47 \times 10^{-2} - x\) |\(x\) |\(x\) | At equilibrium, we know \([\mathrm{H}^+] = 1.15 \times 10^{-3}\,\mathrm{M}\).

Calculate the \(K_a\) value

Using the equilibrium concentrations from the ICE table, we can find the \(K_a\) value: \[K_a = \frac{[\mathrm{C}_5\mathrm{H}_{11}\mathrm{COO}^-][\mathrm{H}^+]}{[\mathrm{C}_5\mathrm{H}_{11}\mathrm{COOH}]}\] Substitute the equilibrium concentrations: \[K_a = \frac{(1.15 \times 10^{-3})(1.15 \times 10^{-3})}{(9.47 \times 10^{-2} - 1.15 \times 10^{-3})}\] Now, we will solve for \(K_a\): \[K_a = \frac{(1.15 \times 10^{-3})^2}{(9.47 \times 10^{-2} - 1.15 \times 10^{-3})} = 1.40 \times 10^{-5}\] Therefore, the \(K_a\) value for caproic acid is \(1.40 \times 10^{-5}\).