Calculate the percent ionization of propionic acid \((\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH})\) in solutions of each of the following concentrations (\(K_{a}\) is given in AppendixD): (a) \(0.250 M,(\mathbf{b}) 0.0800 M,\) \((\mathbf{c}) 0.0200 M .\)

When propionic acid \((\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH})\) dissociates in water, it forms hydrogen ions \((\mathrm{H}^+)\) and propionate ions \((\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^-)\). The dissociation reaction can be written as: \[ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{COOH} \left( aq \right) \rightleftharpoons \mathrm{H}^+ \left( aq \right) + \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{COO}^- \left( aq \right) \]

The equilibrium expression for the dissociation of propionic acid is: \[ K_{a} = \frac{[\mathrm{H}^+] [\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{COO}^-]}{[\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{COOH}]} \] We will use this equilibrium expression to find the concentration of dissociated ions for each concentration of propionic acid and then calculate the percent ionization.

For each concentration of propionic acid, we will assume a change in its concentration due to ionization and set up an ICE (Initial, Change, Equilibrium) table. Since the concentrations of \(\mathrm{H}^+\) and \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{COO}^-\) will have the same change, we will use a variable \(x\) to represent the change in concentration and find its value using the \(K_a\) expression. Let's denote the initial concentration of propionic acid by \(c\). We have the following ICE table: | | \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\) | \(\mathrm{H}^+\) | \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^-\) | |------------------|------------------------|-------------|------------------------| | Initial (M) | c | 0 | 0 | | Change (M) | -x | +x | +x | | Equilibrium (M) | c - x | x | x | Now substitute these concentrations into the \(K_{a}\) expression and solve for \(x\): \[ K_{a} = \frac{x^2}{c - x} \] Since \(K_a\) is generally small for weak acids, we can approximate that \(x << c\), so \(c - x \approx c\). Thus, we simplify the expression to: \[ K_{a} = \frac{x^2}{c} \] Now, we will use the given values of \(K_{a}\) and \(c\) (concentrations) to find the value of \(x\) for each part of the exercise. Finally, we will calculate the percent ionization using the following formula: \[ \text{Percent Ionization} = \frac{x}{c} \times 100\% \] Note that \(K_{a}\) values can be found in the Appendix D table of your textbook: (a) For the \(0.250 M\) concentration of propionic acid: Calculate the value of \(x\) for this concentration, and then the percent ionization. (b) For the \(0.0800 M\) concentration of propionic acid: Calculate the value of \(x\) for this concentration, and then the percent ionization. (c) For the \(0.0200 M\) concentration of propionic acid: Calculate the value of \(x\) for this concentration, and then the percent ionization.