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

The first step is to write the ionization equilibrium for propionic acid. Propionic acid is a weak acid that dissociates partially when dissolved in water as follows: \[ \text{C}_{2}\text{H}_{5}\text{COOH}(a{\text q}) \rightleftarrows \text{C}_{2}\text{H}_{5}\text{COO}^{-} (a{\text q}) + \mathrm{H}^{+}(a{\text q})\]

The next step is to write the Ka expression for the propionic acid ionization. This can be written as the concentrations of the products divided by the concentration of the reactants: \[K_{a} = \frac{[\text{C}_{2}\text{H}_{5}\text{COO}^{-}][\mathrm{H}^{+}]}{[\text{C}_{2}\text{H}_{5}\text{COOH}]}\]

Now we can set up an ICE (Initial, Change, Equilibrium) table for each given concentration. Since the initial concentrations of \(\text{C}_{2}\text{H}_{5}\text{COO}^{-}\) and \(\mathrm{H}^{+}\) are zero, we can write the table as follows for each concentration: | | C2H5COOH | -> | C2H5COO- | + | H+ | |-------|---------|-----|----------|---|------| | Initial | c_i | - | 0 | + | 0 | | Change | -x | - | +x | + | +x | | Equilibrium | c_i-x | - | x | + | x | Here, c_i represents the initial concentration of propionic acid and x represents the change in its concentration due to ionization.

Now, we will solve for x using the Ka expression with the given value of Ka for propionic acid, which is 1.3 x 10^-5. Let's plug the equilibrium concentrations from the ICE table into the Ka expression: \[K_{a} = \frac{x^2}{c_i - x}\] This equation can be solved using the quadratic formula or by the x<<c_i approximation. We'll use the approximation method for simplicity. We can disregard x in the denominator because x<<c_i, which simplifies the equation to: \[K_{a} = \frac{x^2}{c_i}\] Solve for x: \[x = \sqrt{K_{a} \cdot c_i}\]

We can now calculate the percent ionization for each concentration by using the equation: \[\text{Percent Ionization} = \frac{x}{c_i} \times 100\% \] Now, we'll calculate the percent ionization for each given concentration: a) For 0.250 M: \[x = \sqrt{K_{a} \cdot c_i} = \sqrt{1.3 \times 10^{-5} \cdot 0.250}\] \[x = 7.2 \times 10^{-4}\] \[\text{Percent Ionization} = \frac{7.2 \times 10^{-4}}{0.250} \times 100\% \approx 0.29\%\] b) For 0.0800 M: \[x = \sqrt{K_{a} \cdot c_i} = \sqrt{1.3 \times 10^{-5} \cdot 0.0800}\] \[x = 4.1 \times 10^{-4}\] \[\text{Percent Ionization} = \frac{4.1 \times 10^{-4}}{0.0800} \times 100\% \approx 0.51\%\] c) For 0.0200 M: \[x = \sqrt{K_{a} \cdot c_i} = \sqrt{1.3 \times 10^{-5} \cdot 0.0200}\] \[x = 2.6 \times 10^{-4}\] \[\text{Percent Ionization} = \frac{2.6 \times 10^{-4}}{0.0200} \times 100\% \approx 1.3\%\] The percent ionization for the different concentrations of propionic acid are approximately 0.29%, 0.51%, and 1.3% for 0.250 M, 0.0800 M, and 0.0200 M, respectively.