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 \(\left(K_{a}\right.\) is given in Appendix \(\left.D\right):\) (a) \(0.250 \mathrm{M}\), (b) \(0.0800 \mathrm{M}\), (c) \(0.0200 \mathrm{M}\).

Short Answer

Expert verified
The percent ionization of propionic acid in solutions with concentrations (a) \(0.250 \mathrm{M}\), (b) \(0.0800 \mathrm{M}\), and (c) \(0.0200 \mathrm{M}\) are approximately \(1.3\%\), \(2.3\%\), and \(5.2\%\), respectively.

Step by step solution

01

Write the Equilibrium Expression for Propionic Acid

Write the equilibrium expression for the dissociation of propionic acid in water: \[C_2H_5COOH \leftrightharpoons C_2H_5COO^- + H^+\]
02

Set up the ICE table

Set up an ICE (Initial, Change, Equilibrium) table. Let \(x\) be the concentration of \(C_2H_5COO^-\) and \(H^+\) formed at equilibrium. \[ \begin{array}{c|c|c|c} & [C_2H_5COOH] & [C_2H_5COO^-] & [H^+] \\ \hline \text{Initial} & C_0 & 0 & 0 \\ \text{Change} & -x & +x & +x \\ \text{Equilibrium} & C_0 - x & x & x \\ \end{array} \]
03

Write the \(K_a\) Expression and Rearrange

Write the \(K_a\) expression for the dissociation of propionic acid and rearrange it for \(x\) and substitute equilibrium concentrations: \[K_a = \frac{[C_2H_5COO^-][H^+]}{[C_2H_5COOH]} = \frac{x^2}{C_0 - x}\] Since the acid is weak, \(x\) is small compared to \(C_0\), so we can approximate that \(C_0 - x \approx C_0\): \[x^2 \approx K_a C_0\] Solve for \(x\): \[x = \sqrt{K_a C_0}\]
04

Calculate Percent Ionization

Now use the formula for percent ionization: Percent Ionization \(= \frac{x}{C_0} \times 100\%\) Substitute \(x\) with the expression from Step 3: Percent Ionization \(= \frac{\sqrt{K_a C_0}}{C_0} \times 100\%\)
05

Solve for Each Concentration

From Appendix D, we find that the \(K_a\) value for propionic acid is \(1.34 \times 10^{-5}\). Now we can plug in the given concentrations and calculate the percent ionization for each case. (a) For \(C_0 = 0.250\ \mathrm{M}\): Percent Ionization \(= \frac{\sqrt{1.34 \times 10^{-5} \times 0.250}}{0.250} \times 100\% \approx 1.3\%\) (b) For \(C_0 = 0.0800\ \mathrm{M}\): Percent Ionization \(= \frac{\sqrt{1.34 \times 10^{-5} \times 0.0800}}{0.0800} \times 100\% \approx 2.3\%\) (c) For \(C_0 = 0.0200\ \mathrm{M}\): Percent Ionization \(= \frac{\sqrt{1.34 \times 10^{-5} \times 0.0200}}{0.0200} \times 100\% \approx 5.2\%\) The percent ionization of propionic acid in solutions with concentrations (a) \(0.250 \mathrm{M}\), (b) \(0.0800 \mathrm{M}\), and (c) \(0.0200 \mathrm{M}\) are approximately \(1.3\%\), \(2.3\%\), and \(5.2\%\), respectively.

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