Question

Calculate the \(\mathrm{pH}\) of each of the following solutions \(\left(K_{a}\right.\) and \(K_{b}\) values are given in Appendix \(D\) ): (a) \(0.095 M\) propionic acid \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\right),\) (b) \(0.100 \quad M\) hydrogen chromate ion \(\left(\mathrm{HCrO}_{4}^{-}\right),\left(\right.\) c) \(0.120 \mathrm{M}\) pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\)

Short answer

The pH of the given solutions can be calculated as follows: 1. For the $0.095 M$ propionic acid solution, after solving for x in the equilibrium equation, we find the pH of the solution. 2. Similarly, for the $0.100 M$ hydrogen chromate ion solution, we find the pH using the given $K_a$ and initial concentration. 3. For the $0.120 M$ pyridine solution, we find the pH using the $K_b$ value and converting the $\mathrm{OH}^{-}$ ion concentration to $\mathrm{H}^{+}$ concentration.

Step-by-step solution

1. Finding the pH of the propionic acid solution

Write the ionization equilibrium for propionic acid: \(C_2H_5COOH \rightleftharpoons H^+ + C_2H_5COO^-\) Set up an ICE table to find the equilibrium concentrations of \(\mathrm{H}^{+}\) and \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COO}^{-}\): \[ \begin{array}{c|c c c} & [C_2H_5COOH] & [H^+] & [C_2H_5COO^-] \\ \hline I & 0.095 & 0 & 0 \\ C & -x & x & x \\ E & 0.095-x & x & x \\ \end{array} \] Use the given \(K_a\) value from Appendix D for propionic acid to find x: \(K_a = \frac{[H^+][C_2H_5COO^-]}{[C_2H_5COOH]} = \frac{x^2}{0.095-x}\) Solve for x, which represents the \(\mathrm{H}^{+}\) ion concentration, and then use the pH formula to find the pH of the solution.

2. Finding the pH of the hydrogen chromate ion solution

Write the ionization equilibrium for hydrogen chromate ion: \(HCrO_4^- \rightleftharpoons H^+ + CrO_4^{2-}\) Follow the same ICE table and equilibrium steps as in the propionic acid calculation with the initial concentration of hydrogen chromate ion given as 0.100 M and the \(K_a\) value for hydrogen chromate from Appendix D.

3. Finding the pH of the pyridine solution

Write the base ionization equilibrium for pyridine: \(C_5H_5N + H_2O \rightleftharpoons OH^- + C_5H_5NH^+\) Set up an ICE table to find the equilibrium concentrations of \(\mathrm{OH}^{-}\) and \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\), with the initial concentration of pyridine given as 0.120 M: \[ \begin{array}{c|c c} & [OH^-] & [C_5H_5NH^+] \\ \hline I & 0 & 0 \\ C & +x & +x \\ E & x & x \\ \end{array} \] Use the given \(K_b\) value from Appendix D for pyridine to find x: \(K_b = \frac{[\mathrm{OH}^{-}][\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}]}{[\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}]} = \frac{x^2}{0.120}\) Solve for x, which represents the \(\mathrm{OH}^{-}\) ion concentration. Then, convert this to \(\mathrm{H}^{+}\) concentration by using the relation: \([\mathrm{H}^{+}][\mathrm{OH}^{-}] = K_w\), where \(K_w = 1.0 \times 10^{-14}\) Finally, use the pH formula to find the pH of the pyridine solution.

Key concepts

Acid-Base Equilibrium

Understanding acid-base equilibrium is key when calculating the pH of a solution. Each acid or base, when dissolved in water, reaches a balance between its ionized and non-ionized forms. This state is known as equilibrium. For example, consider the propionic acid solution given in the exercise. Treated as a weak acid, it only partially ionizes in water, leading to the equilibrium:\[\text{C}_2\text{H}_5\text{COOH} \rightleftharpoons \text{H}^+ + \text{C}_2\text{H}_5\text{COO}^-\]The position of this equilibrium can be described by the ion concentrations at the equilibrium state. Recognizing this duality in acid-base equilibrium helps to predict how an acid or base behaves in solution.

Ionization Constant

The ionization constant, represented as \(K_a\) or \(K_b\) for acids and bases respectively, quantifies the strength of an acid or base in solution. For weak acids and bases, these constants are particularly important as they indicate the extent of ionization.For example, in the case of the propionic acid exercise, the \(K_a\) provides the tendency of the acid to donate its \(H^+\) ion:\[K_a = \frac{[\text{H}^+][\text{C}_2\text{H}_5\text{COO}^-]}{[\text{C}_2\text{H}_5\text{COOH}]}\]A similar calculation applies to bases with \(K_b\), such as pyridine, which measures the production of \([\text{OH}^-]\) in solution. Knowing the ionization constant allows us to calculate the concentrations of ions at equilibrium, which is a precursor to finding the pH.

ICE Table

An ICE table, which stands for Initial, Change, Equilibrium, is a valuable tool that helps students systematically calculate the concentration changes in a chemical equilibrium. It simplifies the setup and resolution of equilibrium problems by breaking them into smaller, manageable parts.Consider the hydrogen chromate ion solution in the original exercise. The ICE table begins with the initial concentrations:

• \(I\): Initial concentrations before the equilibrium
• \(C\); Changes in concentrations as equilibrium is reached
• \(E\); Equilibrium concentrations

ICE tables clarify the relationship between initial conditions, changes that occur, and the final state of compounds in solution. This step-by-step structure is essential for correctly solving equilibrium concentration problems.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of reactants and products after a reaction has reached its equilibrium state. This is the key to calculating pH after applying the ICE table strategy.In the discussed pyridine solution, the equilibrium concentration of \(\text{OH}^-\) is pivotal. Once the values are determined using \(K_b\):\[\text{Equilibrium Concentration of \(OH^-\)} = x\]Following the relationship \([\text{H}^+][\text{OH}^-] = K_w\), where \(K_w = 1.0 \times 10^{-14}\), enables conversion from hydroxide to hydronium ions, ultimately leading to the solution's pH.Understanding equilibrium concentrations allows us to transition from calculated ion levels to a universal measure of acidity or alkalinity, the pH.