Problem 59 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

The pH values for the three solutions are: (a) For the 0.095 M propionic acid (C2H5COOH) solution, the pH is approximately 2.87. (b) For the 0.100 M hydrogen chromate ion (HCrO4-) solution, the pH is approximately 1.34. (c) For the 0.120 M pyridine (C5H5N) solution, the pH is approximately 9.38.

Step 1: (a) Calculating pH of propionic acid solution

First, write the ionization reaction and the corresponding equilibrium expression. \(C_{2}H_{5}COOH \rightleftharpoons C_{2}H_{5}COO^{-} + H^{+}\) \(K_{a} = \frac{[C_{2}H_{5}COO^{-}][H^+]}{[C_{2}H_{5}COOH]}\) We are given that the initial concentration of propionic acid is 0.095 M, and the initial concentrations of the other ions are 0 (since they have not yet been produced). Let \(x\) be the equilibrium concentration of \(H^+\) ions. Then: \([C_{2}H_{5}COO^{-}] = x\) \([H^{+}] = x\) \([C_{2}H_{5}COOH] = 0.095 - x\) Now substitute these concentrations into the equilibrium expression, and solve for \(x\): \(K_{a} = \frac{x^{2}}{0.095-x}\) Next, check if the approximation method can be applied (\(x<<0.095\)). If so, we can simplify the equation by assuming that \(0.095 - x = 0.095\). Plug in the given value of \(K_a\) for propionic acid from Appendix D and solve for \(x\). Finally, calculate the pH using the formula pH = -log(\([H^{+}]\)).

Step 2: (b) Calculating pH of hydrogen chromate ion solution

Since hydrogen chromate ion is a conjugate weak acid, first write the ionization reaction and the corresponding equilibrium expression. \(HCrO_{4}^{-} \rightleftharpoons CrO_{4}^{2-} + H^{+}\) \(K_{a} = \frac{[CrO_{4}^{2-}][H^+]}{[HCrO_{4}^{-}]}\) Given the initial concentration of hydrogen chromate ion is 0.100 M and the initial concentrations of the other ions are 0. Let the equilibrium concentration of \(H^+\) ions be \(x\). Then: \([CrO_{4}^{2-}] = x\) \([H^{+}] = x\) \([HCrO_{4}^{-}] = 0.100 - x\) Now substitute these concentrations into the equilibrium expression and solve for \(x\). Apply the approximation method if possible. Use given \(K_{a}\) value for hydrogen chromate ion to solve for \(x\). Finally, calculate the pH using the formula pH = -log(\([H^{+}]\)).

Step 3: (c) Calculating pH of pyridine solution

Pyridine is a weak base. Write the reaction and the corresponding equilibrium expression. \(C_{5}H_{5}N + H_2O \rightleftharpoons C_{5}H_{5}NH^+ + OH^-\) \(K_{b} = \frac{[C_{5}H_{5}NH^{+}][OH^{-}]}{[C_{5}H_{5}N]}\) Initial concentration of pyridine is 0.120 M, and initial concentrations of the other ions are 0. Let the equilibrium concentration of \(OH^-\) ions be \(x\). Then: \([C_{5}H_{5}NH^{+}] = x\) \([OH^{-}] = x\) \([C_{5}H_{5}N] = 0.120 - x\) Now substitute these concentrations into the equilibrium expression and solve for \(x\). Apply the approximation method if possible. Use given \(K_{b}\) value for pyridine to solve for \(x\). Then, calculate the \(pOH\) using the formula \(pOH\) = -log(\([OH^{-}]\)). Finally, find the pH using the relationship pH = 14 - pOH.

Key Concepts

Equilibrium expression

When dealing with acid-base reactions, the equilibrium expression is a crucial concept. It represents the balance between reactants and products at equilibrium. For weak acids and bases, such as propionic acid or pyridine, we set up an equilibrium expression using their respective equilibrium constants, the acid dissociation constant \(K_a\) or base dissociation constant \(K_b\).

For propionic acid, the equilibrium expression is:

• \[K_a = \frac{[C_{2}H_{5}COO^{-}][H^+]}{[C_{2}H_{5}COOH]}\]

This tells us how the acid ionizes into its conjugate base \([C_{2}H_{5}COO^-]\) and hydrogen ions \([H^+]\).

Similarly, for pyridine, which acts as a weak base:

• \[K_b = \frac{[C_{5}H_{5}NH^{+}][OH^{-}]}{[C_{5}H_{5}N]}\]

This reflects how pyridine accepts a proton from water to form \([C_{5}H_{5}NH^+]\) and hydroxide ions \([OH^-]\). An understanding of these expressions helps determine the concentration of ions at equilibrium, which is vital for calculating the pH of a solution.

Acid-base ionization

Acid-base ionization refers to the process by which an acid or base releases its ions in solution. For weak acids like propionic acid \(C_{2}H_{5}COOH\), it ionizes slightly in water to form \([C_{2}H_{5}COO^-]\) and \([H^+]\).

This partial ionization is what distinguishes weak acids and bases from strong ones. Strong acids and bases ionize completely in solution. Weak bases, such as pyridine \(C_{5}H_{5}N\), follow a similar partial ionization pattern, reacting with water to produce its conjugate acid \([C_{5}H_{5}NH^+]\) and \([OH^-]\).

This ionization supports calculations for pH, as it helps to predict how much acid or base is present in its ionized form. In cases where the ionization constant \(K_a\) or \(K_b\) is small, it indicates that the substance is not very strong in donating or accepting protons.

Approximation method

The approximation method is a helpful mathematical tool in solving equilibrium calculations for weak acids and bases. When the concentration change \(x\) due to ionization is very small compared to the initial concentration, this method greatly simplifies calculations.

For example, if \(x\ll 0.095\) for propionic acid, we can approximate \(0.095 - x \approx 0.095\). This means you can ignore the change in the initial concentration caused by ionization, allowing you to solve the equation more efficiently.

After applying the approximation, you substitute back to find the concentration of hydrogen ions \([H^+]\) or hydroxide ions \([OH^-]\). With this, calculating the pH or pOH becomes straightforward with the log formulas. However, after obtaining results, always verify whether the approximation was valid by checking if \(x\) is indeed small enough compared to the initial concentration. This maintains the accuracy of your results.

Weak acid

A weak acid is one that does not completely ionize in solution. Propionic acid is a prime example. Upon dissolving, only a small fraction of the acid molecules donate hydrogen ions to form \([H^+]\).

The equilibrium expression for its dissociation, \(K_a\), is quite essential as it reflects the strength of the acid. Generally, a smaller \(K_a\) value indicates a weaker acid.

Characteristics of weak acids include:

• Partial ionization in water.
• Lower conductivity than strong acids.
• Equilibria strongly favor the direction of the unionized form.

Understanding weak acids is crucial for solving their pH because their partial ionization greatly affects the hydrogen ions present, thus affecting the pH.

Weak base

A weak base, such as pyridine, behaves similarly to weak acids in that it only partially ionizes in solution. When pyridine reacts with water, it partially accepts protons to form its conjugate acid \([C_{5}H_{5}NH^+]\). This reaction also produces hydroxide ions \([OH^-]\).

The equilibrium constant \(K_b\) for weak bases is lower than that of strong bases, reflecting their limited extent of ionization. For example, pyridine's \(K_b\) will be lower than that of ammonia, indicating it is a weaker base.

Characteristics of weak bases:

• Partial ionization in water.
• Lower \(pH\) increase compared to strong bases.
• Equilibria heavily favor the reactants, or the non-ionized form.

Understanding weak bases helps in calculating pOH and subsequently pH since the limited production of hydroxide ions impacts the overall solution pH.