Problem 82 - Chemistry The Central Science

Question

Pyridinium bromide \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHBr}\right)\) is a strong electrolyte that dissociates completely into \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) and \(\mathrm{Br}^{-} .\) An aqueous solution of pyridinium bromide has a pH of \(2.95 .\) (a) Write out the reaction that leads to this acidic pH. (b) Using Appendix D, calculate the \(K_{a}\) for pyridinium bromide. (c) A solution of pyridinium bromide has a pH of 2.95 . What is the concentration of the pyridinium cation atequilibrium, in units of molarity?

Step-by-Step Solution

Answer

The dissociation reaction of pyridinium bromide leading to the acidic pH is: C₅H₅NH⁺ + H₂O → C₅H₅N + H₃O⁺ The \(K_{a}\) value for pyridinium bromide is approximately 1.12 × 10^(-3), and the concentration of the pyridinium cation in a solution with a pH of 2.95 is approximately 1.12 × 10^(-3) M.

Step 1: 1. Write the dissociation reaction of pyridinium bromide

In an aqueous solution, pyridinium bromide dissociates into pyridinium cation and bromide ion, which will lead to an acidic pH. The dissociation reaction can be written as: C₅H₅NHBr → C₅H₅NH⁺ + Br⁻

Step 2: 2. Write the reaction that shows the acidic pH

Since the pyridinium cation (C₅H₅NH⁺) is acidic, it will donate a proton to a water molecule, forming hydronium ions (H₃O⁺) and dissociated pyridine (C₅H₅N), which results in the acidic pH of the solution. The reaction can be written as: C₅H₅NH⁺ + H₂O → C₅H₅N + H₃O⁺

Step 3: 3. Calculate the concentration of H₃O⁺ ions using pH

We know that the pH of the solution is 2.95. The relationship between pH and the concentration of H₃O⁺ ions can be expressed using the following equation: pH = -log[H₃O⁺] Now, to find the concentration of H₃O⁺ ions at equilibrium, we can use the given pH value: [H₃O⁺] = 10^(-pH) [H₃O⁺] = 10^(-2.95) [H₃O⁺] ≈ 1.12 × 10^(-3) M

Step 4: 4. Calculate the \(K_{a}\) value for pyridinium bromide

Since pyridinium bromide is a strong electrolyte, the concentration of pyridinium cation (C₅H₅NH⁺) is equal to the concentration of H₃O⁺ ions at equilibrium. Therefore, we can write the equilibrium expression for the \(K_{a}\) value using the relationship between the concentrations of the reactants and products: \(K_{a} = \frac{[C_{5}H_{5}N][H_{3}O^{+}]}{[C_{5}H_{5}NH^{+}]}\) As we know that [C₅H₅NH⁺] = [H₃O⁺], \(K_{a} = \frac{[C_{5}H_{5}N][H_{3}O^{+}]}{[H_{3}O^{+}]} = [C_{5}H_{5}N]\) By stoichiometry, at equilibrium, [C₅H₅N] = [H₃O⁺], So, \(K_{a} = [H_{3}O^{+}] ≈ 1.12 × 10^(-3)\)

Step 5: 5. Determine the concentration of the pyridinium cation

Since the pyridinium cation and hydronium ions have equal concentration in the solution, the concentration of the pyridinium cation is the same as the concentration of H₃O⁺ ions in the solution: [C₅H₅NH⁺] ≈ 1.12 × 10^(-3) M To summarize, the dissociation reaction of pyridinium bromide leading to the acidic pH is: C₅H₅NH⁺ + H₂O → C₅H₅N + H₃O⁺ The \(K_{a}\) value for pyridinium bromide is approximately 1.12 × 10^(-3), and the concentration of the pyridinium cation in a solution with a pH of 2.95 is approximately 1.12 × 10^(-3) M.

Key Concepts

Pyridinium Bromide Dissociation

In any solution containing pyridinium bromide (\(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHBr}\)), the salt dissociates completely as it behaves as a strong electrolyte. This means it breaks into its ions, the pyridinium cation (\(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\)) and the bromide anion (\(\mathrm{Br}^{-}\)), once dissolved in water. Due to the complete dissociation characteristic of strong electrolytes, the concentration of pyridinium bromide initially introduced into the solution becomes equal to the concentration of the pyridinium cation and bromide anion at equilibrium.

It is this pyridinium cation that plays a key role in creating an acidic environment. It can protonate with water, donating a proton and forming hydronium ions (\(\mathrm{H}_{3}\mathrm{O}^{+}\)) and free pyridine (\(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\)). This reaction can be written as:

• \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{NH}^{+} + \mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N} + \mathrm{H}_{3}\mathrm{O}^{+}\)

Thus, the release of these hydronium ions into the solution is why pyridinium bromide results in an acidic pH.

pH Calculation

The pH of a solution measures the acidity or concentration of hydrogen ions. In the context of pyridinium bromide, we know the solution has a pH of 2.95. The pH is related to the concentration of hydronium ions by the formula:

• \(pH = -\log[\mathrm{H}_{3}\mathrm{O}^{+}]\)

Given this pH, we can rearrange this formula to find the concentration of hydronium ions (\([\mathrm{H}_{3}\mathrm{O}^{+}]\)). Using algebraic manipulation and the definition of the logarithm, the hydronium ion concentration is calculated as:

• \([\mathrm{H}_{3}\mathrm{O}^{+}] = 10^{-pH}\)

Substituting the known pH value, the formula becomes:

• \([\mathrm{H}_{3}\mathrm{O}^{+}] = 10^{-2.95}\)
• \([\mathrm{H}_{3}\mathrm{O}^{+}] \approx 1.12 \times 10^{-3} \text{ M}\)

This means that the equilibrium concentration of hydronium ions, a key factor in determining the acidity of the solution, is approximately \(1.12 \times 10^{-3} \text{ M}\).

Equilibrium Concentration

When examining the equilibrium concentration in a reaction equilibrium context, it is important to identify the concentrations of the reactants and products at equilibrium. For the pyridinium bromide in water, the equilibrium is characterized by the concentration of pyridinium ions, hydronium ions, and pyridine.

As the solution is acidic, it indicates the presence of hydronium ions, which are at the same concentration as the pyridinium ions because of their complete dissociation and simple 1:1 reaction stoichiometry:

• \(C_{5}H_{5}NH^{+} + H_{2}O \rightarrow C_{5}H_{5}N + H_{3}O^{+}\)

At equilibrium, the concentration of \(C_{5}H_{5}NH^{+}\) is equal to the concentration of \([\mathrm{H}_{3}\mathrm{O}^{+}]\), calculated previously to be \(1.12 \times 10^{-3} \text{ M}\). Therefore, the equilibrium concentration of the pyridinium ions is also \(1.12 \times 10^{-3} \text{ M}\).

Understanding this balance of concentrations at equilibrium helps explain how acidic the solution is, depending on how much pyridinium bromide is initially dissolved and how it dissociates into ions.