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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.

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 \(1.12 \times 10^{-3} \text{ M}\).

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.