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 at equilibrium, in units of molarity?

Step-by-Step Solution

Answer

The acid dissociation reaction of pyridinium bromide in water is: \(C_5H_5NH^+ + H_2O \to C_5H_5N + H_3O^+\). The acid dissociation constant (Ka) for pyridinium bromide is calculated as \(10^{-2.95}\), and the equilibrium concentration of the pyridinium cation (\(C_5H_5NH^+\)) in the solution is also \(10^{-2.95}\) M.

Step 1: Write the acid dissociation reaction of pyridinium bromide in water

Pyridinium bromide (\(C_5H_5NHBr\)) dissociates into pyridinium cation (\(C_5H_5NH^+\)) and bromide ion (\(Br^-\)) in water. Further, the pyridinium cation acts as an acid, donating its proton (H⁺) and leading to the acidic pH: \[C_5H_5NH^+ + H_2O \to C_5H_5N + H_3O^+\]

Step 2: Calculate the acid dissociation constant (Ka) using the pH and data from Appendix D

Given the pH value of the solution is 2.95, we can find the concentration of \(H_3O^+\) as: \[H_3O^+ = 10^{(-pH)}\] \[H_3O^+ = 10^{(-2.95)}\] Now, let's use the information given in Appendix D. It is mentioned that pyridinium bromide is a strong electrolyte and dissociates completely in water. This suggests that the initial concentration of pyridinium cation and bromide ion in the solution are equal, before undergoing any reaction with water. Since the pyridinium cation (\(C_5H_5NH^+\)) acts as an acid and donates a proton to the water, we can write the expression of acid dissociation constant (Ka) at equilibrium: \[K_a = \frac{[C_5H_5N][H_3O^+]}{[C_5H_5NH^+]}\] We are given that the pH is 2.95, so we can assume that at equilibrium, the concentration of pyridinium cation and pyridine base, the conjugate base, \(C_5H_5N\) are equal. So, substituting the equilibrium concentrations in the above equation, we get: \[K_a = \frac{[H_3O^+][H_3O^+]}{[H_3O^+]}\]

Step 3: Find Ka and equilibrium concentration of pyridinium cation

Now, let's simplify and find the Ka and equilibrium concentration of pyridinium cation using the equation we derived in step 2: \[K_a = [H_3O^+] = 10^{(-2.95)}\] So, the Ka of pyridinium bromide is calculated as \(10^{(-2.95)}\). At equilibrium, the concentration of pyridinium cation in the solution is: \[C_5H_5NH^+ = [H_3O^+] = 10^{-2.95}\] Hence, the equilibrium concentration of pyridinium cation (\(C_5H_5NH^+\)) in the solution is \(10^{-2.95}\) M.