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Chapter 16: Problem 82

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?

Short Answer

Expert verified
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 by step solution

01

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⁻
02

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⁺
03

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
04

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, \(K_{a} = [H_{3}O^{+}] ≈ 1.12 × 10^(-3)\)
05

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.

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Most popular questions from this chapter

Write the chemical equation and the \(K_{b}\) expression for the reaction of each of the following bases with water: (a) propylamine, \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2} ;\) (b) monohydrogen phosphate ion, \(\mathrm{HPO}_{4}^{2-} ;(\mathbf{c})\) benzoate ion, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}.\)

If a solution of \(\mathrm{HF}\left(K_{a}=6.8 \times 10^{-4}\right)\) has a pH of \(3.65,\) calculate the concentration of hydrofluoric acid.

How many milliliters of concentrated hydrochloric acid solution \((36.0 \%\) HCl by mass, density \(=1.18 \mathrm{g} / \mathrm{mL})\) are required to produce 10.0 \(\mathrm{L}\) of a solution that has a pH of 2.05\(?\)

The acid-dissociation constant for chlorous acid \(\left(\mathrm{HClO}_{2}\right)\) is \(1.1 \times 10^{-2} .\) Calculate the concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{ClO}_{2}^{-}\) and \(\mathrm{HClO}_{2}\) at equilibrium if the initial concentration of \(\mathrm{HClO}_{2}\) is 0.0125 \(\mathrm{M} .\)

At the freezing point of water \(\left(0^{\circ} \mathrm{C}\right), K_{w}=1.2 \times 10^{-15}\) Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for a neutral solution at this temperature.
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