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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 at equilibrium, in units of molarity?

Short Answer

Expert verified
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 by step solution

01

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鈦�) and leading to the acidic pH: \[C_5H_5NH^+ + H_2O \to C_5H_5N + H_3O^+\]
02

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: \[K_a = \frac{[H_3O^+][H_3O^+]}{[H_3O^+]}\]
03

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.

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Most popular questions from this chapter

Which member of each pair produces the more acidic aqueous solution: (a) \(\mathrm{ZnBr}_{2}\) or \(\mathrm{CdCl}_{2}\), (b) \(\mathrm{CuCl}\) or \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\), (c) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) or \(\mathrm{NiBr}_{2}\) ?

Succinic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{4}\right)\), which we will denote \(\mathrm{H}_{2} \mathrm{Suc}\), is a biologically relevant diprotic acid with the structure shown below. It is closely related to tartaric acid and malic acid (Figure 16.1). At \(25^{\circ} \mathrm{C}\), the acid-dissociation constants for succinic acid are \(K_{a 1}=6.9 \times 10^{-5}\) and \(K_{a 2}=2.5 \times 10^{-6}\). (a) Determine the pH of a \(0.32 \mathrm{M}\) solution of \(\mathrm{H}_{2} \mathrm{Suc}\) at \(25^{\circ} \mathrm{C}\), assuming that only the first dissociation is relevant. (b) Determine the molar concentration of \(\mathrm{Suc}^{2-}\) in the solution in part (a). (c) Is the assumption you made in part (a) justified by the result from part (b)? (d) Will a solution of the salt NaHSuc be acidic, neutral, or basic?

Write the chemical equation and the \(K_{a}\) expression for the acid dissociation of each of the following acids in aqueous solution. First show the reaction with \(\mathrm{H}^{+}(a q)\) as a product and then with the hydronium ion: (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\), (b) \(\mathrm{HCO}_{3}^{-}\)

Indicate whether each of the following statements is correct or incorrect. (a) Every Br酶nsted-Lowry acid is also a Lewis acid. (b) Every Lewis acid is also a Bronsted-Lowry acid. (c) Conjugate acids of weak bases produce more acidic solutions than conjugate acids of strong bases. (d) \(\mathrm{K}^{+}\)ion is acidic in water because it causes hydrating water molecules to become more acidic. (e) The percent ionization of a weak acid in water increases as the concentration of acid decreases.

Calculate the molar concentration of \(\mathrm{OH}^{-}\)in a \(0.724 \mathrm{M}\) solution of hypobromite ion \(\left(\mathrm{BrO}^{-} ; K_{b}=4.0 \times 10^{-6}\right)\). What is the \(\mathrm{pH}\) of this solution?
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