Problem 62 - General Chemistry: Principles and Modern Applications

Question

Pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\left(\mathrm{p} K_{\mathrm{b}}=8.82\right),\) forms a salt, pyridinium chloride, as a result of a reaction with HCl. Write an ionic equation to represent the hydrolysis of the pyridinium ion, and calculate the \(\mathrm{pH}\) of \(0.0482 \mathrm{M} \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} \mathrm{Cl}^{-}(\mathrm{aq})\).

Step-by-Step Solution

Answer

The hydrolysis of the pyridinium ion can be represented as follows: C5H5NH+(aq) + H2O(l) ⇌ C5H5N(aq) + H3O+(aq). The pH of 0.0482 M C5H5NH+Cl-(aq) is 6.60.

Step 1: Write the Hydrolysis of Pyridinium Ion

The hydrolysis of the pyridinium ion (C5H5NH+) can be represented as follows: \[\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{C}_{5}\mathrm{H}_{5} \mathrm{N}(\mathrm{aq}) + \mathrm{H}_{3} \mathrm{O}^{+} (\mathrm{aq})\] This equation shows that the pyridinium ion reacts with water to form pyridine and a hydronium ion.

Step 2: Calculate Concentration of Hydronium Ions

We know that \[K_{a}=K_{w}/K_{b}\], where \(K_{w}\) is the autoprotolysis constant of water, \(10^{-14} \mathrm{M^{2}}\) at 25°C. Substitute given \[K_{b}=10^{-8.82}\] into the equation getting \[K_{a}=10^{-14}/10^{-8.82} = 10^{-5.18}\]. Also, we can use \(K_{a}\) as follows: \[K_a = [C_5H_5N][H_3O^+]/[C_5H_5NH^+] \rightarrow [H_3O^+]= K_a[C_5H_5NH^+]/[C_5H_5N]\]. We have no free \(C_5H_5N\) so \([H_3O^+] = Ka[C_5H_5NH^+]= 10^{-5.18} * 0.0482 M = 2.50 * 10^{-7}\]

Step 3: Calculate pH

The \(pH\) of the solution can be calculated using the following formula: \[pH = -log[H_3O^+]\]. Substitute \([H_3O^+]\) calculated in previous step getting \(pH = -log[2.50*10^{-7}] = 6.60\)

Key Concepts

Pyridinium Ion

Pyridinium ion, designated as \( \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^+ \), forms when pyridine \((\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N})\) reacts with hydrogen chloride (HCl). This reaction results in the formation of the salt pyridinium chloride. The pyridinium ion is the protonated form of pyridine, where an additional hydrogen ion \((\mathrm{H}^+)\) is attracted to the nitrogen atom in the pyridine ring.

• Pyridine is a weak base, and its conjugate acid is the pyridinium ion.
• The additional proton in the pyridinium ion results from the acid-base reaction of pyridine with HCl.
• The presence of pyridinium ions indicates that a weak base reacted with a strong acid.

Understanding the role of the pyridinium ion is pivotal for predicting how it will behave in an aqueous solution, especially when it comes to hydrolysis reactions. As pyridinium ions interact with water, they have the potential to modify the solution's pH.

Hydrolysis Reaction

Hydrolysis reactions are processes where chemical bonds are broken through the reaction with water. In the case of the pyridinium ion \((\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^+)\), hydrolysis is a key reaction that affects the pH of the solution. For the pyridinium ion, the hydrolysis reaction can be written as follows: \\[ \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^+ (\mathrm{aq}) + \mathrm{H}_2\mathrm{O} (\mathrm{l}) \rightleftharpoons \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N} (\mathrm{aq}) + \mathrm{H}_3\mathrm{O}^+ (\mathrm{aq}) \]

• The hydrolysis involves the conversion of pyridinium ions into pyridine \((\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N})\) and hydronium ions \((\mathrm{H}_3\mathrm{O}^+)\).
• This reaction helps regenerate the weak base (pyridine) and produces \( \mathrm{H}_3\mathrm{O}^+ \), impacting the acidity of the solution.

By understanding the hydrolysis of pyridinium ions, we can predict how the balance between acids and bases in solution will shift. This knowledge is crucial when determining the pH of a solution containing pyridinium ions.

pH Calculation

Calculating the pH of a solution involves determining the concentration of hydronium ions \([\mathrm{H}_3\mathrm{O}^+]\) present. For pyridinium chloride in water, the pH calculation takes into account the equilibrium set by the hydrolysis reaction.Using the potassium constant, \(K_a\), derived from the water ion product \(K_w = 10^{-14}\) and knowing that \(K_b = 10^{-8.82}\) for pyridine, we can find \(K_a\): \[ K_a = \frac{K_w}{K_b} = 10^{-14}/10^{-8.82} = 10^{-5.18} \]To find the concentration of \( \mathrm{H}_3\mathrm{O}^+ \), use the equation: \[ \mathrm{[H_3O^+]} = K_a \times \mathrm{[C_5H_5NH^+]} = 10^{-5.18} \times 0.0482 \mathrm{M} \]

• This results in \([\mathrm{H}_3\mathrm{O}^+] = 2.50 \times 10^{-7} \; \mathrm{M} \).

Finally, the pH is computed using the relation \[ pH = -\log \mathrm{[H_3O^+]} \]Substituting the hydronium ion concentration gives:\[ pH = -\log(2.50 \times 10^{-7}) = 6.60 \]Understanding pH calculations is essential for chemists and students to predict how the acidity or basicity of a solution impacts chemical reactions.