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Chapter 16: Problem 62

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 \(0.0482 \mathrm{M} \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} \mathrm{Cl}^{-}(\mathrm{aq})\).

Short Answer

Expert verified
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 by step solution

01

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.
02

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}\]
03

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 \(pH = -log[2.50*10^{-7}] = 6.60\)

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Most popular questions from this chapter

What is the \(\mathrm{pH}\) of the solution obtained when \(125 \mathrm{mL}\) of \(0.606 \mathrm{M} \mathrm{NaOH}\) is diluted to \(15.0 \mathrm{L}\) with water?

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