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Chapter 42: Problem 12

Complete and balance the following reaction for the weak base pyridine: $$\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons$$ Write the equilibrium expression for the \(K_{\mathrm{b}}\) of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\).

Short Answer

Expert verified
The completed and balanced reaction is \(\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N}(\mathrm{aq}) + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N}\mathrm{H}^{+} + \mathrm{OH}^{-}\). The equilibrium expression for the \( K_{\mathrm{b}} \) of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\) is \( K_{\mathrm{b}} = \frac{[\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N}\mathrm{H}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N}]} \).

Step by step solution

01

Equation Completion and Balancing

The given starting chemicals are pyridine (\( \mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N} \)) and water. In this reaction, water will act as an acid, donating a proton to the base, pyridine. This results in the formation of the pyridinium cation (\( \mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N}\mathrm{H}^{+} \)) and the hydroxide ion (\( \mathrm{OH}^{-} \)). So, the balanced equation is: \( \mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N}(\mathrm{aq}) + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N}\mathrm{H}^{+} + \mathrm{OH}^{-} \).
02

Equilibrium Expression Writing

Using the law of mass action, we can write the equilibrium expression for this reaction. The base dissociation constant \( K_{\mathrm{b}} \) is the ratio of the concentrations of the products to the concentration of the reactants, each raised to the power of their stoichiometric coefficients in the balanced chemical equation. Therefore, \( K_{\mathrm{b}} = \frac{[\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N}\mathrm{H}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{5}\mathrm{H}_{5}\mathrm{N}]}. \) Note that we do not include water in this expression, as the concentration of water remains nearly constant.

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Most popular questions from this chapter

For a weak acid, why is \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{A}^{-}\right] ?\)

A \(0.50 \mathrm{M}\) solution of \(\mathrm{CH}_{3} \mathrm{COOH}\) has \(\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]=3.0 \times 10^{-3} \mathrm{M}\). What are the values of: \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)_{0}=\) \(\left[\mathrm{CH}_{3} \mathrm{COOH}\right]=\) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \quad=\) \(K_{\mathrm{a}} \quad=\)

The \(K_{b}\) of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\), at \(25^{\circ} \mathrm{C}\), is \(1.7 \times 10^{-9}\). Find " \(x\) " (CTQ 13), and enter the equilibrium concentration values in the last row of the following table. $$\begin{array}{|l|c|c|c|} \hline & \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N} & \mathrm{OH}^{-} & \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+} \\ \hline \text { initial moles } & 0.30 & 0 & 0 \\ \hline \text { change in moles } & & & \\ \hline \text { equilibrium moles } & & & \\ \hline \begin{array}{l} \text { equilibrium conc. } \\ \text { expression } \end{array} & & & \\ \hline \begin{array}{l} \text { equilibrium conc. } \\ \text { value } \end{array} & & & \\ \hline \end{array}$$ Verify that your equilibrium concentrations are correct.

Complete and balance the following reaction for the weak acid HOCl: $$\mathrm{HOCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons$$ Write the equilibrium expression for the \(K_{\mathrm{a}}\) of HOCl.
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