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Chapter 42: Problem 4

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.

Short Answer

Expert verified
The balanced chemical reaction is \(HOCl(aq) + H_2O(l) \rightleftharpoons Cl^-(aq) + H_3O^+(aq)\) and the simplified equilibrium expression for \(Ka\) of HOCl is: \[Ka = \frac{[Cl^-][H_3O^+]}{[HOCl]}\]

Step by step solution

01

Complete and Balance the Reaction

HOCl, being a weak acid, will donate a proton \(H^+\) to water, acting as a base. This leaves behind a Cl- ion and transforms water to a hydronium ion \(H_3O^+\). The balanced chemical equation becomes: \[HOCl(aq) + H_2O(l) \rightleftharpoons Cl^-(aq) + H_3O^+(aq)\].
02

Write the Equilibrium Expression

The equilibrium constant \(Ka\) for a reaction can be summarized as: \([products]/[reactants]\). We can write this for the given reaction as: \[Ka = \frac{[Cl^-][H_3O^+]}{[HOCl]}\]. Notice that concentration of water \([H_2O]\) is not included because its concentration remains essentially constant in dilute aqueous solutions.
03

Simplify the Equilibrium Expression

Since the coefficients are all 1, there's no need to take powers in the equilibrium expression. Therefore, it remains as: \[Ka = \frac{[Cl^-][H_3O^+]}{[HOCl]}\]

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Most popular questions from this chapter

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

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 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 \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\).

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: \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)_{0}=\) \(\left[\mathrm{CH}_{3} \mathrm{COOH}\right]=\) \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \quad=\) \(K_{\mathrm{a}} \quad=\)
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