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Chapter 42: Problem 18

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

Short Answer

Expert verified
[H3O+] = [A-] because when a weak acid ionizes in water, it produces an equal amount of H3O+ and A-. This is a simplification but is generally accurate for dilute solutions of a single weak acid.

Step by step solution

01

Understand Weak Acid Dissociation

The formula HA ↔ H+ + A- represents the partial ionization of a weak acid HA in solution. This is an equilibrium reaction, meaning the reaction can proceed in both forward and backward directions. In the forward direction, the weak acid HA donates one of its protons (H+) to a water molecule, forming a hydronium ion, H3O+, and the corresponding conjugate base, A-.
02

Equal Concentrations

Since the weak acid is not fully ionized in solution, the concentration of the hydronium ions, [H3O+], is the same as that of the conjugate base, [A-], because every molecule of HA that ionizes produces one H3O+ and one A-. Therefore, we can write [H3O+] = [A-]. This is a simplified version of the actual scenario but is usually accurate for dilute solutions.
03

Understand the simplified scenario

This scenario is simplified as it does not take into account if there are other sources of H3O+ or A- in the solution that could affect their concentrations. However, in a scenario where we focus on a single weak acid dissolving in pure water, this simplification helps us estimate the approximate concentrations of these ions and is generally applicable for dilute solutions.

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Most popular questions from this chapter

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

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.

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.
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