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Chapter 15: Problem 9

The common ion effect for weak acids is to significantly decrease the dissociation of the acid in water. Explain the common ion effect.

Short Answer

Expert verified
The common ion effect is the suppression of the dissociation of a weak electrolyte, like a weak acid, due to the presence of a common ion in the solution. A weak acid partially dissociates in water, reaching an equilibrium between undissociated acid and produced ions. For example, adding sodium acetate (CH₃COONa) to a solution of acetic acid (CH₃COOH) increases the concentration of the common ion CH₃COO⁻. According to Le Châtelier's principle, the system adjusts to reestablish equilibrium by shifting towards the formation of undissociated CH₃COOH molecules, which decreases the dissociation of the weak acid.

Step by step solution

01

Define the Common Ion Effect

The common ion effect refers to the suppression of the dissociation of a weak electrolyte (like a weak acid or weak base) when another electrolyte containing a common ion is added to the solution. In other words, the presence of a common ion in the solution reduces the solubility of the weak electrolyte.
02

Describe Weak Acids

A weak acid is an acid that does not fully dissociate in water, meaning not all the molecules break apart into ions. Instead, a partial equilibrium is reached between the undissociated acid molecules and the produced ions. The level of dissociation in water for a given weak acid can be represented by an equilibrium constant, known as the acid dissociation constant (K_a). For example, considering a generic weak acid, HA: \[HA \rightleftharpoons H^+ + A^-\] The acid dissociation constant (K_a) is given by: \[K_a = \frac{[H^+][A^-]}{[HA]}\] Where [H^+], [A^-], and [HA] represent the equilibrium concentrations of the respective species.
03

Illustrate the Common Ion Effect with an Example

Consider the weak acid acetic acid, CH₃COOH, which dissociates according to the following equation: \[CH_3COOH \rightleftharpoons H^+ + CH_3COO^-\] Now let's add a common ion, for example, sodium acetate (CH₃COONa), which is a strong electrolyte and completely dissociates into its ions in water: \[CH_3COONa \rightarrow CH_3COO^- + Na^+\] As a result, the concentration of the common ion CH₃COO⁻ increases. According to Le Châtelier's principle, the system will now adjust to counteract this change and reestablish the equilibrium. In this case, the reaction will shift towards the left, favoring the formation of undissociated CH₃COOH molecules and decreasing the dissociation of the weak acid.
04

Explain the Role of Equilibrium and Le Châtelier's Principle

The common ion effect can be explained using the concepts of equilibrium and Le Châtelier's principle. In a weak acid dissociation reaction, an equilibrium exists between the undissociated acid and its dissociated ions. Le Châtelier's principle states that if an external stress, e.g., a change in concentration, is applied to a system at equilibrium, the system will adjust to minimize the effect of the stress and restore equilibrium. In the case of the common ion effect, the increased concentration of the common ion causes the equilibrium to shift towards the undissociated acid, thus reducing the dissociation.

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Most popular questions from this chapter

A \(0.400 M\) solution of ammonia was titrated with hydrochloric acid to the equivalence point, where the total volume was \(1.50\) times the original volume. At what \(\mathrm{pH}\) does the equivalence point occur?

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Calculate the \(\mathrm{pH}\) of a solution that is \(1.00 \mathrm{M} \mathrm{HNO}_{2}\) and \(1.00 \mathrm{M}\) \(\mathrm{NaNO}_{2}\)

Consider the titration of \(100.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}\left(K_{\mathrm{b}}=\right.\) \(3.0 \times 10^{-6}\) ) by \(0.200 \mathrm{M} \mathrm{HNO}_{3} .\) Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of \(\mathrm{HNO}_{3}\) have been added. a. \(0.0 \mathrm{~mL}\) d. \(40.0 \mathrm{~mL}\) b. \(20.0 \mathrm{~mL}\) e. \(50.0 \mathrm{~mL}\) c. \(25.0 \mathrm{~mL}\) f. \(100.0 \mathrm{~mL}\)

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