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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

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The common ion effect occurs when a weak electrolyte solution is combined with a solution containing a common ion, which suppresses the extent of ionization of the weak electrolyte. In the case of weak acids, it decreases their dissociation in water. According to Le Chatelier's principle, when the concentration of the common ion increases, the system shifts to counteract the change, decreasing the amount of H+ and A- produced. This ultimately leads to a significant decrease in the dissociation of the weak acid in water, demonstrating the common ion effect.

Step by step solution

01

Understanding the common ion effect

The common ion effect is a phenomenon that occurs when a weak electrolyte solution, like a solution of a weak acid, is combined with a solution containing an ion that is also present in the weak electrolyte. The presence of this common ion suppresses the extent of ionization of the weak electrolyte because of the established equilibrium in weak electrolyte solutions.
02

Demonstrating weak acids' dissociation

Weak acids partially dissociate in water, resulting in an equilibrium between the undissociated acid, the free hydrogen ions (H+), and the conjugate base (A-), which is produced when the weak acid loses a proton. This equilibrium can be represented by the equation: \(HA \rightleftharpoons H^+ + A^-\) Here, HA represents the weak acid, H+ represents the hydrogen ion, and A- is the conjugate base. The equilibrium constant (Ka) of this reaction is given by: \(K_a = \frac{[H^+][A^-]}{[HA]}\) Where [H+], [A-], and [HA] are the concentrations of the hydrogen ions, conjugate base, and undissociated weak acid, respectively.
03

Introducing the common ion

When a salt containing an ion common with the weak acid solution is introduced, the concentration of the common ion increases. For example, in the case of a weak acid with the conjugate base A- as its ions, if we add a salt like NaA, the concentration of the common ion, A-, will significantly increase. This increase in the [A-] will result in the shift of the equilibrium towards the left, as predicted by Le Chatelier's principle, decreasing the dissociation of the weak acid (HA).
04

Explaining dissociation decrease

Le Chatelier's principle states that when a system in equilibrium is disturbed by a change in concentration of the species involved, the system shifts to counteract the change and re-establish the equilibrium. In this case, when the concentration of the common ion (A-) is increased, the system counteracts this change by reducing the ionization of the weak acid (HA), decreasing the amount of H+ and A- produced. This shift in equilibrium ultimately leads to a significant decrease in the dissociation of the weak acid in water, which is the common ion effect.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Weak Acids
Weak acids are fascinating chemical compounds that do not fully ionize in solution. Imagine them as having a reserved nature, where only a small portion separates into their ions. When dissolved in water, a weak acid forms an equilibrium between its undissociated form and its ions:
  • Undissociated molecules (HA): These are the weak acid molecules that have not ionized.
  • Hydrogen ions (H^+): These are the free ions that result from the acid losing a proton.
  • Conjugate base (A^-): This ion forms when the acid loses a hydrogen ion.
For example, acetic acid, a common weak acid, only partially dissociates in water. The equation for this equilibrium looks like this:\[HA \rightleftharpoons H^+ + A^-\] The constant that describes this balance is the acid dissociation constant, K_a. It's a measure of how much the acid dissociates. With weak acids, K_a values are relatively small, which means they do not release many H^+ ions in solution.
Dissociation
Dissociation is the process by which an acid splits into its ions in solution. For weak acids, dissociation is only partial. Unlike strong acids that completely break apart, weak acids delicately maintain an equilibrium. This part-dissociation is key to understanding why weak acids behave as they do in different conditions. The dissociation can be expressed with the same equation:\[HA \rightleftharpoons H^+ + A^-\] The degree to which the acid dissociates is essential in many chemical processes. It affects:
  • pH levels, as only some H^+ ions are freed to affect the environment.
  • Chemical reactivity, as the concentration of ionized acid impacts reactions.
Understanding the balance in dissociation is vital, as it can change due to environmental factors like temperature or the introduction of a common ion, as seen in the common ion effect.
Chemical Equilibrium
Chemical equilibrium refers to a state where the forward and reverse reactions occur at the same rate, causing the concentrations of reactants and products to remain constant over time. For weak acids in solution, this equilibrium is particularly delicate. When a weak acid dissociates, it forms products between its ions and undissociated molecules:\[HA \rightleftharpoons H^+ + A^-\]At equilibrium, the rate of the acid dissociating into ions and re-associating into the whole molecule is balanced. This means the concentrations of HA, H^+, and A^- will remain steady unless the system is affected by external changes. Such stability is important, especially in solutions that require precise pH levels. The interplay between dissociation and equilibrium is essential for understanding how weak acids behave under different conditions and in various chemical reactions.
Le Chatelier's Principle
Le Chatelier's Principle is an essential concept in chemistry, particularly when examining how systems at equilibrium respond to changes. This principle states that if an equilibrium system is disturbed, it will adjust to counteract the disturbance and restore a new equilibrium. In the context of weak acids and the common ion effect, consider this: If you introduce an ion common to the weak acid's dissociation products, like adding A^- from a salt such as NaA to an HA solution, more A^- ions will be present. Le Chatelier's Principle predicts that the equilibrium will shift to reduce this disturbance by decreasing the dissociation of HA , lessening its production of H^+ ions and thus reducing A^- production.
  • This principle helps explain why, in solutions, adding a common ion typically results in decreased ionization of weak acids.
  • It also plays a vital role in controlling reactions and processes in chemical labs and industrial applications.
Understanding Le Chatelier's Principle provides insight into the dynamic nature of chemical reactions, showcasing how systems adapt to maintain stability in response to changes.

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Most popular questions from this chapter

A buffer is prepared by dissolving HONH_ and \(\mathrm{HONH}_{3} \mathrm{NO}_{3}\) in some water. Write equations to show how this buffer neutralizes added \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\)

Calculate the pH after 0.010 mole of gaseous HCl is added to 250.0 \(\mathrm{mL}\) of each of the following buffered solutions. $$ \begin{array}{l}{\text { a. } 0.050 M \mathrm{NH}_{3} / 0.15 \mathrm{MNH}_{4} \mathrm{Cl}} \\ {\text { b. } 0.50 \mathrm{M} \mathrm{NH}_{3} / 1.50 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}}\end{array} $$ Do the two original buffered solutions differ in their pH or their capacity? What advantage is there in having a buffer with a greater capacity?

You have a solution of the weak acid HA and add some of the salt NaA to it. What are the major species in the solution? What do you need to know to calculate the pH of the solution, and how would you use this information? How does the pH of the solution of just the HA compare with that of the final mixture? Explain

Consider the titration of 40.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M} \mathrm{HClO}_{4}\) by 0.100 \(\mathrm{M}\) KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 80.0 \mathrm{mL}} \\ {\text { b. } 10.0 \mathrm{mL}} & {\text { e. } 100.0 \mathrm{mL}} \\ {\text { c. } 40.0 \mathrm{mL}}\end{array} $$

Calculate the volume of \(1.50 \times 10^{-2} M \mathrm{NaOH}\) that must be added to 500.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M} \mathrm{HCl}\) to give a solution that has \(\mathrm{pH}=2.15 .\)

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