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Chapter 17: Problem 13

(a) What is the common-ion effect? (b) Give an example of a salt that can decrease the ionization of \(\mathrm{HNO}_{2}\) in solution.

Short Answer

Expert verified
(a) The common-ion effect is the phenomenon in which the solubility of a weak electrolyte is reduced by the presence of a common ion in the solution, causing the equilibrium to shift away from ionization due to the increased concentration of the common ion. (b) An example of a salt that can decrease the ionization of \(\mathrm{HNO}_{2}\) in solution is \(\mathrm{KNO}_{2}\) (potassium nitrite).

Step by step solution

01

Define the common-ion effect

The common-ion effect is a phenomenon in which the solubility of a weak electrolyte is reduced by the presence of a common ion in the solution. This occurs because, according to Le Châtelier's principle, the equilibrium shifts away from the ionization of the weak electrolyte to counteract the increase in the concentration of the common ion.
02

Apply the common-ion effect to weak acids

In the context of weak acids, the common-ion effect is particularly relevant. When a weak acid (such as \(\mathrm{HNO}_{2}\)) ionizes in water, it donates a proton and forms a conjugate base, as shown in the following equilibrium reaction: \[\mathrm{HNO}_{2} \leftrightarrows \mathrm{H}^{+} + \mathrm{NO}_{2}^{-}\] If a salt containing a common ion with the weak acid is added to the solution, it will increase the concentration of that ion. As a result, the equilibrium of the ionization reaction will shift to the left, reducing the ionization of the weak acid.
03

Provide an example of a salt that decreases the ionization of \(\mathrm{HNO}_{2}\)

Now, let's find a salt that can decrease the ionization of \(\mathrm{HNO}_{2}\) in solution. We need to look for a salt that shares a common ion with the weak acid \(\mathrm{HNO}_{2}\). In this case, the common ion is \(\mathrm{NO}_{2}^{-}\). A suitable example of such a salt is \(\mathrm{KNO}_{2}\) (potassium nitrite). When dissolved in water, \(\mathrm{KNO}_{2}\) dissociates into its constituent ions: \[\mathrm{KNO}_{2} \rightarrow \mathrm{K}^{+} + \mathrm{NO}_{2}^{-}\] With the increased concentration of the \(\mathrm{NO}_{2}^{-}\) ion in the solution, the equilibrium of the dissociation reaction of \(\mathrm{HNO}_{2}\) shifts to the left, reducing its ionization. Thus, the answer to this exercise is as follows: (a) The common-ion effect is the phenomenon in which the solubility of a weak electrolyte is reduced by the presence of a common ion in the solution. This happens because the equilibrium shifts away from ionization due to the increased concentration of the common ion. (b) An example of a salt that can decrease the ionization of \(\mathrm{HNO}_{2}\) in solution is \(\mathrm{KNO}_{2}\) (potassium nitrite).

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium refers to a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, leading to constant concentrations of products and reactants over time. This balance is dynamic, meaning that the reactions continue to occur, but without changing the overall amounts of compounds present. It is represented by the equilibrium constant, denoted by K, a value that reflects the ratio of the concentrations of products to the reactants for a reaction at equilibrium.

Understanding chemical equilibrium is critical when studying the common-ion effect as it explains why the addition of a common ion to a solution impacts the ionization of a weak electrolyte. The concept helps us to predict how the system will respond to changes in concentration, temperature, or pressure, by shifting the equilibrium position to either favor the formation of products or reactants, a principle known as Le Châtelier's principle.
Solubility of Weak Electrolytes
Solubility refers to the extent to which a substance, known as a solute, can dissolve in a solvent to form a homogeneous solution. For weak electrolytes, which only partially ionize in solution, their solubility is of keen interest because it determines the concentration of ions that are available for chemical reactions.

Weak electrolytes have a unique characteristic in that they establish an equilibrium between the undissociated molecules and the ions they form upon dissolving. This equilibrium is influenced by various factors, including the presence of other ions in the solution. The common-ion effect is an essential aspect of this, as it leads to a shift in the established equilibrium of a weak electrolyte when a salt supplying a common ion is added to the solution. As a result, the solubility product of the weak electrolyte, indicated by Ksp, is seemingly lowered, leading to a decreased solubility.
Le Châtelier's Principle
Le Châtelier's principle is a central theme in chemical equilibrium, providing a qualitative explanation for how systems at equilibrium respond to external changes. According to this principle, if a stress is applied to a system at equilibrium, the system will adjust to minimize the stress and restore a new state of equilibrium.

Examples of stresses include changes in concentration, temperature, or pressure. In the context of the common-ion effect, the 'stress' is the addition of a common ion. The principle predicts that the equilibrium will shift to accommodate this change by favoring the reaction that consumes the added ion, thus reducing its ionization in the case of weak electrolytes. This shift reduces the solubility of the weak electrolyte, a practical concern in various applications ranging from medicinal chemistry to environmental sciences.

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Most popular questions from this chapter

Two buffers are prepared by adding an equal number of moles of formic acid (HCOOH) and sodium formate (HCOONa) to enough water to make \(1.00 \mathrm{~L}\) of solution. Buffer \(\mathrm{A}\) is prepared using \(1.00 \mathrm{~mol}\) each of formic acid and sodium formate. Buffer B is prepared by using \(0.010 \mathrm{~mol}\) of each. (a) Calculate the \(\mathrm{pH}\) of each buffer, and explain why they are equal. (b) Which buffer will have the greater buffer capacity? Explain. (c) Calculate the change in \(\mathrm{pH}\) for each buffer upon the addition of \(1.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). (d) Calculate the change in \(\mathrm{pH}\) for each buffer upon the addition of \(10 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). (e) Discuss your answers for parts (c) and (d) in light of your response to part (b).

(a) Consider the equilibrium \(\mathrm{B}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\mathrm{HB}^{+}(a q)+\mathrm{OH}^{-}(a q) .\) Using Le Châtelier's principle, explain the effect of the presence of a salt of \(\mathrm{HB}^{+}\) on the ionization of B. (b) Give an example of a salt that can decrease the ionization of \(\mathrm{NH}_{3}\) in solution.

A buffer contains a weak acid, HX, and its conjugate base. The weak acid has a \(\mathrm{p} K_{a}\) of \(4.5,\) and the buffer has a \(\mathrm{pH}\) of \(4.3 .\) Without doing a calculation, predict whether \([\mathrm{HX}]=\left[\mathrm{X}^{-}\right]\) \([\mathrm{HX}]>\left[\mathrm{X}^{-}\right],\) or \([\mathrm{HX}]<\left[\mathrm{X}^{-}\right] .\) Explain. \([\) Section 17.2\(]\)

A \(35.0-\mathrm{mL}\) sample of \(0.150 \mathrm{M}\) acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) is titrated with \(0.150 \mathrm{M} \mathrm{NaOH}\) solution. Calculate the \(\mathrm{pH}\) after the following volumes of base have been added: (a) \(0 \mathrm{~mL},(\mathbf{b})\) \(17.5 \mathrm{~mL},(\mathrm{c}) 34.5 \mathrm{~mL},(\) d) \(35.0 \mathrm{~mL},\) (e) \(35.5 \mathrm{~mL},\) (f) \(50.0 \mathrm{~mL} .\)

To what final concentration of \(\mathrm{NH}_{3}\) must a solution be adjusted to just dissolve \(0.020 \mathrm{~mol}\) of \(\mathrm{NiC}_{2} \mathrm{O}_{4}\left(K_{s p}=4 \times 10^{-10}\right)\) in \(1.0 \mathrm{~L}\) of solution? (Hint: You can neglect the hydrolysis of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) because the solution will be quite basic.)
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