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Chapter 16: Problem 12

The \(\mathrm{pH}\) of \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{2}\) (methylamine) is \(11.8\). When the chloride salt of methylamine, \(\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\), is added to this solution, does the pH increase or decrease? Explain, using Le Ch芒telier's principle and the common-ion effect.

Short Answer

Expert verified
The pH decreases due to the common-ion effect and Le Ch芒telier's principle.

Step by step solution

01

Identify the chemical equilibrium

The chemical equilibrium in this situation involves the weak base methylamine, \(\mathrm{CH}_3\mathrm{NH}_2\), which partially ionizes in water to form \(\mathrm{CH}_3\mathrm{NH}_3^+\) and \(\mathrm{OH}^-\): \[ \mathrm{CH}_3\mathrm{NH}_2 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{CH}_3\mathrm{NH}_3^+ + \mathrm{OH}^- \]
02

Determine the effect of adding a salt

Adding \(\mathrm{CH}_3\mathrm{NH}_3\mathrm{Cl}\) introduces more \(\mathrm{CH}_3\mathrm{NH}_3^+\) ions into the solution, as the chloride ion \(\mathrm{Cl}^-\) is the spectator ion and does not participate actively in the equilibrium.
03

Apply Le Ch芒telier's principle

According to Le Ch芒telier鈥檚 principle, when an external change (such as the addition of a reactant or product) is applied to a system at equilibrium, the system will adjust to offset that change. Thus, adding \(\mathrm{CH}_3\mathrm{NH}_3^+\) will shift the equilibrium to the left, reducing the concentration of \(\mathrm{OH}^-\) ions in the solution.
04

Consider the common-ion effect

The common-ion effect refers to the decrease in solubility of a compound in a solution that already contains one of the ions in the compound. Here, \(\mathrm{NH}_3^+\) is the common ion, and its increase shifts the equilibrium left, reducing \(\mathrm{OH}^-\) concentration. A decrease in hydroxide ion concentration means the solution becomes less basic.
05

Conclude the effect on pH

Since the concentration of \(\mathrm{OH}^-\) ions decreases, the pH will decrease, moving closer to 7 from its initial 11.8, indicating a less basic (lower) pH.

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Key Concepts

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

Chemical Equilibrium
In the context of chemical reactions, equilibrium is a state where the rate of the forward reaction equals the rate of the reverse reaction. This means the concentrations of reactants and products remain constant over time. In the system involving methylamine (\(\mathrm{CH}_3\mathrm{NH}_2\)), equilibrium is established when it partially ionizes in water:

\[\mathrm{CH}_3\mathrm{NH}_2 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{CH}_3\mathrm{NH}_3^+ + \mathrm{OH}^-\]This chemical reaction highlights that the methylamine acts as a weak base, accepting a proton (\(\mathrm{H}^+\)) from water to form its conjugate acid (\(\mathrm{CH}_3\mathrm{NH}_3^+\)) and hydroxide ions (\(\mathrm{OH}^-\)). Reaching a state of equilibrium means any disturbance will provoke the system to adjust itself to restore balance.
Common-Ion Effect
The common-ion effect plays a significant role when a salt of a weak acid or base is added to its solution. This effect involves suppressing the ionization of a weak electrolyte by the presence of a common ion from a strong electrolyte. In our methylamine example, adding \(\mathrm{CH}_3\mathrm{NH}_3\mathrm{Cl}\) introduces more \(\mathrm{CH}_3\mathrm{NH}_3^+\) ions from the salt.

The chloride ion (\(\mathrm{Cl}^-\)) acts as a spectator ion and does not participate in the equilibrium reaction. However, the added \(\mathrm{CH}_3\mathrm{NH}_3^+\) ions increase this ion's concentration in the solution. Because \(\mathrm{CH}_3\mathrm{NH}_3^+\) is already a product of the equilibrium reaction, its increased concentration pushes the equilibrium to favor the reactants side 鈥� reducing \(\mathrm{OH}^-\) concentration due to less ionization of \(\mathrm{CH}_3\mathrm{NH}_2\).
pH Change
pH is a scale used to specify the acidity or basicity of an aqueous solution. It is determined by the concentration of \(\mathrm{H}^+\) ions in a solution, with lower pH indicating higher acidity, and higher pH indicating higher basicity.

In systems involving weak bases like methylamine, the addition of \(\mathrm{CH}_3\mathrm{NH}_3\mathrm{Cl}\) affects the pH due to changes in the equilibrium. As the equilibrium shifts to favor the reactants due to the common-ion effect, \(\mathrm{OH}^-\) concentration reduces:- Lower \(\mathrm{OH}^-\) concentration means fewer hydroxide ions are available to contribute to the basicity.- Consequently, the pH decreases from its original state.Therefore, when methylamine鈥檚 chloride salt is added, despite the solution starting with a pH of 11.8, this decrease causes the pH to move closer to a neutral pH of 7.
Weak Bases
Weak bases only partially ionize in solution, meaning they do not completely dissociate into ions. This characteristic differentiates them from strong bases, which fully dissociate. In the case of methylamine (\(\mathrm{CH}_3\mathrm{NH}_2\)), a weak base, it partially accepts protons from water:

- Because of this partial ionization, the concentration of hydroxide ions (\(\mathrm{OH}^-\)) is much less compared to strong bases.- It is crucial because the chemical equilibrium is sensitive to changes, including the introduction of additional ions from an external source like \(\mathrm{CH}_3\mathrm{NH}_3\mathrm{Cl}\).The weak nature of methylamine means adding more \(\mathrm{CH}_3\mathrm{NH}_3^+\) ions shifts the equilibrium back to the formation of the base (reactants), leading to the aforementioned decrease in \(\mathrm{OH}^-\) concentration and thus a lowered pH.

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Most popular questions from this chapter

Write the equation for the acid ionization of the \(\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) ion.

a. When \(0.10\) mol of the ionic solid NaX, where \(X\) is an unknown anion, is dissolved in enough water to make \(1.0 \mathrm{~L}\) of solution, the \(\mathrm{pH}\) of the solution is \(9.12 .\) When \(0.10 \mathrm{~mol}\) of the ionic solid \(\mathrm{ACl}\), where \(\mathrm{A}\) is an unknown cation, is dissolved in enough water to make \(1.0 \mathrm{~L}\) of solution, the \(\mathrm{pH}\) of the solution is \(7.00 .\) What would be the \(\mathrm{pH}\) of \(1.0 \mathrm{~L}\) of solution that contained \(0.10\) mol of AX? Be sure to document how you arrived at your answer. b. In the AX solution prepared above, is there any \(\mathrm{OH}^{-}\) present? If so, compare the \(\left[\mathrm{OH}^{-}\right]\) in the solution to the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) c. From the information presented in part a, calculate \(K_{b}\) for the \(\mathrm{X}^{-}(a q)\) anion and \(K_{a}\) for the conjugate acid of \(\mathrm{X}^{-}(a q)\). d. To \(1.0 \mathrm{~L}\) of solution that contains \(0.10 \mathrm{~mol}\) of \(\mathrm{AX}\), you add \(0.025\) mol of \(\mathrm{HCl}\). How will the \(\mathrm{pH}\) of this solution compare to that of the solution that contained only NaX? Use chemical reactions as part of your explanation; you do not need to solve for a numerical answer. e. Another \(1.0 \mathrm{~L}\) sample of solution is prepared by mixing \(0.10\) mol of \(\mathrm{AX}\) and \(0.10 \mathrm{~mol}\) of \(\mathrm{HCl}\). The \(\mathrm{pH}\) of the resulting solution is found to be \(3.12\). Explain why the \(\mathrm{pH}\) of this solution is \(3.12\). f. Finally, consider a different \(1.0\) -L sample of solution that contains \(0.10\) mol of \(A X\) and \(0.1\) mol of \(\mathrm{NaOH}\). The \(\mathrm{pH}\) of this solution is found to be \(13.00\). Explain why the \(\mathrm{pH}\) of this solution is \(13.00\). g. Some students mistakenly think that a solution that contains \(0.10 \mathrm{~mol}\) of \(\mathrm{AX}\) and \(0.10 \mathrm{~mol}\) of \(\mathrm{HCl}\) should have a \(\mathrm{pH}\) of \(1.00\). Can you come up with a reason why students have this misconception? Write an approach that you would use to help these students understand what they are doing wrong.

A solution made up of \(1.0 \mathrm{M} \mathrm{NH}_{3}\) and \(0.50 \mathrm{M}\) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) has a pH of \(9.26\). a. Write the net ionic equation that represents the reaction of this solution with a strong acid. b. Write the net ionic equation that represents the reaction of this solution with a strong base. c. To \(100 . \mathrm{mL}\) of this solution, \(10.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\) is added. How many moles of \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4}{ }^{+}\) are present in the reaction system before and after the addition of the HCl? What is the \(\mathrm{pH}\) of the resulting solution? d. Why did the \(\mathrm{pH}\) change only slightly upon the addition of \(\mathrm{HCl} ?\)

Calculate the \(\mathrm{pH}\) of a solution obtained by mixing 456 \(\mathrm{mL}\) of \(0.10 M\) hydrochloric acid with \(285 \mathrm{~mL}\) of \(0.15 M\) sodium hydroxide. Assume the combined volume is the sum of the two original volumes.

Calculate the pH of a solution obtained by mixing \(500.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{NH}_{3}\) with \(200.0 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{HCl}\).
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