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Chapter 16: Problem 49

Write the chemical equation for the base ionization of methylamine, \(\mathrm{CH}_{3} \mathrm{NH}_{2} .\) Write the \(K_{b}\) expression for methylamine.

Short Answer

Expert verified
The equation is \(\mathrm{CH}_{3} \mathrm{NH}_{2} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{CH}_{3}\mathrm{NH}_{3}^{+} + \mathrm{OH}^{-}\), with \(K_{b} = \frac{[\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}][\mathrm{OH}^{-}]}{[\mathrm{CH}_{3} \mathrm{NH}_{2}]} \).

Step by step solution

01

Identify the Ionization Reaction

Methylamine, \(\mathrm{CH}_{3} \mathrm{NH}_{2}\), is a weak base that will ionize in water by accepting a proton from water to form its conjugate acid, \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\), and hydroxide ions (\(\mathrm{OH}^{-}\)). The reaction can be written as: \[\mathrm{CH}_{3} \mathrm{NH}_{2} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{CH}_{3}\mathrm{NH}_{3}^{+} + \mathrm{OH}^{-}.\]
02

Write the Equilibrium Expression

The equilibrium expression for the base ionization constant, \(K_{b}\), is derived from the concentrations of the products divided by the concentration of the reactants, excluding pure water. For the reaction, the expression is: \[K_{b} = \frac{[\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}][\mathrm{OH}^{-}]}{[\mathrm{CH}_{3} \mathrm{NH}_{2}]} .\] This represents the ratio of the product concentrations to the methylamine concentration at equilibrium.

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Key Concepts

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

Weak Base
A weak base is a substance that partially ionizes in solution to accept protons. Unlike strong bases, which fully ionize, weak bases only produce a small amount of hydroxide ions \( \mathrm{(OH^{-})} \) in aqueous solution. This partial ionization is a key feature of weak bases.
Methylamine \( \mathrm{(CH_{3}NH_{2})} \) serves as a classic example of a weak base.
When dissolved in water, methylamine slightly accepts protons to form its conjugate acid while producing hydroxide ions.
It's important to understand that weak bases establish an equilibrium between the base, its conjugate acid, and hydroxide ions, instead of fully ionizing in the solution.
This is why we need to calculate an equilibrium expression for its ionization.
Equilibrium Expression
The equilibrium expression, often represented as \(K_b \) for bases, is a way to describe the extent of ionization of a weak base in equilibrium conditions.
For methylamine, this involves expressing the relationship between the concentrations of the reactants and products of its ionization.
The balanced chemical reaction for the ionization of methylamine in water is given as: \[ \mathrm{CH}_{3} \mathrm{NH}_{2} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CH}_{3} \mathrm{NH}_{3}^{+} + \mathrm{OH}^{-}. \]
In this expression, water (\( \mathrm{H}_{2} \mathrm{O} \)) is not included in the \(K_b \) expression because pure liquids do not affect the equilibrium constant.
Thus, the equation for the base ionization constant is: \[K_b = \frac{[\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}][\mathrm{OH}^{-}]}{[\mathrm{CH}_{3} \mathrm{NH}_{2}]}. \]
This formula helps calculate the equilibrium concentrations and understand the degree to which methylamine functions as a weak base.
Conjugate Acid
In acid-base chemistry, the conjugate acid of a base is what is formed when the base gains a proton.
For methylamine, the conjugate acid is methylammonium (\( \mathrm{CH}_{3} \mathrm{NH}_{3}^{+} \)).
When methylamine accepts a proton from water during its ionization process, it transforms into its conjugate acid.
Understanding conjugate acids helps us track and predict the behavior of weak bases in solution.
Methylamine and its conjugate acid establish a reversible equilibrium. This balance plays a crucial role in determining the pH of the solution.
As the conjugate acid concentration increases, the solution becomes less basic, illustrating how conjugate acids can influence the overall properties of weakly basic solutions.

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Most popular questions from this chapter

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.

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 \(0.108 M\) sample of a weak acid is \(4.16 \%\) ionized in solution. What is the hydroxide concentration of this solution?

Chloroacetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{2} \mathrm{ClO}_{2}\), has a greater acid strength than acetic acid, because the electronegative chlorine atom pulls electrons away from the \(\mathrm{O}-\mathrm{H}\) bond and thus weakens it. Calculate the hydronium-ion concentration and the \(\mathrm{pH}\) of a \(0.0020 M\) solution of chloroacetic acid. \(K_{a}\) is \(1.3 \times 10^{-3}\).

Calculate the pH of a \(0.15 M\) aqueous solution of aluminum chloride, \(\mathrm{AlCl}_{3}\). The acid ionization of hydrated aluminum ion is \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)=\) \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}^{2+}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)\) and \(K_{a}\) is \(1.4 \times 10^{-5}\).
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