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Chapter 1: Problem 5

A student titrates 20.0 \(\mathrm{mL}\) of 1.0 \(M \mathrm{NaOH}\) with 2.0 \(\mathrm{M}\), \(\mathrm{HCO}_{2} \mathrm{H}\left(K_{\mathrm{a}}=1.8 \times 10^{-4}\right) .\) Formic acid is a monoprotic acid. \(\mathrm{CH}_{3} \mathrm{NH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \leftrightarrow \mathrm{OH}^{-}(a q)+\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}(a q)\) The above equation represents the reaction between the base methylamine \(\left(K_{\mathrm{b}}=4.38 \times 10^{-4}\right)\) and water. Which of the following best represents the.concentrations of the various species at equilibrium? (A) \(\left[\mathrm{OH}^{-}\right]>\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]\) (B) \(\left[\mathrm{OH}^{-}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]\) (C) \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]>\left[\mathrm{OH}^{-}\right]>\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]\) (D) \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]>\left[\mathrm{OH}^{-}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{3}+\right]\)

Short Answer

Expert verified
The correct representation of the species at equilibrium is (D) \(\[CH3NH2\]\) > \(\[OH^-\]\) = \(\[CH3NH3^+]\)

Step by step solution

01

Understanding the reaction

Methylamine, \(\(CH_3NH_2\)\), is a weak base that undergoes ionization in water to form the methylammonium ion, \(\(CH_3NH_3^+\)\), and hydroxide ion, \(\(OH^-\)\). The initial concentration of methylamine is larger, but some of it is used to form \(\(CH_3NH_3^+\)\) and \(\(OH^-\)\) in the reaction.
02

Analyzing the equilibrium process

At equilibrium, the concentration of \(\(OH^-\)\) and \(\(CH_3NH_3^+\)\) are equal, because they are both produced from the ionization of the same amount of methylamine. However, the concentration of methylamine, \(\(CH_3NH_2\)\), will still be greater than \(\(OH^-\)\) and \(\(CH_3NH_3^+\)\) because it is only partially ionized.
03

Selecting the correct option

Therefore, the option expressing that \(\[CH3NH2]\) is greater than \(\[OH^-]\), and \(\[OH^-]\) equals \(\[CH3NH3^+]\) (Option (D)) is the one correctly representing the concentration of the species at equilibrium.

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Key Concepts

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

Acid-Base Titration
Acid-base titration is a laboratory technique widely used in chemistry to determine the concentration of an unknown acid or base solution. It involves adding a titrant, with a known concentration, drop by drop to a solution of analyte until the reaction reaches the equivalence point. This point is where the number of moles of titrant equals the number of moles of analyte.
  • Indicators: Indicators are chemicals that change color at a certain pH level and are used to signal the equivalence point. For formic acid titration, we might use phenolphthalein, which changes color at around pH 8.2, indicating the endpoint of the titration.
  • Calculations: At the equivalence point for a monoprotic acid and base, like in the titration of formic acid and NaOH, the equation \(M_1V_1 = M_2V_2\) helps find the molarity of the unknown solution.
The formic acid titration given in the original exercise can help a student understand how the volume of titrant (formic acid), required to neutralize the base (NaOH), can calculate its concentration. The process involves knowing the initial concentrations and volumes, using the balanced chemical equation and calculating the changes as the titration progresses.
Equilibrium Concentrations
In a chemical reaction that reaches equilibrium, the concentrations of reactants and products remain constant over time, despite ongoing reactions. When considering equilibrium, the important idea is that the rates of the forward and reverse reactions are equal. Thus, concentrations reach a steady state.
  • Equilibrium Constant (\(K\)): The equilibrium constant, either \(K_a\) for acids or \(K_b\) for bases, indicates the extent of a reaction at equilibrium. For weak acids and bases, the value is much less than 1, meaning the reaction doesn't proceed much towards the products.
  • Analysing Equilibrium: In the solution's context, we need to analyze the equilibrium concentrations of species, which involves comparing the initial and equilibrium states. The use of ICE (Initial, Change, Equilibrium) tables can help track these changes.

    • For example, in the exercise's context, at equilibrium the concentration of \(OH^-\) and \(CH_3NH_3^+\) are equal because they are produced in equal amounts from the weak base methylamine. This understanding allows us to determine the equilibrium concentrations and explain why Option (D) is correct, where methylamine concentration remains highest.

Weak Base Ionization
Weak bases do not ionize completely in water. This incomplete ionization results in an equilibrium being established between the weak base, its ions, and the hydroxide ion produced. Understanding how weak bases behave helps predict the concentrations of different species in a solution.
  • Dissociation: When a weak base like methylamine is dissolved in water, it accepts protons from water to generate hydroxide ions \(OH^-\) and the conjugate acid, \(CH_3NH_3^+\).
  • Base Ionization Constant (\(K_b\)): This constant provides insight into the strength of the base. For methylamine, \(K_b\) is tiny (\(4.38 \times 10^{-4}\)), indicating most of the base remains unionized in solution, resulting in a relatively higher concentration of \(CH_3NH_2\).
Given that only a fraction of the base is ionized, the concentration of the initial base will still be higher compared to its ionic products. This means, at equilibrium, \([CH_3NH_2] > [OH^-] = [CH_3NH_3^+]\), justifying why Option (D) is selected for the problem. Understanding weak base ionization can help students predict behavior in complex reactions.

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Most popular questions from this chapter

$$\begin{array}{|c|c|}\hline \text { Time (Hours) } & {[\mathrm{A}] M} \\\ \hline 0 & {0.40} \\ \hline 1 & {0.20} \\ \hline 2 & {0.10} \\ \hline 3 & {0.05} \\ \hline\end{array}$$ Reactant A underwent a decomposition reaction. The concentration of A was measured periodically and recorded in the chart above. Based on the data in the chart, which of the following is the rate law for the reaction? (A) Rate \(=k[\mathrm{A}]\) (B) Rate \(=k[\mathrm{A}]^{2}\) (C) Rate \(=2 k[\mathrm{A}]\) (D) Rate \(=\frac{1}{2} k[\mathrm{A}]\)

In an experiment 2 moles of \(\mathrm{H}_{2}(g)\) and 1 mole of \(\mathrm{O}_{2}(g)\) were completely reacted, according to the following equation in a sealed container of constant volume and temperature: $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)$$ If the initial pressure in the container before the reaction is denoted as \(P_{i}\) which of the following expressions gives the final pressure, assuming ideal gas behavior? (A) \(P_{i}\) (B) 2\(P_{i}\) (C) \((3 / 2) P_{i}\) (D) \((2 / 3) P_{i}\)

$$4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ The above reaction will experience a rate increase by the addition of a catalyst such as platinum. Which of the following best explains why? (A) The catalyst causes the value for \(\Delta G\) to become more negative. (B) The catalyst increases the percentage of collisions that occur at the proper orientation in the reactant molecules. (C) The catalyst introduces a new reaction mechanism for the reaction. (D) The catalyst increases the activation energy for the reaction.

Use the following information to answer questions 1-5. \(\begin{array}{ll}{\text { Reaction } 1 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=?} \\ {\text { Reaction } 2 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)} & {\Delta H=-37 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 3 : \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=-46 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 4 : \mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{H}_{2}(g)} & {\Delta H=-65 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}}\end{array}\) If reaction 2 were repeated at a higher temperature, how would the reaction's value for \(\Delta G\) be affected? (A) It would become more negative because entropy is a driving force behind this reaction. (B) It would become more positive because the reactant molecules would collide more often. (C) It would become more negative because the gases will be at a higher (D) It will stay the same; temperature does not affect the value for \(\Delta G\) .

A sample of a hydrate of \(\mathrm{CuSO}_{4}\) with a mass of 250 grams was heated until all the water was removed. The sample was then weighed and found to have a mass of 160 grams. What is the formula for the hydrate? (A) \(\mathrm{CuSO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}\) (B) \(\mathrm{CuSO}_{4} \cdot \mathrm{TH}_{2} \mathrm{O}\) (C) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) (D) \(\mathrm{CuSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\)
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