Log out Go to App
Go to App

Learning Materials

Features

Discover

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which of the following pairs of substances would make a good buffer solution? (A) \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)\) and \(\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)\) (B) \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)\) and \(\mathrm{LiOH}(a q)\) (C) \(\mathrm{HCl}(a q)\) and \(\mathrm{KCl}(a q)\) (D) \(\mathrm{HF}(a q)\) and \(\mathrm{NH}_{3}(a q)\)

\(\begin{array}{ll}{\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-390 \mathrm{kJ} / \mathrm{mol}} \\\ {\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-290 \mathrm{kJ} / \mathrm{mol}} \\ {2 \mathrm{C}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)} & {\Delta H^{\circ}=+230 \mathrm{kJ} / \mathrm{mol}}\end{array}\) Based on the information given above, what is \(\Delta H^{\circ}\) for the following reaction? $$ \begin{aligned} \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \\ \text { (A) }-1,300 \mathrm{kJ} \\ \text { (B) }-1,070 \mathrm{kJ} \\ \text { (C) }-840 \mathrm{kJ} \\ \text { (D) }-780 \mathrm{kJ} \end{aligned} $$

$$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ The reaction above came to equilibrium at a temperature of \(100^{\circ} \mathrm{C} .\) At equilibrium the partial pressure due to NOBr was 4 atmospheres, the partial pressure due to NO was 4 atmospheres, and the partial pressure due to \(\mathrm{Br}_{2}\) was 2 atmospheres. What is the equilibrium constant, \(K_{p},\) for this reaction at \(100^{\circ} \mathrm{C}\) ? (A) \(\frac{1}{4}\) (B) \(\frac{1}{2}\) (C) 1 (D) 2

The following reaction is found to be at equilibrium at 25°C: \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g)+2 \mathrm{SO}_{2}(g) \quad \Delta H=-198 \mathrm{kJ} / \mathrm{mol}\) Which of the following would cause the reverse reaction to speed up? (A) Adding more \(\mathrm{SO}_{3}\) (B) Raising the pressure (C) Lowering the temperature (D) Removing some \(\mathrm{SO}_{2}\)

Which expression below should be used to calculate the mass of copper that can be plated out of a 1.0 \(\mathrm{M} \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) , solution using a current of 0.75 A for 5.0 minutes? (A) \(\frac{(5.0)(60)(0.75)(63.55)}{(96500)(2)}\) (B) \(\frac{(5.0)(60)(63.55)(2)}{(0.75)(96500)}\) (C) \(\frac{(5.0)(60)(96500)(0.75)}{(63.55)(2)}\) (D) \(\frac{(5.0)(60)(96500)(63.55)}{(0.75)(2)}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free