Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 1

According to the Bronsted-Lowry theory, label each of the following as an acid or a base. (a) \(\mathrm{HNO}_{2}\) (b) \(\mathrm{OCl}^{-} ;(\mathrm{c}) \mathrm{NH}_{2}^{-} ;\) (d) \(\mathrm{NH}_{4}^{+} ;\) (e) \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\)

Short Answer

Expert verified
(a) \(\mathrm{HNO}_{2}\) is an acid. (b) \(\mathrm{OCl}^{-}\) is a base. (c) \(\mathrm{NH}_{2}^{-}\) is a base. (d) \(\mathrm{NH}_{4}^{+}\) is an acid. (e) \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\) is an acid.

Step by step solution

01

Identify Proton Donors and Acceptors

According to the Bronsted-Lowry theory, we're going to identify if each substance in the exercise can donate a proton (making it an acid) or if it can accept a proton (making it a base).
02

Classify \(\mathrm{HNO}_{2}\)

\(\mathrm{HNO}_{2}\) has a hydrogen atom that it can donate. Thus, \(\mathrm{HNO}_{2}\) is a Bronsted-Lowry acid.
03

Classify \(\mathrm{OCl}^{-}\)

\(\mathrm{OCl}^{-}\) is a negative ion and can accept an extra hydrogen ion. Thus, \(\mathrm{OCl}^{-}\) is a Bronsted-Lowry base.
04

Classify \(\mathrm{NH}_{2}^{-}\)

\(\mathrm{NH}_{2}^{-}\) is a negative ion and can accept a extra hydrogen ion. Thus, \(\mathrm{NH}_{2}^{-}\) is a Bronsted-Lowry base.
05

Classify \(\mathrm{NH}_{4}^{+}\)

\(\mathrm{NH}_{4}^{+}\) can donate one of its hydrogen ions. Thus, \(\mathrm{NH}_{4}^{+}\) is a Bronsted-Lowry acid.
06

Classify \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\)

\(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\) can donate one of its hydrogen ions indicated by its positive charge. Thus, \(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\) is a Bronsted-Lowry acid.

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

Oxalic acid, HOOCCOOH, a weak diprotic acid, has \(\mathrm{p} K_{\mathrm{a}_{1}}=1.25\) and \(\mathrm{p} K_{\mathrm{a}_{2}}=3.81 .\) A related diprotic acid, suberic acid, \(\mathrm{HOOC}\left(\mathrm{CH}_{2}\right)_{8} \mathrm{COOH}\) has \(\mathrm{p} K_{\mathrm{a}_{1}}=4.21\) and \(\mathrm{p} K_{\mathrm{a}_{2}}=5.40 .\) Offer a plausible reason as to why the difference between \(\mathrm{pK}_{\mathrm{a}_{1}}\) and \(\mathrm{pK}_{\mathrm{a}_{2}}\) is so much greater for oxalic acid than for suberic acid.

Of the following, the amphiprotic ion is (a) \(\mathrm{HCO}_{3}^{-}\) (b) \(\mathrm{CO}_{3}^{2-} ;\) (c) \(\mathrm{NH}_{4}^{+} ;\) (d) \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+} ;\) (e) \(\mathrm{ClO}_{4}^{-}\).The \(\mathrm{pH}\) in \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}(\mathrm{aq})\) must be (a) equal to \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) in \(0.10 \mathrm{M} \mathrm{HNO}_{2}(\mathrm{aq}) ;\) (b) less than the \(\mathrm{pH}\) in \(0.10 \mathrm{M} \mathrm{HI}(\mathrm{aq}) ;\) (c) greater than the \(\mathrm{pH}\) in \(0.10 \mathrm{M} \mathrm{HBr}(\mathrm{aq}) ;\) (d) equal to \(1.0.\)

Each of the following is a Lewis acid-base reaction. Which reactant is the acid, and which is the base? Explain. (a) \(\mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}\) (b) \(\operatorname{Zn}(\mathrm{OH})_{2}(\mathrm{s})+2 \mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}(\mathrm{aq})\)

What is the (a) degree of ionization and (b) percent ionization of propionic acid in a solution that is \(0.45 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} ?\) $$\begin{aligned} &\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2}^{-}\\\ &&\mathrm{p} K_{\mathrm{a}}=4.89 \end{aligned}$$

\(\operatorname{In} 0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}),\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) is equal to \((\mathrm{a}) 0.050 \mathrm{M}\) (b) \(0.10 \mathrm{M} ;\) (c) \(0.11 \mathrm{M} ;\) (d) \(0.20 \mathrm{M}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

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