/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 136 Zinc hydroxide is an amphoteric ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 136

Zinc hydroxide is an amphoteric substance. Write equations that describe \(\mathrm{Zn}(\mathrm{OH})_{2}\) acting as a Br酶nsted-Lowry base toward \(\mathrm{H}^{+}\) and as a Lewis acid toward \(\mathrm{OH}^{-}\).

Short Answer

Expert verified
Zinc hydroxide, \(\mathrm{Zn}(\mathrm{OH})_{2}\), can act as a Br酶nsted-Lowry base and a Lewis acid. In the Br酶nsted-Lowry base reaction, it accepts a proton (\(\mathrm{H}^{+}\)): \[\mathrm{Zn}(\mathrm{OH})_{2} + \mathrm{H}^{+} \rightarrow \mathrm{Zn}(\mathrm{OH})_{3}^{+}\] In the Lewis acid reaction, it accepts an electron pair from a hydroxide ion (\(\mathrm{OH}^{-}\)): \[\mathrm{Zn}(\mathrm{OH})_{2} + \mathrm{OH}^{-} \rightarrow \mathrm{Zn}(\mathrm{OH})_{3}^{-}\]

Step by step solution

01

Acting as a Br酶nsted-Lowry base

Zinc hydroxide, \(\mathrm{Zn}(\mathrm{OH})_{2}\), can accept a proton (\(\mathrm{H}^{+}\)) to form a new compound. In this case, the reaction would be: \[\mathrm{Zn}(\mathrm{OH})_{2} + \mathrm{H}^{+} \rightarrow \mathrm{Zn}(\mathrm{OH})_{3}^{+}\]
02

Acting as a Lewis acid

Zinc hydroxide, \(\mathrm{Zn}(\mathrm{OH})_{2}\), can accept an electron pair from a hydroxide ion (\(\mathrm{OH}^{-}\)) to form a new compound. In this case, the reaction would be: \[\mathrm{Zn}(\mathrm{OH})_{2} + \mathrm{OH}^{-} \rightarrow \mathrm{Zn}(\mathrm{OH})_{3}^{-}\] These equations show how zinc hydroxide can act as a Br酶nsted-Lowry base by accepting a proton when it reacts with a hydrogen ion, and as a Lewis acid by accepting an electron pair when it reacts with a hydroxide ion.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

What are the major species present in a \(0.150 \mathrm{M} \mathrm{NH}_{3}\) solution? Calculate the \(\left[\mathrm{OH}^{-}\right]\) and the \(\mathrm{pH}\) of this solution.

Monochloroacetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{2} \mathrm{ClO}_{2}\), is a skin irritant that is used in "chemical peels" intended to remove the top layer of dead skin from the face and ultimately improve the complexion. The value of \(K_{\mathrm{a}}\) for monochloroacetic acid is \(1.35 \times 10^{-3}\). Calculate the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution of monochloroacetic acid.

Use the Lewis acid-base model to explain the following reaction. $$ \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3}(a q) $$

Formic acid \(\left(\mathrm{HCO}_{2} \mathrm{H}\right)\) is secreted by ants. Calculate \(\left[\mathrm{H}^{+}\right]\) and the \(\mathrm{pH}\) of a \(0.025 M\) solution of formic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-4}\right)\).

Calculate the \(\mathrm{pH}\) of a \(1.0 \times 10^{-7} M\) solution of \(\mathrm{NaOH}\) in water.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

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.