/*! 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 35 For the following reactions, lab... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 35

For the following reactions, label each species as an acid or a base. Indicate the species that are conjugates of one another. a. \(\mathrm{HSO}_{4}^{-}+\mathrm{NH}_{3} \rightleftharpoons \mathrm{SO}_{4}^{2-}+\mathrm{NH}_{4}^{+}\) b. \(\mathrm{HPO}_{4}^{2-}+\mathrm{NH}_{4}^{+} \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{4}^{-}+\mathrm{NH}_{3}\) c. \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}^{2+}+\mathrm{H}_{3} \mathrm{O}^{+}\) d. \(\mathrm{SO}_{3}^{2-}+\mathrm{NH}_{4}^{+} \rightleftharpoons \mathrm{HSO}_{3}^{-}+\mathrm{NH}_{3}\)

Short Answer

Expert verified
In (a), HSO鈧勨伝 is an acid, SO鈧劼测伝 is conjugate base; NH鈧� is a base, NH鈧勨伜 is conjugate acid. In (b), HPO鈧劼测伝 is a base, H鈧侾O鈧勨伝 is conjugate acid; NH鈧勨伜 is an acid, NH鈧� is conjugate base. In (c), Al(H鈧侽)鈧喡斥伜 is an acid, Al(H鈧侽)鈧匫H虏鈦� is conjugate base; H鈧侽 is a base, H鈧僌鈦� is conjugate acid. In (d), SO鈧兟测伝 is a base, HSO鈧冣伝 is conjugate acid; NH鈧勨伜 is an acid, NH鈧� is conjugate base.

Step by step solution

01

Analyze Reaction (a)

For the reaction \[ \text{HSO}_4^- + \text{NH}_3 \rightleftharpoons \text{SO}_4^{2-} + \text{NH}_4^+ \]- \(\text{HSO}_4^-\) donates a proton to become \(\text{SO}_4^{2-}\), so \(\text{HSO}_4^-\) is an acid and \(\text{SO}_4^{2-}\) is its conjugate base.- \(\text{NH}_3\) accepts a proton to become \(\text{NH}_4^+\), so \(\text{NH}_3\) is a base and \(\text{NH}_4^+\) is its conjugate acid.
02

Analyze Reaction (b)

For the reaction \[ \text{HPO}_4^{2-} + \text{NH}_4^+ \rightleftharpoons \text{H}_2\text{PO}_4^- + \text{NH}_3 \]- \(\text{HPO}_4^{2-}\) accepts a proton to become \(\text{H}_2\text{PO}_4^-\), so \(\text{HPO}_4^{2-}\) is a base and \(\text{H}_2\text{PO}_4^-\) is its conjugate acid.- \(\text{NH}_4^+\) donates a proton to become \(\text{NH}_3\), so \(\text{NH}_4^+\) is an acid and \(\text{NH}_3\) is its conjugate base.
03

Analyze Reaction (c)

For the reaction \[ \text{Al(H}_2\text{O})_6^{3+} + \text{H}_2\text{O} \rightleftharpoons \text{Al(H}_2\text{O})_5\text{OH}^{2+} + \text{H}_3\text{O}^+ \]- \(\text{Al(H}_2\text{O})_6^{3+}\) donates a proton to become \(\text{Al(H}_2\text{O})_5\text{OH}^{2+}\), so \(\text{Al(H}_2\text{O})_6^{3+}\) is an acid and \(\text{Al(H}_2\text{O})_5\text{OH}^{2+}\) is its conjugate base.- \(\text{H}_2\text{O}\) donates a proton to become \(\text{H}_3\text{O}^+\), so \(\text{H}_2\text{O}\) is a base and \(\text{H}_3\text{O}^+\) is its conjugate acid.
04

Analyze Reaction (d)

For the reaction \[ \text{SO}_3^{2-} + \text{NH}_4^+ \rightleftharpoons \text{HSO}_3^- + \text{NH}_3 \]- \(\text{SO}_3^{2-}\) accepts a proton to become \(\text{HSO}_3^-\), so \(\text{SO}_3^{2-}\) is a base and \(\text{HSO}_3^-\) is its conjugate acid.- \(\text{NH}_4^+\) donates a proton to become \(\text{NH}_3\), so \(\text{NH}_4^+\) is an acid and \(\text{NH}_3\) is its conjugate base.

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!

Key Concepts

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

Understanding Conjugate Acid-Base Pairs
In an acid-base reaction, we often come across the term "conjugate acid-base pairs." What does it mean? A conjugate acid-base pair consists of two species that transform into each other by the gain or loss of a proton (\( ext{H}^+\)). For every acid, there is a conjugate base, and similarly, for every base, a conjugate acid can be identified.
  • Acid to Conjugate Base: An acid donates a proton to become its conjugate base. For example, in the reaction \( ext{HSO}_4^-\) donates a proton to form \( ext{SO}_4^{2-}\). Thus, \( ext{HSO}_4^-\) is the acid and \( ext{SO}_4^{2-}\) is its conjugate base.
  • Base to Conjugate Acid: A base accepts a proton to become its conjugate acid. Consider \( ext{NH}_3\) which accepts a proton to become \( ext{NH}_4^+\), making \( ext{NH}_3\) the base and \( ext{NH}_4^+\) its conjugate acid.
Recognizing these pairs helps in understanding how acids and bases react with each other, showing the dynamic nature of these acids and bases in reactions.
Proton Transfer in Acid-Base Reactions
Proton transfer is the hallmark of acid-base reactions. It involves the movement of protons (\( ext{H}^+\)) from one substance to another.
  • Proton Donors: Acids act as proton donors, giving up an \( ext{H}^+\) ion. For instance, \( ext{Al(H}_2 ext{O})_6^{3+}\) donates a proton to form \( ext{Al(H}_2 ext{O})_5 ext{OH}^{2+}\).
  • Proton Acceptors: Bases are proton acceptors, gaining an \( ext{H}^+\) ion. This can be seen where \( ext{SO}_3^{2-}\) accepts a proton and becomes \( ext{HSO}_3^-\).
Identifying the direction of proton transfer helps clarify which species are acting as acids or bases, and demonstrates the reversible nature of these reactions. By examining proton movement, one can effectively predict reaction outcomes and balance equations.
Basics of Acid-Base Theory
Acid-base theory provides the foundational understanding of how substances interact to form acids and bases. Several models attempt to explain acid-base behavior, but many reactions you'll encounter can be understood through the Br酶nsted-Lowry theory.
  • Br酶nsted-Lowry Acids: Defined as substances that donate protons. For example, in the reaction, \( ext{NH}_4^+\) acts as an acid by donating a proton to become \( ext{NH}_3\).
  • Br酶nsted-Lowry Bases: Substances that accept protons. An example is \( ext{HPO}_4^{2-}\) accepting a proton to become \( ext{H}_2 ext{PO}_4^-\).
This theory allows us to predict and identify acids and bases in a variety of chemical reactions, explaining their reactivity and how they interact in different environments. Acid-base theory thus acts as a guiding principle for understanding chemical equilibria and reaction dynamics, playing a crucial role in the study of chemistry.

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

The following are solution concentrations. Indicate whether each solution is acidic, basic, or neutral. a. \(2 \times 10^{-11} \mathrm{M} \mathrm{OH}^{-}\) b. \(2 \times 10^{-9} \mathrm{M} \mathrm{H}_{3} \mathrm{O}^{+}\) c. \(6 \times 10^{-5} \mathrm{M} \mathrm{OH}^{-}\) d. \(6 \times 10^{-3} \mathrm{M} \mathrm{H}_{3} \mathrm{O}^{+}\)

Natural gas frequently contains hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\). \(\mathrm{H}_{2} \mathrm{~S}\) is removed from natural gas by passing it through aqueous ethanolamine, \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) (an ammonia derivative), which reacts with the hydrogen sulfide. Write the equation for the reaction. Identify each reactant as either a Lewis acid or a Lewis base. Explain how you arrived at your answer.

List the following compounds in order of increasing acid strength: \(\mathrm{HBr}, \mathrm{H}_{2} \mathrm{Se}, \mathrm{H}_{2} \mathrm{~S}\).

Compare the structures of \(\mathrm{HNO}_{2}\) and \(\mathrm{H}_{2} \mathrm{CO}_{3}\). Which would you expect to be the stronger acid? Explain your choice.

Write a reaction for each of the following in which the species acts as a Br酶nsted base. The equilibrium should favor the product side. a. \(\mathrm{H}_{2} \mathrm{O}\) b. \(\mathrm{HCO}_{3}^{-}\) c. \(\mathrm{NH}_{3}\) d. \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Physical 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.