/*! 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 260 One of the following species act... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 260

One of the following species acts as both Bronsted acid and base (a) \(\mathrm{H}_{2} \mathrm{PO}_{2}\) (b) \(\mathrm{HPO}_{3}^{2}\) (c) \(\mathrm{HPO}_{4}^{2}\) (d) all of the above

Short Answer

Expert verified
(c) HPO_{4}^{2-} can act as both an acid and a base.

Step by step solution

01

Understand Bronsted Acid and Base

A Bronsted acid is a species that can donate a proton (H^+) to another species, whereas a Bronsted base is a species that can accept a proton. Therefore, a molecule acting as both must be able to donate and accept a proton.
02

Evaluate H_{2}PO_{2}

The species H_{2}PO_{2} can donate a proton, forming HPO_{2}^{-}, but cannot accept a proton as it has no negative charge or easily accessible lone pairs for accepting protons. Thus, it can only act as an acid.
03

Evaluate HPO_{3}^{2-}

The species HPO_{3}^{2-} is a dianion and likely substantial negative charge could accept a proton, acting as a base. However, donating a proton to form H_{2}PO_{3}^{-} is unlikely because it already has a substantial negative charge. It is primarily a base.
04

Evaluate HPO_{4}^{2-}

This species can both donate and accept a proton. It can donate a proton to form PO_{4}^{3-}, acting as an acid. It can also accept a proton to form H_{2}PO_{4}^{-}, acting as a base. Hence, it can act as both a Bronsted acid and base.
05

Compare All Options

Among the given options, only HPO_{4}^{2-} can act as both a Bronsted acid and base. The others can only predominantly act as either an acid or a base, not both.

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.

Proton Donation
In the context of Bronsted-Lowry theory, proton donation is a core characteristic of acids. When an acid donates a proton, it typically becomes a new species with one less positive charge. For example,
  • The species \(H_{2}PO_{2}^{-}\) can donate a proton, creating \(HPO_{2}^{2-}\).
  • Similarly, \(HPO_{4}^{2-}\) can donate a proton to form \(PO_{4}^{3-}\).
The act of donating a proton is what defines a substance as a Bronsted acid. This transition between species highlights the reversible nature of acid-base reactions, showcasing how a single molecule can change its role in a reaction.
Proton Acceptance
Proton acceptance is the defining action of a Bronsted base. Bases typically have lone pairs of electrons or a negative charge that facilitates the acceptance of protons. Let's consider a few examples:
  • The dianion \(HPO_{3}^{2-}\) has sufficient negative charge and can act as a base by accepting a proton to form \(H_{2}PO_{3}^{-}\).
  • Also, \(HPO_{4}^{2-}\) can accept a proton to create \(H_{2}PO_{4}^{-}\).
This ability to accept protons makes a substance a Bronsted base. Essentially, it transforms the molecule into a more positively charged species, emphasizing the adaptable nature of these reagents.
Amphoteric Species
An amphoteric species is a versatile player in chemistry, capable of acting both as an acid and a base. This adaptability allows such species to donate and accept protons, depending on the surrounding chemical environment:
  • \(HPO_{4}^{2-}\) is an excellent example of an amphoteric ion. It can donate a proton to form \(PO_{4}^{3-}\), behaving as an acid.
  • Conversely, it can accept a proton to become \(H_{2}PO_{4}^{-}\), functioning as a base.
Amphoterism is a valuable characteristic for maintaining balance in chemical reactions and is pivotal in processes where pH conditions vary.
Acid-Base Reactions
Acid-base reactions are pivotal exchanges between proton donors (acids) and acceptors (bases). The interaction primarily results in the transfer of protons, which shapes the reaction dynamics. In the Bronsted-Lowry mode,
  • An acid such as \(HPO_{4}^{2-}\) donates a proton to become \(PO_{4}^{3-}\), aligning with its acidic role.
  • Simultaneously, this ion can accept a proton, transforming into \(H_{2}PO_{4}^{-}\), illustrating its basic side.
These reactions are essential in many natural and industrial processes. Understanding the dualistic roles of ions like \(HPO_{4}^{2-}\) helps in predicting the outcome and direction of chemical reactions.

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

50 litres of \(0.1 \mathrm{M} \mathrm{HCl}\) are mixed with 50 litres of \(0.2\) \(\mathrm{M} \mathrm{NaOH}\). The POH of the resulting solution is (a) \(12.70\) (b) \(12.34\) (c) \(8.7\) (d) \(4.2\)

The solubility of \(\mathrm{AgCl}\) in moles per litre when its solubility product is \(1.56 \times 10^{-10}\) at \(25^{\circ} \mathrm{C}\) is (a) \(0.576 \times 10^{-8} \mathrm{~mol} /\) litre (b) \(1.056 \times 10^{-4} \mathrm{~mol} /\) litre (c) \(1.249 \times 10^{-5} \mathrm{~mol} /\) litre (d) \(1.478 \times 10^{-6} \mathrm{~mol} /\) litre

Hydrolysis constant \(\mathrm{K}_{\mathrm{A}}\) and \(\mathrm{K}_{\mathrm{B}}\) of two salts of weak acids HA and \(\mathrm{HB}\) are \(10^{-8}\) and \(10^{-6}\) respectively. If the dissociation constant of third acid \(\mathrm{HC}\) is \(10^{-2} .\) The order of acidic strengths of three acids will be (a) \(\mathrm{HA}>\mathrm{HB}>\mathrm{HC}\) (b) \(\mathrm{HB}>\mathrm{HA}>\mathrm{HC}\) (c) \(\mathrm{HC}>\mathrm{HA}>\mathrm{HB}\) (d) \(\mathrm{HA}=\mathrm{HB}=\mathrm{HC}\)

Which of the following solution(s) have \(\mathrm{pH}\) between 6 and \(7 ?\) 1\. \(2 \times 10^{-6} \mathrm{M} \mathrm{NaOH}\) 2\. \(2 \times 10^{-6} \mathrm{M} \mathrm{HCl}\) 3\. \(10^{-8} \mathrm{M} \mathrm{HCl}\) 4\. \(10^{-13} \mathrm{M} \mathrm{NaOH}\) (a) 1,2 (b) 2,3 (c) 3,4 (d) \(2,3,4\)

An acid-base indicator has \(\mathrm{K}=3.0 \times 10^{-5} .\) The acid form of the indicator is red and the basic form is blue. The \(\left[\mathrm{H}^{+}\right]\)required to change the indicator from \(75 \%\) red to \(75 \%\) blue is (a) \(8 \times 10^{-5} \mathrm{M}\) (b) \(9 \times 10^{-5} \mathrm{M}\) (c) \(1 \times 10^{-5} \mathrm{M}\) (d) \(3 \times 10^{-4} \mathrm{M}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

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.