Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 22

Write chemical equations that show how the hydrogen carbonate ion, HCO \(_{3}^{-}\) acts as an amphoteric ion.

Short Answer

Expert verified
Hydrogen carbonate ion (HCO3-) acts as an amphoteric ion, meaning it can act both as an acid and a base. As a base, it reacts with an acid (like HCl) to form CO3^2-: HCO3- (aq) + HCl (aq) -> H2O (l) + CO3^2- (aq). As an acid, it reacts with a base (like water) to form H2CO3: HCO3- (aq) + H2O (l) -> H2CO3 (aq) + OH- (aq).

Step by step solution

01

Reaction as a base

To show how the hydrogen carbonate ion (HCO3-) reacts as a base, it must accept a proton (H+) from an acid, such as hydrochloric acid (HCl). During this process, it forms water (H2O) and the carbonate ion (CO3^2-). The equation is: HCO3- (aq) + HCl (aq) -> H2O (l) + CO3^2- (aq) In this reaction, the HCO3- ion acts as a base because it accepts a proton from the hydrochloric acid.
02

Reaction as an acid

To show how the hydrogen carbonate ion (HCO3-) acts as an acid, it must donate a proton (H+) to a base such as water (H2O). In this process, it forms the bicarbonate ion (H2CO3) and hydroxide ion (OH-). The equation is: HCO3- (aq) + H2O (l) -> H2CO3 (aq) + OH- (aq) The reaction is showing how HCO3- ion acts as an acid as it donates a proton to water.

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

A small volume of indicator solution is usually added to the titration flask right before the titration. As a result, the sample is diluted slightly. Does this matter? Why or why not?

At the end point of a titration of 25 \(\mathrm{mL}\) of 0.300 \(\mathrm{M} \mathrm{NaOH}\) with \(0.200 \mathrm{M} \mathrm{HNO}_{3},\) what would the concentration of sodium nitrate in the titration flask be?

How much \(\mathrm{HCl}\) would you need to dissolve in 1.0 \(\mathrm{L}\) of water so that \(\left[\mathrm{OH}^{-}\right]=\) \(6.0 \times 10^{-12} \mathrm{M} ?\)

Why is water considered amphoteric?

If 25 \(\mathrm{mL}\) of 1.00 \(\mathrm{M} \mathrm{HCl}\) is mixed with 75 \(\mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{NaOH},\) what are the final amounts and concentrations of all ions present?
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

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