Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 22: Problem 82

Sulfuric acid is a weaker acid than hydrochloric acid. Yet hydrogen chloride is evolved when concentrated sulfuric acid is added to sodium chloride. Explain.

Short Answer

Expert verified
Sulfuric acid, despite being weaker than hydrochloric acid, is strong enough to donate a proton to Sodium Chloride. This results in the formation and release (evolution) of Hydrogen Chloride gas.

Step by step solution

01

Understanding the properties of Sulfuric Acid and Sodium Chloride

Sulfuric acid (H2SO4) is a strong acid, and Sodium Chloride (NaCl) is a salt. When they react, Sulfuric acid donates a proton (H+) to Sodium Chloride.
02

Explanation of the reaction

In this reaction, the Sulfuric Acid (H2SO4), being a strong acid, can donate a proton (H+) to the chloride ion (Cl-) of Sodium Chloride (NaCl). This is because the chloride ion (Cl-) is a better nucleophile than the sulfate ion (SO4^2-). The reaction will then proceed as follows: H2SO4 + NaCl -> NaHSO4 + HCl. The HCl gas is then evolved, i.e., the gas is released from the solution.
03

Summary of the process

Despite Sulfuric Acid being a weaker acid than hydrochloric acid, it is still strong enough to donate a proton to Sodium Chloride (NaCl). The reaction results in the formation and evolution of Hydrogen Chloride gas.

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

Metal chlorides can be prepared in a number of ways: (a) direct combination of metal and molecular chlorine, (b) reaction between metal and hydrochloric acid, (c) acid-base neutralization, (d) metal carbonate treated with hydrochloric acid, (e) precipitation reaction. Give an example for each type of preparation.

How are cyanide ions used in metallurgy?

How is hydrogen sulfide generated in the laboratory?

What is the change in oxidation number for the following reaction? $$3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{O}_{3}$$

Aqueous copper(II) sulfate solution is blue. When aqueous potassium fluoride is added to the \(\mathrm{CuSO}_{4}\) solution, a green precipitate is formed. If aqueous potassium chloride is added instead, a bright-green solution is formed. Explain what happens in each case.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Physical 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