Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 142

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

Short Answer

Expert verified
In the given reaction, CO2 acts as the Lewis acid, accepting a lone pair of electrons from H2O, which is the Lewis base. The electron pair donation from H2O forms a covalent bond between the carbon atom in CO2 and an oxygen atom in H2O. An unstable molecule is formed, which undergoes a proton transfer reaction, resulting in the formation of carbonic acid (H2CO3). Thus, the reaction demonstrates the Lewis acid-base model with CO2 accepting an electron pair from H2O: \[ \mathrm{CO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{H}_{2}\mathrm{CO}_{3}(aq) \]

Step by step solution

01

Identify the Lewis acid and base in the reaction

According to the Lewis acid-base model, a Lewis acid is a species that can accept an electron pair, while a Lewis base is a species that can donate an electron pair. In this reaction, CO2 is the Lewis acid, as it accepts a lone pair of electrons from H2O. The H2O molecule is the Lewis base, as it donates a lone pair of electrons to CO2.
02

Explain the interaction between the Lewis acid and base

The lone pair of electrons on one of the oxygen atoms in the H2O molecule forms a covalent bond with the carbon atom in the CO2 molecule. This creates a new molecule with a carbon atom bonded to three oxygen atoms and one hydrogen atom.
03

Explain the formation of the other hydrogen bond

The new molecule formed in Step 2 is unstable, and it undergoes a proton transfer reaction. The remaining hydrogen atom from the original water molecule gets attracted to the negatively charged oxygen atom within the new molecule, forming another covalent bond. This results in the formation of the carbonic acid (H2CO3) molecule.
04

Summarize the reaction according to the Lewis acid-base model

In this reaction, CO2 (the Lewis acid) accepts an electron pair from H2O (the Lewis base). This forms an unstable molecule, which undergoes a proton transfer reaction to produce the final product H2CO3. The interaction between the electron pair donor (H2O) and electron pair acceptor (CO2) demonstrates the Lewis acid-base model in the given reaction: $$ \mathrm{CO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{H}_{2}\mathrm{CO}_{3}(aq) $$

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

Phosphoric acid is a common ingredient in traditional cola drinks. It is added to provide the drinks with a pleasant tart taste. Assuming that in cola drinks the concentration of phosphoric acid is \(0.007 M,\) calculate the \(\mathrm{pH}\) of this solution.

A typical vitamin \(\mathrm{C}\) tablet (containing pure ascorbic acid, \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}\) ) weighs $500 . \mathrm{mg}\( . One vitamin \)\mathrm{C}$ tablet is dissolved in enough water to make 200.0 \(\mathrm{mL}\) of solution. Calculate the \(\mathrm{pH}\) of this solution. Ascorbic acid is a diprotic acid.

Calculate the pH of a \(0.010-M\) solution of iodic acid (HIO $_{3}, K_{\mathrm{a}}\( \)=0.17 )$

Saccharin, a sugar substitute, has the formula $\mathrm{HC}_{7} \mathrm{H}_{4} \mathrm{NSO}_{3}$ and is a weak acid with \(K_{\mathrm{a}}=2.0 \times 10^{-12} .\) If 100.0 \(\mathrm{g}\) of saccharin is dissolved in enough water to make 340 \(\mathrm{mL}\) of solution, calculate the \(\mathrm{pH}\) of the resulting solution.

For propanoic acid $\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right),$ determine the concentration of all species present, the \(\mathrm{pH},\) and the percent dissociation of a \(0.100-M\) solution.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

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