Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 85

When acetic acid is treated with aqueous sodium bicarbonate, \(\mathrm{CO}_{2}\) gas is evolved. The 'Carbon' of \(\mathrm{CO}_{2}\) comes from (a) carboxyl group (b) methyl group (c) bicarbonate (d) both carboxyl group and bicarbonate

Short Answer

Expert verified
Answer: (c) bicarbonate

Step by step solution

01

Identify the structure of reactants

Acetic acid has the formula \(\mathrm{CH}_{3}\mathrm{COOH}\), which has a carboxyl group (-\(\mathrm{COOH}\)) and a methyl group (-\(\mathrm{CH}_{3}\)). Aqueous sodium bicarbonate has the formula \(\mathrm{NaHCO}_{3}\) and has a bicarbonate ion (\(\mathrm{HCO}_{3}^{-}\)).
02

Write down the balanced chemical equation

The balanced chemical equation for the reaction can be written as: $$\mathrm{CH}_{3}\mathrm{COOH} + \mathrm{NaHCO}_{3} \rightarrow \mathrm{CH}_{3}\mathrm{COONa} + \mathrm{H}_{2}\mathrm{O} + \mathrm{CO}_{2}$$ This equation shows the products formed during the reaction: sodium acetate (\(\mathrm{CH}_{3}\mathrm{COONa}\)), water (\(\mathrm{H}_{2}\mathrm{O}\)), and carbon dioxide (\(\mathrm{CO}_{2}\)).
03

Identify the origin of the Carbon

In this step, we will track the origin of the carbon atom in the \(\mathrm{CO}_{2}\) molecule. The carbon dioxide molecule has one carbon atom, which can be traced back to the molecules react from the reactants. From the balanced chemical equation, it is evident that the carbon atom in \(\mathrm{CO}_{2}\) comes from the bicarbonate ion. The bicarbonate ion loses hydrogen and oxygen in the form of water, leaving behind the carbon atom to form a molecule of carbon dioxide. The remaining atoms from the carboxyl group and the methyl group form the sodium acetate. So, the 'Carbon' in \(\mathrm{CO}_{2}\) comes from: (c) bicarbonate

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 compound (A) gives an yellow precipitate with iodine and sodium hydroxide. On treating with a Grignard reagent, it gives (B). (B) gives a blue coloured pseudonitrole (C) in the Victor Meyer reaction which does not react with \(\mathrm{NaOH}\). (a) is (a) acetone (b) acetaldehyde (c) benzaldehvde (d) propanal

CC(C)=O (A) $\stackrel{\text { dil. } \mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow}(\mathrm{B})$ The product (B) in the above reaction is (a) Hexan-2-one (b) 3, 3-Dimethyl butan-2-one (c) 2, 3-Dimethyl butan-2-ol (d) 4-Hydroxy-4-methyl pentan-2-one

When distilled with concentrated sulphuric acid, propanone gives (a) monomethyl benzene (b) dimethyl benzene (c) trimethyl benzene (d) propanoic acid

2 -phenyl ethanol on oxidation with alkaline \(\mathrm{KMnO}_{4}\) gives (a) benzaldehyde (b) 2 -phenyl ethanoic acid (c) benzoic acid (d) 2 -phenyl ethanol

Identify the product (c) in the following sequence of reaction \(\mathrm{PhCH}_{2} \mathrm{Cl} \underset{\text { DMF }}{\longrightarrow}(\mathrm{A}) \stackrel{\mathrm{KCN}}{\text { PhCHO/A }}\) (B) \(\stackrel{\mathrm{H}_{3} \mathrm{O}^{+}}{\longrightarrow}(\mathrm{C})\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Physical 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