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What is the anhydride for each of the following acids: (a) \(\mathrm{H}_{2} \mathrm{SO}_{4}\), (b) \(\mathrm{HClO}_{3}\), (c) \(\mathrm{HNO}_{2}\), (d) \(\mathrm{H}_{2} \mathrm{CO}_{3}\), (e) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) ?

Short Answer

Expert verified
The anhydrides for the given acids are: (a) \(\mathrm{H}_{2}\mathrm{S}_{2}\mathrm{O}_{7}\), (b) \(\mathrm{Cl}_{2}\mathrm{O}_{6}\), (c) \(\mathrm{N}_{2}\mathrm{O}_{3}\), (d) \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{5}\), and (e) \(\mathrm{H}_{4}\mathrm{P}_{2}\mathrm{O}_{7}\).

Step by step solution

01

(a) Find the anhydride of \(\mathrm{H}_{2} \mathrm{SO}_{4}\)

To find the anhydride for sulfuric acid, \(\mathrm{H}_{2}\mathrm{SO}_{4}\), combine two molecules of the acid and remove one \(\mathrm{H}_{2}\mathrm{O}\) molecule: \(2 \times \mathrm{H}_{2} \mathrm{SO}_{4} - \mathrm{H}_{2}\mathrm{O} = \mathrm{H}_{2}\mathrm{S}_{2}\mathrm{O}_{7}\) The anhydride for sulfuric acid is \(\mathrm{H}_{2}\mathrm{S}_{2}\mathrm{O}_{7}\).
02

(b) Find the anhydride of \(\mathrm{HClO}_{3}\)

To find the anhydride for chloric acid, \(\mathrm{HClO}_{3}\), combine two molecules of the acid and remove one \(\mathrm{H}_{2}\mathrm{O}\) molecule: \(2 \times \mathrm{HClO}_{3} - \mathrm{H}_{2}\mathrm{O} = \mathrm{Cl}_{2}\mathrm{O}_{6}\) The anhydride for chloric acid is \(\mathrm{Cl}_{2}\mathrm{O}_{6}\).
03

(c) Find the anhydride of \(\mathrm{HNO}_{2}\)

To find the anhydride for nitrous acid, \(\mathrm{HNO}_{2}\), combine two molecules of the acid and remove one \(\mathrm{H}_{2}\mathrm{O}\) molecule: \(2 \times \mathrm{HNO}_{2} - \mathrm{H}_{2}\mathrm{O} = \mathrm{N}_{2}\mathrm{O}_{3}\) The anhydride for nitrous acid is \(\mathrm{N}_{2}\mathrm{O}_{3}\).
04

(d) Find the anhydride of \(\mathrm{H}_{2} \mathrm{CO}_{3}\)

To find the anhydride for carbonic acid, \(\mathrm{H}_{2}\mathrm{CO}_{3}\), combine two molecules of the acid and remove one \(\mathrm{H}_{2}\mathrm{O}\) molecule: \(2 \times \mathrm{H}_{2}\mathrm{CO}_{3} - \mathrm{H}_{2}\mathrm{O} = \mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{5}\) The anhydride for carbonic acid is \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{5}\).
05

(e) Find the anhydride of \(\mathrm{H}_{3} \mathrm{PO}_{4}\)

To find the anhydride for phosphoric acid, \(\mathrm{H}_{3}\mathrm{PO}_{4}\), combine two molecules of the acid and remove one \(\mathrm{H}_{2}\mathrm{O}\) molecule: \(2 \times \mathrm{H}_{3}\mathrm{PO}_{4} - \mathrm{H}_{2}\mathrm{O} = \mathrm{H}_{4}\mathrm{P}_{2}\mathrm{O}_{7}\) The anhydride for phosphoric acid is \(\mathrm{H}_{4}\mathrm{P}_{2}\mathrm{O}_{7}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sulfuric Acid and Its Anhydride
Sulfuric acid, denoted as \(\mathrm{H}_2\mathrm{SO}_4\), is a highly important industrial chemical, known for its various uses in battery acid, fertilizer, and as a dehydrating agent. When it comes to its anhydride, the process involves combining two sulfuric acid molecules and removing a water molecule \(\mathrm{H}_2\mathrm{O}\), resulting in \(\mathrm{H}_2\mathrm{S}_2\mathrm{O}_7\), also known as pyrosulfuric acid or oleum. This is a useful reaction for understanding the relationship between acids and their respective anhydrides.

Oleum is used in various industrial applications, including the production of detergents and explosives. By understanding the creation of anhydrides, students can get insights into chemical synthesis and reactivity, both key concepts in chemistry.
Chloric Acid and Its Anhydride
Chloric acid, with the formula \(\mathrm{HClO}_3\), is known for its oxidative properties and is a part of a family of oxoacids of chlorine. To form its anhydride, you will combine two molecules of chloric acid and eliminate a single water molecule. This yields \(\mathrm{Cl}_2\mathrm{O}_6\), which is a powerful oxidizing agent itself.

Although less commonly encountered outside the laboratory, understanding chloric acid and its anhydride helps students grasp the varied oxidative states chlorine can exist in, offering a peek into the versatile nature of halogens in chemistry.
Nitrous Acid and Its Anhydride
Nitrous acid, \(\mathrm{HNO}_2\), is a weaker acid often used in the preparation of diazonium salts in organic synthesis. Its anhydride is formed by combining two nitrous acid molecules and removing one molecule of water, resulting in \(\mathrm{N}_2\mathrm{O}_3\), known as dinitrogen trioxide. This compound is important as it exemplifies the behavior of nitrogen oxides, which are significant in the study of environmental chemistry and atmospheric reactions.

It's critical to understand the properties and reactions of nitrous acid and its anhydride as they play a significant role in nitrogen cycles and pollution.
Carbonic Acid and Its Anhydride
Carbonic acid \(\mathrm{H}_2\mathrm{CO}_3\) plays an essential role in biological and geological processes. Like others, its anhydride is constructed by removing a water molecule from two carbonic acid molecules, which gives \(\mathrm{H}_2\mathrm{C}_2\mathrm{O}_5\), though it's more commonly recognized in the form of its gaseous counterpart, carbon dioxide \(\mathrm{CO}_2\).

This transformation is pivotal in understanding the carbon cycle and the impact of \(\mathrm{CO}_2\) in environmental science, as well as the role of acids and anhydrides in biochemical systems.
Phosphoric Acid and Its Anhydride
Phosphoric acid \(\mathrm{H}_3\mathrm{PO}_4\) is widely used in agriculture and food industries. Its anhydride, \(\mathrm{H}_4\mathrm{P}_2\mathrm{O}_7\), or pyrophosphoric acid, results from dehydrating two phosphoric acid molecules. This occurs naturally when phosphoric acid is heated, leading to the loss of a water molecule.

The anhydride form is significant in biochemistry, especially in DNA synthesis and energy transfer through ATP. For students, understanding the process of forming pyrophosphoric acid can help illustrate the concept of condensation reactions and their importance in both industrial chemistry and life sciences.

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Most popular questions from this chapter

Ultrapure germanium, like silicon, is used in semiconductors. Germanium of "ordinary" purity is prepared by the high-temperature reduction of \(\mathrm{GeO}_{2}\) with carbon. The Ge is converted to \(\mathrm{GeCl}_{4}\) by treatment with \(\mathrm{Cl}_{2}\) and then purified by distillation; \(\mathrm{GeCl}_{4}\) is then hydrolyzed in water to \(\mathrm{GeO}_{2}\) and reduced to the elemental form with \(\mathrm{H}_{2}\). The element is then zone refined. Write a balanced chemical equation for each of the chemical transformations in the course of forming ultrapure Ge from \(\mathrm{GeO}_{2}\).

(a) Draw the Lewis structures for at least four species that have the general formula $$ [: X \equiv Y:]^{\pi} $$ where \(X\) and \(Y\) may be the same or different, and \(n\) may have a value from \(+1\) to \(-2\). (b) Which of the compounds is likely to be the strongest Bronsted base? Explain. [Sections 22.1, \(22.7\), and \(22.91\) eriodic Trends and Chemical Reactions Section 22.1)

Chemists tried for a long time to make molecular compounds containing silicon- silicon double bonds, they finally succeeded in 1981 . The trick is having large, bulky R groups on the silicon atoms to make \(\mathrm{R}_{2} \mathrm{Si}=\mathrm{SiR}_{2}\) compounds. What experiments could you do to prove that a new compound has a silicon-silicon double bond rather than a silicon-silicon single bond?

Account for the following observations: (a) \(\mathrm{H}_{3} \mathrm{PO}_{3}\) is a diprotic acid. (b) Nitric acid is a strong acid, whereas phosphoric acid is weak. (c) Phosphate rock is ineffective as a phosphate fertilizer. (d) Phosphorus does not exist at room temperature as diatomic molecules, but nitrogen does. (e) Solutions of \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) are quite basic.

Write the Lewis structure for each of the following species, describe its geometry, and indicate the oxidation state of the nitrogen: (a) \(\mathrm{HNO}_{2}\), (b) \(\mathrm{N}_{3}^{-}\), (c) \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}\), (d) \(\mathrm{NO}_{3}^{-}\).

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