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What is the conjugate acid of the bisulfate ion, \(\mathrm{HSO}_{4}^{-}\) and what is its conjugate base?

Short Answer

Expert verified
Answer: The conjugate acid of the bisulfate ion, HSO鈧勨伝, is sulfuric acid, H鈧係O鈧, and its conjugate base is the sulfate ion, SO鈧劼测伝.

Step by step solution

01

Find the conjugate acid

First, we need to find the conjugate acid of the bisulfate ion by adding a proton to it. The conjugate acid will have one more proton (H\(^+\)) than the original ion. The bisulfate ion is \(\mathrm{HSO}_{4}^{-}\). Adding a proton (H\(^+\)) to it, we get the conjugate acid: \(\mathrm{H}_{2}\mathrm{SO}_{4}\). So, the conjugate acid of \(\mathrm{HSO}_{4}^{-}\) is sulfuric acid, \(\mathrm{H}_{2}\mathrm{SO}_{4}\).
02

Find the conjugate base

Next, we need to find the conjugate base of bisulfate ion by removing a proton from it. The conjugate base will have one less proton than the original ion. The bisulfate ion is \(\mathrm{HSO}_{4}^{-}\). Removing a proton (H\(^+\)) from it, we get the conjugate base: \(\mathrm{SO}_{4}^{2-}\). So, the conjugate base of \(\mathrm{HSO}_{4}^{-}\) is sulfate ion, \(\mathrm{SO}_{4}^{2-}\). In conclusion, the conjugate acid of the bisulfate ion, \(\mathrm{HSO}_{4}^{-}\), is sulfuric acid, \(\mathrm{H}_{2}\mathrm{SO}_{4}\) and its conjugate base is the sulfate ion, \(\mathrm{SO}_{4}^{2-}\).

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Key Concepts

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

Bisulfate Ion
The bisulfate ion (\(\mathrm{HSO}_{4}^{-}\) ) plays a crucial role in acid-base chemistry. It is a unique species because it can act both as an acid and a base. This characteristic is due to its ability to either donate or accept a proton depending on the reaction it participates in.
- When the bisulfate ion donates a proton (\(\mathrm{H}^{+} \)), it forms the sulfate ion (\(\mathrm{SO}_{4}^{2-}\) ), acting as an acid. - Conversely, when it accepts a proton, it transforms into sulfuric acid (\(\mathrm{H}_{2}\mathrm{SO}_{4}\) ), acting as a base.
The ability to either accept or donate a proton makes it amphiprotic, one of the less common but fascinating characteristics in chemistry. Understanding the bisulfate ion is essential when discussing chemical reactions in solutions, especially in many industrial and laboratory processes.
In everyday terms, the bisulfate ion is like a multitool that can perform dual roles depending on what the situation requires. This versatility emphasizes the dynamic nature of chemical species and their reactions.
Sulfuric Acid
Sulfuric acid (\(\mathrm{H}_{2}\mathrm{SO}_{4}\) ) is a highly significant chemical in both the industrial sector and scientific research.
- It is known for being a strong acid, which means it has a formidable ability to donate protons to other chemicals. - This capacity makes it useful as a starting reagent or catalyst in reactions.
Sulfuric acid is colorless and has a syrupy consistency. It is highly corrosive, and care must be taken to handle it safely. - In industries, sulfuric acid's applications are vast; from manufacturing fertilizers to refining crude oil, it serves countless purposes. - It works as a key component in the production of chemicals like hydrochloric acid and nitric acid.
When thinking about chemical reactions involving sulfuric acid, it's helpful to visualize it as a powerhouse, ready to release its protons and drive forward chemical processes.
Sulfate Ion
The sulfate ion (\(\mathrm{SO}_{4}^{2-}\) ) holds an important position in both natural and industrial processes. It is the conjugate base of the bisulfate ion, formed after the bisulfate ion donates a proton.
This ion carries a double negative charge, making it significant in equilibrating charges in chemical reactions. - Sulfate ions are commonly encountered in the environment, largely due to their presence in minerals and natural water systems. - They also appear in biological systems, such as in the detoxification processes within the liver.
In the commercial sector, sulfates are used in making detergents and in the processing of wastewater, where they help in controlling the chemical balance. - These ions work as crucial players in various chemical equations and are integral to technologies that rely on such processes, like battery manufacturing.
To sum up, the sulfate ion is like a fundamental building block, influencing a wide array of functions in both nature and industry. Whether in water treatment facilities or in your cleaning supplies, sulfate ions help ensure processes function correctly.

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Most popular questions from this chapter

Why is it unnecessary to publish tables of \(K_{b}\) values of the conjugate bases of weak acids whose \(K_{\mathrm{a}}\) values are known?

Thiosulfuric acid, \(\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{3},\) can be prepared by the reaction of \(\mathrm{H}_{2} \mathrm{S}\) with \(\mathrm{HSO}_{3} \mathrm{Cl}\) : $$\mathrm{HSO}_{3} \mathrm{Cl}(\ell)+\mathrm{H}_{2} \mathrm{S}(g) \rightarrow \mathrm{HCl}(g)+\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\ell)$$ a. Draw a Lewis structure for \(\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{3},\) given that it is isostructural with \(\mathrm{H}_{2} \mathrm{SO}_{4}\) b. Do you expect \(H_{2} S_{2} O_{3}\) to be a stronger or weaker acid than \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\) Explain your answer.

Painkillers Morphine is an effective painkiller but is also highly addictive. Codeine is a popular prescription painkiller because it is much less addictive than morphine. Codeine contains a basic nitrogen atom that can be protonated to give the conjugate acid of codeine. a. Calculate the \(\mathrm{pH}\) of a \(1.8 \times 10^{-3} M\) solution of morphine if its \(\mathrm{p} K_{\mathrm{b}}=5.79\) b. Calculate the pH of a \(2.7 \times 10^{-4} M\) solution of codeine if the \(\mathrm{p} K_{\mathrm{a}}\) of the conjugate acid is 8.21

Calculate the pH of a \(1.25 \times 10^{-2} M\) solution of the decongestant ephedrine hydrochloride if the \(\mathrm{p} K_{\mathrm{b}}\) of ephedrine (its conjugate base) is 3.86.

When \(1,\) 2-diaminoethane, \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2},\) dissolves in water, the resulting solution is basic. Write the formula of the ionic compound that is formed when hydrochloric acid is added to a solution of 1,2 -diaminoethane.

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