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Identify each of the following species as a Brønsted acid, base, or both: (a) HI, (b) \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-},\) (c) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\), (d) \(\mathrm{HSO}_{4}^{-}\).

Short Answer

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(a) Acid; (b) Base; (c) Both; (d) Both.

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01

Understanding Brønsted Acids and Bases

A Brønsted acid is a substance that can donate a proton (H⁺), while a Brønsted base is a substance that can accept a proton. Some substances can act as both, donating or accepting a proton depending on the context.
02

Analyzing HI

HI, or hydrogen iodide, can donate a proton to form I⁻. Since it donates a proton, HI is classified as a Brønsted acid.
03

Analyzing C₂H₃O₂⁻

C₂H₃O₂⁻, acetate ion, can accept a proton to form acetic acid (CH₃COOH). Because it can accept a proton, it functions as a Brønsted base.
04

Analyzing H₂PO₄⁻

H₂PO₄⁻ (dihydrogen phosphate) can donate a proton to become HPO₄²⁻ or accept a proton to form H₃PO₄. Thus, it can act as both a Brønsted acid (donating) and a base (accepting).
05

Analyzing HSO₄⁻

HSO₄⁻ (hydrogen sulfate) can donate a proton to become SO₄²⁻ or accept a proton to return to H₂SO₄. Therefore, it acts as both a Brønsted acid and base.

Key Concepts

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

Proton Donor
In the realm of chemistry, a proton donor is another name for a Brønsted acid. These substances have the ability to donate a proton, specifically a hydrogen ion ( H^+ ), during chemical reactions. The process of donating a proton transforms the original molecule, creating a new product.
For instance, in the case of hydrogen iodide (HI), the substance donates its proton to become iodide ( I^- ). This transformation is a key characteristic of Brønsted acids. Essentially, when you think of proton donors, think of substances eager to give away their extra hydrogen ions to partake in and drive chemical reactions.
Understanding which species can donate protons helps in identifying them as Brønsted acids and predicting their behavior in reactions.
Proton Acceptor
A proton acceptor is a term used to describe a Brønsted base. These entities have the ability to accept a proton from their surroundings, which leads to the formation of a new molecule.
When a substance acts as a proton acceptor, it typically starts with a negative charge or a lone pair of electrons that can easily accommodate an incoming proton.
For example, the acetate ion ( C_2H_3O_2^- ) acts as a proton acceptor by bonding with a proton to form acetic acid ( CH_3COOH ). By accepting this proton, the substance is stabilized and transformed, showcasing typical behavior of Brønsted bases.
Recognizing proton acceptors is crucial for understanding how certain reactions progress, as they guide the transfer and transformation of protons throughout the process.
HI Classification
Hydrogen iodide, written chemically as HI, is a classic example of a Brønsted acid. This classification stems from its capacity to donate a proton, a fundamental requirement for any Brønsted acid.
When HI gives away its proton, it leaves behind an iodide anion ( I^- ). This reaction forms new compounds and is indicative of HI's behavior as a Brønsted acid.
Overall, knowing specific molecules like HI helps further understand the broader concept of acids and their roles in chemical reactions. Knowing it donates protons is essential to classifying it correctly in reactions with other substances.
H₂PO₄⁻ Amphoteric Behavior
The concept of amphoteric behavior refers to a molecule's ability to act both as an acid and a base, depending upon the circumstances. The dihydrogen phosphate ion ( H_2PO_4^- ) brilliantly exemplifies this unique dual behavior.
As an acid, H_2PO_4^- can donate a proton, transforming into HPO_4^{2-} . This ability to lose a proton to another molecule highlights its acidic properties.
On the flip side, when acting as a base, H_2PO_4^- accepts a proton, forming H_3PO_4 (phosphoric acid). This addition of a proton underscores its basic capabilities.
Understanding amphoteric behavior is important in assessing how certain compounds react diversely under varying conditions, providing more dynamic roles in chemical equations.

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Most popular questions from this chapter

The concentration of lead ions \(\left(\mathrm{Pb}^{2+}\right)\) in a sample of polluted water that also contains nitrate ions \(\left(\mathrm{NO}_{3}^{-}\right)\) is determined by adding solid sodium sulfate \(\left(\mathrm{Na}_{2} \mathrm{SO}_{4}\right)\) to exactly \(500 \mathrm{~mL}\) of the water. (a) Write the molecular and net ionic equations for the reaction. (b) Calculate the molar concentration of \(\mathrm{Pb}^{2+}\) if \(0.00450 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) was needed for the complete precipitation of \(\mathrm{Pb}^{2+}\) ions as \(\mathrm{PbSO}_{4}\).

Sulfites (compounds containing the \(\mathrm{SO}_{3}^{2-}\) ions) are used as preservatives in dried fruits and vegetables and in wine making. In an experiment to test for the presence of sulfite in fruit, a student first soaked several dried apricots in water overnight and then filtered the solution to remove all solid particles. She then treated the solution with hydrogen peroxide \(\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)\) to oxidize the sulfite ions to sulfate ions. Finally, the sulfate ions were precipitated by treating the solution with a few drops of a barium chloride \(\left(\mathrm{BaCl}_{2}\right)\) solution. Write a balanced equation for each of the preceding steps.

How many grams of \(\mathrm{KOH}\) are present in \(35.0 \mathrm{~mL}\) of a \(5.50 \mathrm{M} \mathrm{KOH}\) solution?

Is the following reaction a redox reaction? Explain. $$ 3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{O}_{3}(g) $$

For the complete redox reactions given here, break down each reaction into its half-reactions, identify the oxidizing agent, and identify the reducing agent. (a) \(2 \mathrm{Sr}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{Sr} \mathrm{O}\) (b) \(2 \mathrm{Li}+\mathrm{H}_{2} \longrightarrow 2 \mathrm{LiH}\) (c) \(2 \mathrm{Cs}+\mathrm{Br}_{2} \longrightarrow 2 \mathrm{CsBr}\) (d) \(3 \mathrm{Mg}+\mathrm{N}_{2} \longrightarrow \mathrm{Mg}_{3} \mathrm{~N}_{2}\)

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