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Chapter 16: Problem 79

Explain why benzenesulfonic acid is a Bronsted acid.

Short Answer

Expert verified
Benzenesulfonic acid is a Br酶nsted acid because it can donate a proton from its sulfonic acid group (-SO_3H).

Step by step solution

01

Understanding Bronsted Acids

A Br酶nsted acid is defined as a substance that can donate a proton ( ext{H}^+ ). This means that for a compound to be considered a Br酶nsted acid, it must have a hydrogen atom that can be released as a proton to another compound.
02

Structure of Benzenesulfonic Acid

Benzenesulfonic acid is a compound with the chemical formula C_6H_5SO_3H. This structure comprises a benzene ring attached to a sulfonic acid functional group (-SO_3H). The sulfonic acid group contains an ext{S-O-H} bond that plays a critical role in its acidity.
03

Identifying the Acidic Proton

The acidic proton in benzenesulfonic acid is the hydrogen atom in the ext{-SO_3H} group. This hydrogen is bound to the oxygen, a highly electronegative atom. The electronegativity of oxygen helps to polarize the O-H bond, making the release of ext{H}^+ (a proton) favorable in certain conditions.
04

Proton Donation Mechanism

Upon dissociation, the O-H bond in the ext{-SO_3H} group breaks, releasing a proton ( ext{H}^+). The remaining structure becomes the benzenesulfonate ion (C_6H_5SO_3^-). As benzenesulfonic acid can donate a proton, it fits the definition of a Br酶nsted acid.

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Key Concepts

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

Benzenesulfonic Acid
Benzenesulfonic acid is an organic compound with a molecular structure that is quite distinctive. It encompasses a benzene ring, which is a simple aryl unit, and a sulfonic acid group (-SO鈧僅) that clings to it. The formula can be articulated as C鈧咹鈧匰O鈧僅 where the six carbon atoms form a ring, sharing bonds with each other. The sulfonic acid group stands out in this molecule due to its ability to donate a proton, making it acidic.
There are a few reasons why benzenesulfonic acid is particularly interesting:
  • It serves as an example of a sulfonic acid where the -SO鈧僅 group is directly linked to an aromatic ring.
  • This connection offers benzenesulfonic acid unique chemical properties compared to other acids.
  • Its action as a strong acid is well recognized due to its proton-donating capability.
Understanding the arrangement of atoms in benzenesulfonic acid provides insight into why it's a capable proton donor.
Proton Donation
Proton donation is a key characteristic of any Br酶nsted acid, including benzenesulfonic acid. The term 'proton' refers to a hydrogen ion, symbolized as H鈦�. In the case of benzenesulfonic acid, the capacity to release this proton is a central aspect of its behavior as an acid.
When benzenesulfonic acid interacts with a base, it can distinguish itself by performing the following:
  • Displays the ability to donate a proton from its sulfonic group.
  • The proton is transferred to another molecule, favorably a base, initiating a reaction.
  • This process ultimately transforms benzenesulfonic acid into its conjugate base, the benzenesulfonate ion.
Proton donation explains why benzenesulfonic acid can act as a Br酶nsted acid, emphasizing the importance of the -SO鈧僅 group in this process.
Acidic Proton
In benzenesulfonic acid, identifying the acidic proton is pivotal to understanding its acidic nature. This refers to the particular hydrogen atom within the sulfonic acid group that can be liberated as a proton. The -SO鈧僅 group harbors this acidic proton, highlighting the role of oxygen, to which it is attached.
Here are some key points to consider:
  • Oxygen, being electronegative, draws electrons toward itself, weakening the O-H bond.
  • The weakened bond makes it easier for the hydrogen atom to dissociate as a H鈦� ion.
  • This property is what predominantly contributes to the acidity, as it allows the release of the proton in certain conditions.
The concept of the acidic proton helps delineate why benzenesulfonic acid can act as an efficient proton donor.
Polarization of O-H Bond
Understanding the polarization of the O-H bond is essential in grasping the behavior of benzenesulfonic acid. In this context, polarization refers to the uneven distribution of electron density across the O-H bond in the -SO鈧僅 functional group.
Key aspects to understand include:
  • Electrons gravitate more towards the oxygen atom due to its higher electronegativity compared to hydrogen.
  • This results in partial negative charge at oxygen and partial positive charge at hydrogen.
  • The polarized bond is weaker, favoring the breakage of the bond and release of the proton.
The polarization of the O-H bond in benzenesulfonic acid is a crucial factor making the proton release favorable, thereby reinforcing its identity as a strong acid.

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Most popular questions from this chapter

You purchase a bottle of water. On checking its \(\mathrm{pH},\) you find that it is not neutral, as you might have expected. Instead, it is slightly acidic. Why?

You have three solutions labeled \(A, B\), and \(C\). You know only that each contains a different cation \(-\mathrm{Na}^{+}, \mathrm{NH}_{4}^{+},\) or \(\mathrm{H}_{3} \mathrm{O}^{+} .\) Each has an anion that does not contribute to the solution \(\mathrm{pH}\) (eg. \(\mathrm{Cl}^{-}\) ). You also have two other solutions, Y and \(\mathrm{Z}\) each containing a different anion, \(\mathrm{Cl}^{-}\) or \(\mathrm{OH}^{-}\) with a cation that does not influence solution \(\mathrm{pH}\) \(\left(\mathrm{e} \cdot \mathrm{g} \cdot, \mathrm{K}^{+}\right) .\) If equal amounts of \(\mathrm{B}\) and \(\mathrm{Y}\) are mixed, the result is an acidic solution. Mixing A and Z gives a neutral solution, whereas \(\mathrm{B}\) and Z give a basic solution. Identify the five unknown solutions. (Adapted from D. H. Barouch: Voyages in Conceptual Chemistry, Boston, Jones and Bartlett, \(1997 .)\)

Oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) is a diprotic acid. Write a chemical equilibrium expression for each ionization step in water.

What is the pH of a \(1.2 \times 10^{-4} \mathrm{M}\) solution of KOH? What is the hydronium ion concentration of the solution?

The ionization constant of a very weak acid, HA, is \(4.0 \times 10^{-9} .\) Calculate the equilibrium concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{A}^{-},\) and \(\mathrm{HA}\) in a \(0.040 \mathrm{M}\) solution of the acid.
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