Write the formula of the conjugate base of the active ingredient in aspirin, acetylsalicylic acid, C9H8O4
The correct answer and explanation is :
The conjugate base of the active ingredient in aspirin, acetylsalicylic acid (C₉H₈O₄), is formed when the acid loses a proton (H⁺) from its carboxyl group (-COOH). The formula of the conjugate base is C₉H₇O₄⁻.
Explanation:
Acetylsalicylic acid contains both an ester group (-COOCH₃) and a carboxyl group (-COOH). The carboxyl group is responsible for its acidic properties. When acetylsalicylic acid dissociates in water, the hydrogen ion (H⁺) from the carboxyl group is released, resulting in the formation of the conjugate base.
- Structure of Acetylsalicylic Acid:
Acetylsalicylic acid has the chemical structure of a benzene ring (C₆H₅) attached to a carboxyl group (-COOH) and an ester group (-COOCH₃). The carboxyl group is acidic, and it is this group that loses a proton during dissociation. - Reaction in Water:
In aqueous solutions, acetylsalicylic acid can lose a proton from the carboxyl group as follows:
[
\text{C₉H₈O₄} \rightarrow \text{C₉H₇O₄}⁻ + \text{H}⁺
]
When the proton (H⁺) is lost, the negative charge is delocalized over the oxygen atoms in the carboxylate group, stabilizing the conjugate base. This gives the conjugate base C₉H₇O₄⁻. - Conjugate Base Characteristics:
The conjugate base, C₉H₇O₄⁻, retains the ester group (-COOCH₃) and the benzene ring but now has one fewer hydrogen atom, resulting in a negative charge on one of the oxygen atoms in the carboxylate group. This form is less acidic than acetylsalicylic acid itself because the conjugate base is a weaker acid compared to the neutral form.
Thus, the conjugate base of acetylsalicylic acid is C₉H₇O₄⁻, and the process involves the deprotonation of the carboxyl group to stabilize the negative charge.
