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Ch.16 - Acids and Bases
Chapter 16, Problem 75

A 0.148 M solution of a monoprotic acid has a percent ionization of 1.55%. Determine the acid ionization constant (Ka) for the acid.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Monoprotic Acid

A monoprotic acid is an acid that can donate only one proton (H+) per molecule during the ionization process. This characteristic simplifies the calculation of its ionization constant, as the equilibrium expression involves only one dissociation step. Common examples include acetic acid and hydrochloric acid.
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Percent Ionization

Percent ionization is a measure of the extent to which an acid dissociates in solution, expressed as a percentage of the initial concentration that has ionized. It is calculated using the formula: (concentration of ionized acid / initial concentration) × 100%. This concept helps in understanding the strength of the acid and its behavior in solution.
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Percent Ionization Example

Acid Ionization Constant (Ka)

The acid ionization constant (Ka) quantifies the strength of an acid in solution, representing the equilibrium constant for the dissociation of the acid into its ions. It is calculated using the formula: Ka = [H+][A-] / [HA], where [H+] is the concentration of hydrogen ions, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the undissociated acid. A higher Ka value indicates a stronger acid.
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Characteristics of Ka and Kb