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Ch.16 - Aqueous Equilibria: Acids & Bases
Chapter 16, Problem 120

Using values of Kb in Appendix C, calculate values of Ka foreach of the following ions.(a) Propylammonium ion, C3H7NH3+

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Identify the conjugate base of the propylammonium ion, C3H7NH3+. The conjugate base is propylamine, C3H7NH2.
Look up the value of Kb for propylamine (C3H7NH2) in Appendix C of your textbook or a reliable chemistry database.
Use the relationship between Ka and Kb for a conjugate acid-base pair, which is given by the equation Ka * Kb = Kw, where Kw is the ion-product constant of water at 25°C (approximately 1.0 x 10^-14).
Rearrange the equation to solve for Ka: Ka = Kw / Kb.
Substitute the value of Kb for propylamine and the value of Kw into the equation to calculate Ka for the propylammonium ion.

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Key Concepts

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

Acid-Base Equilibrium

Acid-base equilibrium refers to the balance between acids and bases in a solution, characterized by the dissociation of acids into protons (H+) and their conjugate bases. The strength of an acid or base is quantified by its dissociation constant (Ka for acids and Kb for bases), which indicates the extent to which the acid or base ionizes in water.
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Relationship Between Ka and Kb

The relationship between the acid dissociation constant (Ka) and the base dissociation constant (Kb) is given by the equation Ka × Kb = Kw, where Kw is the ion product of water (1.0 x 10^-14 at 25°C). This relationship allows for the calculation of Ka from Kb for a conjugate acid-base pair, facilitating the determination of the strength of acids and bases.
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Conjugate Acid-Base Pairs

Conjugate acid-base pairs consist of an acid and its corresponding base that differ by a single proton. For example, the propylammonium ion (C3H7NH3+) is the conjugate acid of propylamine (C3H7NH2). Understanding these pairs is essential for calculating Ka from Kb, as it highlights the interdependence of acid and base strengths in aqueous solutions.
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