Skip to main content
Ch.16 - Aqueous Equilibria: Acids & Bases
Chapter 16, Problem 122

Nicotine 1C10H14N22 can accept two protons because it has two basic N atoms 1Kb1 = 1.0 * 10-6; Kb2 = 1.3 * 10-112. Calculate the values of Ka for the conjugate acids C10H14N2H+ and C10H14N2H22 + .

Verified Solution

Video duration:
3m
This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?

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 their conjugate bases in a solution. In this context, nicotine can accept protons, acting as a base, and its conjugate acids are formed when it gains protons. Understanding this equilibrium is essential for calculating the acid dissociation constants (Ka) for the conjugate acids.
Recommended video:
Guided course
05:11
Triprotic Acid Equilibrium

Base Dissociation Constant (Kb)

The base dissociation constant (Kb) quantifies the strength of a base in solution, indicating how well it can accept protons. In the given question, Kb1 and Kb2 represent the dissociation constants for nicotine's two basic nitrogen atoms. These values are crucial for determining the corresponding Ka values for the conjugate acids using the relationship between Kb and Ka.
Recommended video:
Guided course
03:50
Characteristics of Ka and Kb

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 × 10^-14 at 25°C). This relationship allows us to calculate the Ka values for the conjugate acids of nicotine from the provided Kb values, facilitating the understanding of their acidity in solution.
Recommended video:
Guided course
03:01
Ka and Kb Relationship