Ch.15 - Acid and Base EquilibriumWorksheetSee all chapters
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Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: Given that Kb for (CH3)3N is 6.3 × 10-5 at 25°C, what is the value of Ka for (CH3)3NH+ at 25 °C?

Solution: Given that Kb for (CH3)3N is 6.3 × 10-5 at 25°C, what is the value of Ka for (CH3)3NH+ at 25 °C?

Problem

Given that Kb for (CH3)3N is 6.3 × 10-5 at 25°C, what is the value of Ka for (CH3)3NH+ at 25 °C?

Solution

We are asked to find Ka for (CH3)3NH+. (CH3)3NH+ is the conjugate acid of the weak base (CH3)3NH.

(CH3)3NH     +     H2O      ⇌      (CH3)3NH+     +     OH-
(weak base)              (acid)             (conjugate acid)     (conjugate base) 

Hence, the Ka of (CH3)3NH+ can be calculated from the Kb of (CH3)3NH. Ka and Kb are connected by the autoionization constant of water (Kw) by the following equation:

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