Ch.16 - Aqueous Equilibrium WorksheetSee 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: How do you calculate the pH at the equivalence point for the titration of .190M methylamine with .190M HCl? The Kb of methylamine is 5.0 x 10-4.

Problem

How do you calculate the pH at the equivalence point for the titration of .190M methylamine with .190M HCl? The Kb of methylamine is 5.0 x 10-4.

Solution

We’re being asked to calculate the pH of a solution at equivalence point for the titration of 0.190 M methylamine with 0.190 M HCl. 


HCl (strong acid) will react with CH3NH2 (base).

• CH3NH2 is a base and based on Bronsted-Lowry definition it is a proton acceptor.
• HCl is an acid and based on Bronsted-Lowry definition it is a proton donor.


Reaction:

CH3NH2(aq) + HCl(aq) CH3NH3+(aq) + Cl-(aq)


We will calculate the pH of the solution using the following steps:

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