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: Morphine is a weak base. A 0.150 M solution of morphine has a pH of 10.5.What is Kb for morphine?

Solution: Morphine is a weak base. A 0.150 M solution of morphine has a pH of 10.5.What is Kb for morphine?

Problem

Morphine is a weak base. A 0.150 M solution of morphine has a pH of 10.5.

What is Kb for morphine?

Solution

We are being asked to calculate the Kb for a 0.150 M solution of morphine (C17H19NO3)

Morphine (C17H19NO3) is a neutral amine, therefore we know that it is a weak base.


Since we’re dealing with a weak base and Kb is an equilibrium expression, we will have to create an ICE chart to determine the equilibrium concentration of each species:

C17H19NO3weak baseproton acceptor
H2O → will act as the weak acidproton donor


Equilibrium reaction:       C17H19NO3(aq) + H2O(l)  C17H19NO3H+(aq) + OH-(aq)

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