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: The pH of a 1.00 M solution of urea, a weak organic base, is 7.050.Calculate the Ka of protonated urea.


The pH of a 1.00 M solution of urea, a weak organic base, is 7.050.

Calculate the Ka of protonated urea.


We’re being asked to calculate the Ka of protonated urea (The pH of 1.00 M urea, a weak organic base is 7.050). 

Urea ((NH2)2CO) is a weak base. 

Remember that weak bases partially dissociate in water and that bases accept H+ from the acid (water in this case). 

The dissociation of Urea is as follows:

(NH2)2CO (aq) + H2O(l)  (NH2)2COH+ (aq) + OH-(aq)

From this, we can construct an ICE table. 

Remember that liquids are ignored in the ICE table.

Kb expression for (NH2)2CO  is:

Kb = productsreactants= [(NH2)2COH+][OH-][(NH2)2CO]

Liquids are ignored in the Kb expression.

Note that each concentration is raised by the stoichiometric coefficient: [(NH2)2CO], [OH] and [(NH2)2COH+] are raised to 1.

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