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
Problem

What is the pH of a solution in which 0.250 mol of NH3 (Kb = 1.8 × 10-5) is dissolved in sufficient water to make 1.00 L of solution?

a. 2.67

b. 11.33

c. 4.93

d. 13.4

e. 9.07

Solution

We are being asked to calculate the pH for a 0.250 mol NH3 in 1 L solution

NH3 is a neutral amine, therefore we know that it is a weak base.

NH3(aq) + H2O(aq)  OH(aq) + NH4+(aq)Kb = 1.8 × 10-5


Calculate the molarity:

Molarity = moles soluteL solutionMolarity = 0.250 mol1 L

Molarity =  0.250 M


From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table.




The Kb expression for NH3 is:


Kb = productsreactants = [OH-][NH4+][NH3]


Note that each concentration is raised by the stoichiometric coefficient: [NH3], [OH] and [NH4+] are raised to 1.


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