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: Determine the pH of a 0.62 M NH 4NO3 solution at 25°Celsius. The Kb for NH3 is 1.76 x 10-5

Problem

Determine the pH of a 0.62 M NH 4NO3 solution at 25°Celsius. The Kb for NH3 is 1.76 x 10-5

Solution

We’re being asked to calculate the pH of a 0.62 M NH4NO3 solution.


• NH4NO3 is an ionic compound that dissociates into ions when in a solution. 

Recall: All nitrates (NO3-) are soluble

• The dissociation of NH4NO3 is given by the equation:

NH4NO3 → NH4+(aq) + NO3-(aq)

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