Ch.15 - Acid and Base EquilibriumWorksheetSee all chapters
All Chapters
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: Hydroxylamine, NH2OH, is a weak base. The following is the equilibrium equation for its reaction with water:NH2OH(aq) + H2O(l) ⇌ NH3OH+(aq) + OH -       Kb = 9.1 x 10 -9What is the pOH of a 2.37 M NH 

Solution: Hydroxylamine, NH2OH, is a weak base. The following is the equilibrium equation for its reaction with water:NH2OH(aq) + H2O(l) ⇌ NH3OH+(aq) + OH -       Kb = 9.1 x 10 -9What is the pOH of a 2.37 M NH 

Problem

Hydroxylamine, NH2OH, is a weak base. The following is the equilibrium equation for its reaction with water:

NH2OH(aq) + H2O(l) ⇌ NH3OH+(aq) + OH -       Kb = 9.1 x 10 -9

What is the pOH of a 2.37 M NH 2OH solution?

a) 1.15

b) 3.83

c) 4.99

d) 6.72

e) 8.13