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: 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 2OH solution?a) 1.15b) 3.83c) 4.99d) 6.72e) 8.13

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

Solution

We’re being asked to calculate the pOH of a 2.37 M NH2OH solution.


Since NH2OH has a low Kb value, it’s 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 NH2OH is as follows:


NH2OH(aq) + H2O(l)  OH(aq) + NH3OH+(aq); Kb = 9.1 × 10–9


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




The Kb expression for NH2OH is:



Note that each concentration is raised by the stoichiometric coefficient: [NH2OH], [OH] and [NH3OH+] are raised to 1.


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