Ch.16 - Aqueous Equilibrium WorksheetSee 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

Calculate the molar solubility of AgCl (K sp = 1.8 x 10 -10 at 25 degrees C) in 3.0 M NH3

Ag+(aq) + 2NH3(aq) ⇌ Ag(NH3)2+(aq)

Kf = 1.6 x 107


For this problem, we’re being asked to calculate the molar solubility (in mol/L) of AgCl(s) in 3.0 M NH3.

• Since the AgCl ionic, it will form ions when dissociating in a solution. The dissociation of AgCl in a solution is as follows:

AgCl(s) Ag+(aq) + Cl(aq)                             Ksp = 1.8x10-10

Ksp =[products][reactants]=Ag+Cl-

Note that each concentration is raised by the stoichiometric coefficient: both [Ag+] and [Cl-] are raised to 1. Also, solids are ignored in Ksp expression.

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