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Chapter 18: Solubility and Complex-Ion EquilibriaPowerPoint Presentation

Chapter 18: Solubility and Complex-Ion Equilibria

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Chapter 18: Solubility andComplex-Ion Equilibria

Chemistry 1062: Principles of Chemistry II

Andy Aspaas, Instructor

Solubility product constant

- Excess of a slightly-soluble ionic compound is mixed with water
- An equilibrium occurs between the solid ionic compound and the dissociated ions
CaC2O4(s) = Ca2+(aq) + C2O42-(aq)

- Equilibrium constant for this process is called solubility product constant, Ksp
Ksp = [Ca2+][C2O42-]

(since only aqueous components are included in an equilibrium expression)

- In the Kspequation, raise an ion’s concentration to the power of its stoichiometric number in the chemical equation

Solubility and Ksp

- The solubility of silver chloride is 1.9 x 10-3 g/L. What is Ksp?
- First convert the solubility to molar solubility (mol/L)
- Use an ICE table to find equilibrium molar concentrations of the ions (using 0 as initial) – ignore the solid
- Substitute concentrations into a Ksp expression

Solubility and Ksp

- Remember to account for the correct number of ions forming in their molar concentrations
- The solubility of Pb3(AsO4)2 is 3.0 x 10-5 g/L. What is Ksp?
- Since a single formula unit of lead arsenate forms 3 Pb2+ ions when dissolved, be sure to multiply 3 by the molar concentration in the C row of the ICE table

Calculating solubility

- What is the solubility of calcium phosphate, in g/L? Ksp = 1 x 10-26
- Create an equilibrium expression with correct stoichiometry
- In an ICE table, use x as the unknown change of molar concentrations, multiplying stoichiometric numbers by x
- Solve for x in Ksp expression
- Convert mol/L to g/L

Solubility and the common-ion effect

- Addition of extra ion to a solubility equilibrium solution will shift the equilibrium according to Le Chatelier’s principle
- Ex. Adding additional Ca2+ to this equilibrium:
CaC2O4(s) = Ca2+(aq) + C2O42+(aq)

- This will cause the equilibrium to shift to the left, and the solid will become less soluble

Common-ion effect calculation

- Compare the molar solubilities of BaF2 in pure water and in 0.15 M NaF. Ksp = 1.0 k 10-6.
- Set up the water solution as before and solve for molar solubility (x)
- In the NaF solution, you have an initial concentration of 0.15 M of F-
- You can most likely assume that x << 0.15 + x

Predicting precipitation

- Recall reaction quotient, Qc
- Same calc as equilibrium constant, but system is not necessarily at equilibrium

- If Qc < Kc, reaction goes forward
- If Qc = Kc, reaction is at equilibrium
- If Qc > Kc, reaction goes reverse
- Ion product is Qcfor a solubility reaction
- Reverse means the mixture will precipitate, since the solid dissociates in the forward direction

Precipitation prediction

- [Ca2+] = 0.0025 M
- [C2O42-] = 1.0 x 10-7M
- Will calcium oxalate precipitate? Ksp = 2.3 x 10-9
- Calculate ion product
- Compare to solubility product constant

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