# Chapter 18: Solubility and Complex-Ion Equilibria - PowerPoint PPT Presentation

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Chapter 18: Solubility and Complex-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

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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

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