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 CaC 2 O 4 (

s

) = Ca 2+ (

aq

) + C 2 O 4 2 (

aq

) • Equilibrium constant for this process is called solubility product constant,

K sp K sp

= [Ca 2+ ][C 2 O 4 2 ] (since only aqueous components are included in an equilibrium expression) • In the

K sp

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

Solubility and

K sp

• The solubility of silver chloride is 1.9 x 10 -3 g/L. What is

K sp

?

– 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

K sp

expression

Solubility and

K sp

• Remember to account for the correct number of ions forming in their molar concentrations • The solubility of Pb 3 (AsO 4 ) 2 is 3.0 x 10 -5 g/L. What is 3 Pb 2+

K sp

?

• Since a single formula unit of lead arsenate forms 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?

K sp

= 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

K sp

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 Ca 2+ to this equilibrium: CaC 2 O 4 (

s

) = Ca 2+ (

aq

) + C 2 O 4 2+ (

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 BaF 2 and in 0.15

M

NaF.

K sp

= 1.0 k 10 -6 .

in pure water • 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,

Q c

– Same calc as equilibrium constant, but system is not necessarily at equilibrium • If

Q c

• If

Q c

<

K c

, reaction goes forward =

K c

, reaction is at equilibrium • If

Q c

>

K c

, reaction goes reverse • Ion product is

Q c

for a solubility reaction – Reverse means the mixture will precipitate, since the solid dissociates in the forward direction

Precipitation prediction • [Ca 2+ ] = 0.0025

M

• [C 2 O 4 2 ] = 1.0 x 10 -7

M

• Will calcium oxalate precipitate?

K sp

• Calculate ion product = 2.3 x 10 -9 • Compare to solubility product constant