Transcript Metathesis Problems (and Some Solutions) Identified
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