Solubility
Equilibria
AP Chemistry
Ms. Grobsky
Introducing Solubility Equilibria
 For
the last couple of weeks, we have
considered equilibria involving acids and
bases

Homogeneous equilibria systems
 All
species have been in the same physical
state
 Now,
we will consider the equilibria
involved in the dissolution or precipitation
of ionic compounds

Heterogeneous equilibria systems
Equilibrium of Dissolution and
Precipitation Reactions


A saturated solution is one in which the solution is in
contact with undissolved solute
For example, when solid barium sulfate - an ionic
compound that is a weak electrolyte - is added to
water, the solid will dissolve and yield Ba2+ (aq) and
SO42- (aq), readily establishing the equilibrium:
BaSO4 s ↔ Ba2+ aq + SO2−
4 (aq)

As with any other equilibrium, the extent to which this
dissolution reaction occurs is expressed by the
magnitude of the equilibrium constant, Ksp
Ksp – The Solubility Product
Constant
 Ksp
represents the equilibrium associated with ionic
solids dissolving to form ions in aqueous solutions

sp stands for solubility product
 This
equilibrium constant indicates how soluble the
solid is in water
 But wait…isn’t this just about using the solubility
rules?

The solubility rules only qualitatively predict whether
an ionic compound has a low or high solubility in
water
Solubility Equilibria
 As
it turns out, insoluble products according to the
solubility rules are actually a bit soluble after all

“Soluble” is often defined as 3 grams of solid
dissolving in 100 mL
 Some
ionic compounds do not dissolve completely
and they exist in a dynamic equilibrium state

Some of the solid dissolves and some remains in the
solid form
Ksp – The Solubility Constant
•
•
No ions are
initially present
•
As the solid dissolves,
the concentration of
ions increase until
equilibrium
is
established
•
The solution is then
saturated – no more
solid forms
The rate of solid
dissolution is equal to
solid
formation
(precipitation)
–
EQUILIBRIUM!
Let’s Take a Look At What’s
Happening at the Molecular
Level…
Solubility Animation
The Solubility Product Expression

General form:
AaBb(s)  aAb+ (aq) + bBa- (aq)
Ksp = [Ab+]a [Ba-]b

Example:

Some notes on Ksp
AgCl (s)  Ag+ + ClKsp = [Ag+][Cl-] = 1.6 x 10-6 @ 25°C


Ksp is constant at a given temperature
The greater the Ksp, the more soluble the solid is in H2O
Solubility and Ksp

It is important to distinguish carefully between
solubility and Ksp:

Difference
is the units!
Solubility is the quantity of a substance in grams per
liter of solution that dissolves to form a saturated
solution

Molar solubility is the maximum number of moles of
the solute that dissolves to form a liter of saturated
solution


Solution is “saturated” and thus, equilibrium is
established between solid and hydrated ions
Ksp(solubility product) is the equilibrium constant for
the equilibrium between an ionic solid and its
saturated solution

Unitless number and is a measure of how much of
the solid dissolves to form a saturated solution
Solubility Product Calculations Determining Ksp Given Molar
Solubilities
 Plan:



Write the dissolution equation
Setup ICE table to calculate molar solubility
for ions
Substitute values into solubility product
constant expression and solve for Ksp
Practice!
#
1 on page
Solubility Product Calculations Determining Solubility Given Ksp
 Plan:



Write the dissolution equation
Setup ICE table to calculate molar solubility for
ions
Substitute values into solubility product constant
expression and solve for “S”
 “S”
is molar solubility and acts the same way as “X”
in all ICE tables!
Practice!
 #3
on page 143
Comparing Solubilities
 As
discussed, Ksp values provide valuable
information about a compound’s solubility
 Solubility comparisons can be made between
compounds, but you must take into account the
number of ions present in solution!
 If the salts being compared have the same
number of total ions, direct comparison of Ksp is
okay!
 If the salts have different number of ions, use
ICE table to solve for the molar solubility “S”!
Practice!
 #4
on page 143
Factors That Affect
Solubility
The Effect of Temperature on
Solubility
Solubility
(g/100ml water)
300
SO2 (g)
KCl (s)
200
Glycine (s)
NaBr (s)
KNO3 (s)
100
Sucrose (s)
0
0
20
40
60
80
Temperature (oC)
100
The Common-Ion Effect and
Solubility
Remember from the acid-base equilibria section,
a common ion is defined as the ion in a mixture of
ionic substances that is common to the formulas of
at least two
 So, how does the molar solubility of a solid
change when the solid is not dissolved in pure
water?



In other words, how does the molar solubility of the
solid change if it is dissolved in a solution
containing a second ionic substance containing a
common ion?
To answer this question, you must remember
LeChâtelier’s Principle!
The Common-Ion Effect

As with other equilibria we’ve discussed, adding a ‘common’ ion
will result in a shift of a solubility equilibrium
CaF2 (s) ↔ Ca2+(aq) + 2F- (aq)
KSP = [Ca2+] [F-]2


Adding either Ca2+or F- (from NaF, say) to our system will result in a
shift away from the increase (i.e. driving it to the left)
 CaF2 is formed and precipitation occurs
 The solubility of CaF2 DECREASES!
In general, the solubility of one salt is reduced by the presence of
another having a common ion

Ksp will not change; however, the concentrations of ions and mass of solid
will!
How to Solve Problems Involving
Solubility and the Common Ion
Effect
 Plan:



Write the dissolution equation
Setup ICE table to calculate molar solubility
for ions
Substitute values into solubility product
constant expression and solve for “S”
 Remember,
“S” is molar solubility!
Practice!
 #5
on page 143
Explanation of Previous
Example using LeChâtelier’s
Principle
 Added
silver ions and sulfate ions stress
the system and the system responds by
favoring the reverse reaction

Shifts to the left
 More

solid silver sulfate will be produced
Solubility is decreased
 Less
solid is dissolved
Effect of pH on Solubility


In general, the solubility of a compound containing a basic
anion (i.e. conjugate base of a weak acid) increases as pH
decreases (acidity increases)
Let’s look at an example:
2+
CaF2(s)


Ca (aq) + 2F (aq)
If the F- is removed, then the equilibrium shifts towards the
decrease and CaF2 dissolves
F- can be removed by adding a strong acid
-
+
F (aq) + H (aq)


-
HF(aq)
As pH decreases, [H+] increases and solubility increases
The effect of pH on solubility is dramatic
The Common Ion Effect and
Predicting Precipitation
Precipitation Conditions
 So
far, we have considered the dissolution
process of an ionic solid – the forward
reaction
 Now, let’s consider the conditions under
which a solid will precipitate out of a
solution - the reverse reaction!
Determining Whether
Precipitation will Occur
 To
answer this, calculate the reaction quotient, Q,
and compare to Ksp

Remember, Q is the same as the equilibrium-constant
expression for a reaction, but instead of only
equilibrium concentrations, you can use whatever
concentrations are being considered
 If
Q > Ksp, the reaction will proceed to the left,
towards the solid


Precipitation occurs until Q = Ksp
This is because solution is already supersaturated with
ions
 Supersaturated
solution means it is an unstable solution
in which more solute is dissolved than in a saturated
solution
Determining Whether
Precipitation will Occur

If Q = Ksp, equilibrium exists


The solution is saturated which is the highest
concentration the solution can have without
precipitating
If Q < Ksp, the reaction will proceed to the right,
towards the soluble ions


Solid dissolves until Q = Ksp
This is because you haven’t reached saturation yet

Saturation is a solution that contains the maximum
amount of dissolved solute at a given temperature in
the presence of undissolved solute
Predicting Whether a
Precipitate Forms
 Plan:



Write dissolution equation for solid that is formed
from the two solutions
Calculate initial moles of ions that are a part of
dissolution equation using the volumes and
molarities given in the problem
Calculate the concentration of all ions just after the
solutions are mixed
 Remember,

volumes are additive!
Calculate Q and compare to Ksp
 If
Q > Ksp, precipitate will form
Practice!
 #1
on page ________
Using Solubility for Selective
Precipitation
 Supposed
you have a mixture of metal ions in
solution
 We can selectively remove one of them based
on the solubilities of their salts!
 To do so, think about solubility rules!

Separation of ions in an aqueous solution can be
done using a reagent that forms a precipitate with
one or more (but not all) ions
Performing Selective
Precipitation of Ions
 Select
an anion that selectively precipitates
only one metal ion OR…
 Choose an anion that selectively precipitates
both metal ions but one precipitates at a
lower concentration



Based on the common ion effect!
The salt with the lower Ksp will precipitate first
Example:

Consider a mixture of Zn2+(aq) and Cu2+(aq)
CuS (Ksp= 610-37) is less soluble than ZnS(Ksp=210-25)
 CuS will be removed from solution before ZnS

Solving Selective Precipitation
Problems

Plan:




Determine what ion is necessary for precipitation of BOTH
species
Write dissolution equations and equilibrium expressions for
the two solids that will form
Calculate the concentration of each cation (or anion)
already present in solution
Calculate the minimum concentration of the added ion
that is needed to precipitate the solid using the
appropriate equilibrium expression


You will do this step twice
The salt that requires a lower concentration will
precipitate first!
Practice!
#
4 on page ________