Solubility Product Constant
Ksp Chapter 17
I. BaSO4(s) 
2+
Ba (aq)+
2SO4 (aq)
Barium Sulfate has a certain solubility in
water. At some point when enough is added
the solution will become saturated, and some
solid will remain.
 Ksp = [Ba2+] [SO42-]

solubility product
constant
Even though the concentration
the solid is omitted for the equilibrium
to exist some must exist.

II. Solubility and Ksp
The smaller the value of Ksp the lower the
solubility
 Solubility – the amount of a substance that
dissolves when producing a saturated
solution is often expressed in grams of solute
/liter of solution at saturation point.
 Solubility product constant (Ksp) – describes
the concentrations) of dissolved ions, or
substances at saturated equilibrium and is a
unit less expression.

Solubility and Ksp

Solubility of a substance may change as the
concentrations of various ions change (
including H+), however the value of Ksp is
unique for a given solute at a specific
temperature.
III. Calculating Ksp from the
solubility
Ex. A liter of a solution saturated at 25OC with
calcium oxalate, is evaporated to dryness,
giving a 0.0061 g residue. Calculate the
solubility product constant for this salt.
 CaC2O4(s)  Ca+2(aq) + C2O42-(aq)
0.0061g CaC2O4 (1 mole/128 g) = 4.8x10-5 mol
1mol CaC2O4 = 1 mol Ca+2 = 1 mol C2O42Ksp = [Ca+2] [C2O42-] = [4.8x10-5]2 = 2.3x10-9

More complicated example
By experiment, it is found that 1.2 X10 -3
moles of lead II iodide dissolves in 1 liter of
aqueous solution at 25oC. What is the Ksp of
PbI2 at this temperature?
 PbI2(s)  Pb2+(aq) + 2I(Aq)
 Ksp = [Pb2+] [ I ] 2 [1.2 X10 -3] [1.2 X10 -3]2

IV. Calculating Solubility from Ksp

Calculate the solubility
(in grams per liter) of
calcium fluoride in
water Ksp = 3.4 X 10-11

Calcium fluoride is
commonly used as a
window material for
both infrared and
ultraviolet wavelengths,
since it is transparent
in these regions (about
0.15 µm to 9 µm) and
exhibits extremely
weak birefringence.
Calculating Solubility from Ksp
CaF2(s)  Ca+2(aq) + 2FSt
0
0
Ch
+x
+2X
Eq
+x
+2X
Ksp 3.9 X 10-11 = [Ca+2] [2F-]2
3.9 X 10-11 = [x] [2x]2
3.9 X 10-11 = 4x3
x = 2.1 x 10-4(78.1g/mol) = 1.6X10-2g/liter

V. Factors Effecting Solubility

There are five main factors that control
solubility of a solute.
(1) Temperature
(2) Nature of solute or solvent
(3) Pressure
(4) common ion effect
(5) pH
Temperature
Generally solubility increases with the rise in
temperature and decreases with the fall of
temperature but not in all cases.
 In endothermic process solubility increases
with the increase in temperature and vice
versa.
 In exothermic process solubility decrease
with the increase in temperature.
 Gases are more soluble in cold solvent than
in hot solvent.

NATURE OF SOLUTE AND
SOLVENT
A polar solute dissolved in polar solvent.
 Solubility of a non-polar solute in a solvent is
large.
 A polar solute has low solubility or insoluble
in a non-polar solvent.
 In general like dissolves like

EFFECT OF PRESSURE
The effect of pressure is observed only in the
case of gases.
 An increase in pressure increases of
solubility of a gas in a liquid.
 For example carbon dioxide is filled in cold
drink bottles (such as coca cola, Pepsi 7up
etc.) under pressure.

Common Ion Effect
The presence of other solutes can also
influence solubility – although they do not
effect Ksp
 CaF2(s)  Ca+2(aq) + 2Faddition of calcium or fluoride ion will shift the
equilibrium left favoring solid formation and
decreasing solubility of the compound

Cal the sol of CaF2 at 25oc in
a) 0.01 M Ca(NO3)2 b) 0.01M NaF
Ksp 3.4 X 10-11 = [Ca+2] [2F-]2
 CaF2(s)  Ca+2(aq) + 2FSt
0
0
Ch
+x
+2X
Eq
0.10+x +2X
3.9 X 10-11 = [0.010+x] [2X]2
assume that X is very small compared to 0.01
so

3.9 X 10-11 = (0.010) 4x2
3.9 X 10-11/ (0.010) 4 = x2 x = 3.1x10-5 mol/L

In pure water the molar sol of CaF2 was 2.0 x
10-4 mol/L so we see the sol has decreased
with the addition of a common ion.
Learning check calc part b). Answer 3.9 x 10-7
mol/L
b) 0.01M NaF
CaF2(s)  Ca+2(aq) + 2FSt
0
0
Ch
+x
+2X
Eq
+x
0.1+2X
 Assume x is small ignore 2X
 3.9 X 10-11 = x(0.010) 2
 X = 3.9 X 10-11 / (0.010) 2
 X = 3.9 x 10-7 seems correct because F ion
conc. is squared and has more of an effect
