Download Solubility Product Constants We have been looking at how

Solubility Product Constants
We have been looking at how equilibrium constants can be used in chemical
reactions. The concept of equilibrium also applies to saturated solutions of
ionic solids. A saturated solution is one that is holding the maximum amount
of solute possible at a given temperature. Even though a solution is saturated,
there is still dissolving taking place. Solute particles are still dissolving and
dissolved particles are settling out of the solution at the same time. In a
saturated solution the rate of dissolving is equal to the rate of settling out and
an equilibrium is established. An equilibrium expression can be written for the
dissolving of an ionic solid just like one can be written for a reversible
chemical reaction. Below is an example of the equation and equilibrium
expression for the dissolving of an ionic solid.
AgCl(s) <--------------------------> Ag+(aq) + Cl-(aq)
[Ag+] [Cl-]
Keq = ------------------[AgCl]
In the equilibrium expression for the dissolving of an ionic solid there are
actually two constants, the Keq and the concentration of the solid. If you
change the number of moles of solid, the volume of the solid changes
proportionally, so the concentration is a constant. Chemists leave out the
concentration of the solid since it is a constant and we have a modified
expression for the dissolving of an ionic solid. It is the dissolved ion (products)
times each other and is called the solubility product expression. The modified
expression is shown below.
Ksp = [Ag+] [Cl-]
The concentrations of the products times each other equal a constant that is
called the solubility product constant (Ksp). A Ksp value gives you an idea of
the solubility of an ionic solid. Substances with a very small Ksp value are not
very soluble since the Ksp value was determined by multiplyiing the
concentrations of the dissolved ions. Substances with a large Ksp value are
very soluble. Ksp values are usually only listed for solids that are slightly to
almost insoluble. Most chemistry textbooks include tables of solubility product
constants (Ksp) measured at 25 °C. Ksp values are measured at a specific
temperature since solubility of most solids change with temperature. These
Ksp values can be used to compare the solubilities of ionic solids or to calculate
the concentrations of the dissolved ions in a saturated solution. The examples
below will show how Ksp values can be used.
Example 1
Rank the following solids in order of increasing solubility based on their Ksp
values.
CaSO4
BaSO4
Ksp = 2.4 x 10-5
Ksp = 1.5 x 10-9
Most Soluble
Least Soluble
Example 2
Calculate the Ag+ and Cl- concentrations in a saturated solution at 25 °C. The
Ksp for AgCl is 1.7 x 10-10. The equation for the dissolving of silver chloride
is:
AgCl(s) <---------------------------> Ag+(aq) + Cl-(aq)
According to the balanced equation, for every mole of silver chloride that
dissolves, we should get one mole of silver ions and one mole of chloride ions.
In other words, the silver and chloride ion concentrations should be the same.
We can assign these concentrations X since they will both be the same. Next
we set up the solubility product constant expression and solve for X.
Ksp = 1.7 x 10-10 = [Ag+] [Cl-]
1.7 x 10-10 = [X] [X]
1.7 x 10-10 = X2
X = (1.7 x 10-10)1/2 = 1.3 x 10-5 M = [Ag+] = [Cl-]
Example 3
Calculate the Mg2+ and OH- concentrations in a saturated solution at 25 °C.
The Ksp for Mg(OH)2 is 8.8 x 10-12. The first step in solving this problem is to
write out the equation for the dissolving of magnesium hydroxide.
Mg(OH)2(s) <-----------------------> Mg2+(aq) + 2OH-(aq)
For every one mole of magnesium hydroxide that dissolves, we will get one
mole of Mg2+ and two moles of OH-. Therefore, for every X moles of Mg2+
that form we will get 2X moles of OH-. Now set up the solubility product
expression and solve for X.
Ksp = 8.8 x 10-12 = [Mg2+] [OH-]2
8.8 x 10-12 = [X] [2X]2
8.8 x 10-12 = 4X3
X = (8.8 x 10-12 / 4 )1/3 = 1.3 x 10-4
Mg2+ = 1.3 x 10-4 M
OH- = 2.6 x 10-4 M (2X)
Example 4
What is the Ksp for barium sulfate if its molar solubility is 3.9 x 10-5 moles per
liter at 25 °C? The first step in solving this problem is to write out the equation
for the dissolving of barium sulfate.
BaSO4(s) <---------------------> Ba2+(aq) + SO42-(aq)
Since everything in the equation is a 1:1 ratio, for every 3.9 x 10-5 moles of
barium sulfate that dissolve per liter we will get equal amounts of barium ions
and sulfate ions. Now we set up the solubility product expression and plug in
our concentrations.
Ksp = [Ba2+] [SO42-]
Ksp = [3.9 x 10-5] [3.9 x 10-5]
Ksp = 1.5 x 10-9