Review of Ksp
This document reviews the concepts and calculations of Ksp.
Previously, we learned the terms “soluble” and “insoluble” to mean a salt that dissolves in
water and one that does NOT dissolve in water, respectively. This leads to an idea that all
“soluble” salts dissolve readily and completely and all “insoluble” salts do not dissolve at all.
In truth, the picture is not so black and white. Some salts do dissolve completely; see image
(a) below. Some salts can dissolve to some extent. Some salts can only dissolve to such a
small extent that it might as well be none at all (image (c)). Insoluble salts would be more
accurately described as “relatively insoluble” salts.
For those salts we label as “insoluble”, the small amount of solid that IS able to dissolve into
solution quickly reaches a maximum, then the ions in solution achieve equilibrium with the
solid. For example, silver chloride is so insoluble in water that a saturated solution1 contains
only 1.3 x 10-5 M of each ion. At that point, the small amount of ions in solution are now in
equilibrium with the solid at the bottom:
AgCl(s)  Ag +(aq) + Cl-(aq)
The equilibrium constant for the dissolution of an ionic compound is referred to as the
solubility product constant, Ksp. Said another way, Ksp refers to the equilibrium between a
relatively insoluble solid and its ions in solution. The Ksp expression for the reaction shown
above is
Ksp = [Ag+][Cl-]
(Notice, as discussed previously, the solid does not appear in the expression.)
1
saturated solution – a solution that contains the maximum amount of solute particles for the given
temperature
Ksp is an equilibrium constant, specifically referring to the dissolution of a relatively insoluble
salt. As an equilibrium constant, the numerical value reflects the position of the equilibrium.
And the larger the value, the further toward products rests the equilibrium. In the case of
solubility equilibria, the products in the reaction are the ions that DID dissolve into solution.
A larger Ksp value, then, represents a system that has more dissolved ions in solution, a
(relatively) more soluble salt. Ksp values are always small values, but one can compare the
Ksp values of multiple compounds to determine their solubility relative to one another.
The value of Ksp indicates the amount to which the solid dissociates in
water. The larger the Ksp value, the more soluble the salt.
Consider the following compounds and their Ksp values:
Compound
CaCrO4
CoCO3
PbS
Ag2S
Ksp
7.1 x 10-4
1.0 x 10-10
9.04 x 10-29
6 x 10-51
At identical conditions, more calcium chromate will dissolve into solution than cobalt
carbonate, and much more than lead (II) sulfide or silver sulfide. While lead (II) sulfide does
not dissolve appreciably, it still dissolves noticeably more than silver sulfide.
The numerical value of Ksp is NOT the molarity of ions in solution at
equilibrium.
The amount of ions present in solution expressed as moles per liter is
called the molar solubility.
It is imperative that you understand the difference between the two numerical values
involved (Ksp and molar solubility). Consider lead (II) iodide:
PbI2  Pb2+ + 2I-
Ksp = 9.8 x 10-9
molar solubility = 1.3 x 10-3 M
The Ksp value tells us where the equilibrium rests. How far toward the products was the
system able to move? The molar solubility tells us the concentration of each ion in a solution
that is at equilibrium. The lead concentration is 1.3 x 10-3 M and the iodide concentration is
2.6 x 10-3 M (1.3 x 10-3 M for each iodide; there’s two iodides formed).
Given either value, one can calculate the other using an ICE Table. Consider the following
example problems.
Example 1:
The molar solubility of silver phosphate in pure water is 4.26 x 10-5 M. Calculate its Ksp
value.
Solution:
1.
Write the solubility reaction and Ksp expression.
Ag3PO4  3Ag+ + PO43-
Ksp = [Ag+]3[PO43-]
NOTE: You must be skilled at nomenclature to reliably write the correct compound
formula. On the product side, don’t forget the coefficient representing the correct number
of ions in solution. Understand that polyatomic ions do not internally break apart in
water (PO43- does not become P + 4 O ).
2. Molar solubility was given in the question. This is the concentration of each ion present
in solution at equilibrium. Use this to complete an ICE Table.
NOTE: When an ion has a coefficient, molar solubility is the concentration for each ONE
of them, not all of them total. Here, silver phosphate ionizes to produce THREE silver
ions, each with the molar solubility listed, and one phosphate ion with the molar
solubility listed.
4.26 x 10-5 M
4.26 x 10-5 M
4.26 x 10-5 M
4.26 x 10-5 M
Think of the molar solubility as the number for the change step of the ICE table, plus any
associated coefficient.
Ag3PO4
I
C
E

3Ag+
+
0
+(3 x 4.26 x 10-5)
1.28 x 10-4
PO430
+4.26 x 10-5
4.26 x 10-5
3. Substitute these numbers into the Ksp expression from above and solve.
Ksp =
[Ag+]3
[PO43-]
= (1.28 x 10-4)3(4.26 x 10-5)
= 8.89 x 10-17
Example 2:
Calculate the molar solubility of copper (II) hydroxide in pure water.
Ksp = 2.2 x 10-20
Solution:
1. Write the solubility reaction and Ksp expression.
Cu(OH)2  Cu2+ + 2OH-
Ksp = [Cu2+][OH-]2
2. Ksp was given, not the molar solubility. We know no equilibrium numbers to write into
our ICE Table. Complete an ICE Table with a variable.
Cu(OH)2  Cu2+ + 2OHI
0
0
C
+s
+2s
E
s
2s
3. Substitute these equilibrium expressions into the Ksp expression to solve for s.
Ksp
= [Cu2+][OH-]2
2.2 x 10-20 = (s)
(2s)2 = 4s3
s = 1.8 x 10-7 M
This, then, is the molar solubility for copper(II) hydroxide in pure water. That is the
concentration for each individual ion in the saturated solution. That is the concentration
of copper (II) ions in solution. That is the concentration of each one (of two) of the
hydroxide ions in solution.
Molar solubility refers to the amount of dissolved ions in moles per liter. Solubility
represents the amount of dissolved ions in the unit grams per liter.
Only molar solubility can be used in an ICE Table to calculate Ksp.
Ksp is an equilibrium constant. It has only one value at a given temperature. The molar
solubility, however, depends on the solution conditions. The presence of a common ion, for
example, lowers the molar solubility value. A common ion does NOT change the Ksp value
for the compound. At a given temperature, the equilibrium will always come to rest in the
same place. The individual concentrations of the ions present at equilibrium can be different
values depending on the solution conditions.
Example 3:
Calculate the molar solubility of barium sulfate in (a) pure water, and in (b) a 0.50M solution
of potassium sulfate.
Ksp = 1.07 x 10-10
Solution:
Part (a):
1. Write the solubility reaction and Ksp expression.
BaSO4  Ba2+ + SO42-
Ksp = [Ba2+][SO42-]
2. Complete an ICE Table with a variable.
BaSO4  Ba2+ + SO42I
0
0
C
+s
+s
E
s
s
3. Substitute these equilibrium expressions into the Ksp expression to solve for s.
Ksp
= [Ba2+][SO42-]
1.07 x 10-10 = (s)
(s)
-5
s = 1.03 x 10 M
Part (b):
4. Use the same reaction and Ksp expression, but set up a new ICE Table. This time, sulfate
ions are present initially.
BaSO4  Ba2+ + SO42I
0
0.50
C
+s
+s .
E
s
0.50 +s
5. Can you ignore the “+s” portion of the SO42- expression? Is the number s is added to
more than three decimal places off from the Ksp value? 0.50 ( or 5.0 x 10-1) vs. 1.07 x 10-10.
YES, you can ignore the “+s”.
6. Substitute into the Ksp expression and solve for s.
Ksp
= [Ba2+][SO42-]
1.07 x 10-10 = (s) (0.50)
s = 2.14 x 10-10 M
Notice the difference in the two answers from Example 3. In pure water, the molar solubility
of barium sulfate is 1.03 x 10-5 M. This is the molarity of each ion in solution at equilibrium.
When a common ion was present in part (b), the molar solubility dropped to 2.14 x 10-10 M.
This is a much lower amount of ions dissolved in solution.
In both cases, the Ksp value for barium sulfate is the same; both equilibria come to rest in the
same place. In the second scenario, the (relatively) high existing sulfate concentration
prevents barium sulfate from dissolving in solution to any great extent.
The presence of a common ion lowers the molar solubility value. A
common ion does NOT change the Ksp value for the compound,
though.