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Relating Solubility and Ksp
The relation between solubility and Ksp is quite important when describing the solubility
of slightly ionic compounds. I mention ionic compounds because most ionic compounds
are soluble in water or other solvents as to where molecular compounds are not soluble.
However, in this case, we are discussing the ionic compounds that are difficult to
dissolve; they are known as slightly soluble or almost insoluble. Therefore solubility
product constants (Ksp) are given to those solutes, and by using these constants we can
find the molar solubility of the compounds that make the solute. This relation also allows
us to find the Ksp of a slightly soluble solute from its solubility.
Introduction
1. What is Solubility?
Solubility is the ability of a substance to dissolve. In the process of dissolving something, there is
a solute and a solvent. The solute is the substance that is being dissolved and the solvent is the
substance that is doing the dissolving. For example, sugar is a solute and water is a solvent.
With this concept, we can understand solubility on a more specific level. Solubility is known to
be the maximum amount of solute that can be dissolved in a solvent at equilibrium. Equilibrium
is when the concentrations of products and reactant have become equal after the reaction has
taken place.
2. What is Ksp?
Ksp (Solubility product constant) is the equilibrium between a solid and its respective ions in a
solution. The value of the constant identifies the degree of which the compound can dissociate in
water. For example the higher the Ksp the more soluble the compound is. Ksp is defined in terms
of activity rather than concentration because it is a measure of a concentration that depends on
certain conditions such as temperature, pressure, and composition. It is influenced by
surroundings.
3. What are Solubility Product Constants used for?
Ksp is used to describe the saturated solution of ionic compounds. (A saturated solution is when
there is a state of equilibrium between the dissolved, dissociated, undissolved solid, and the ionic
compound).
For example let's take the compound Barium Carbonate BaCO3 (this is an ionic compound that is
not very soluble)
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BaCO3(s)<=> Ba2+ (aq)+CO32- (aq)
First we can write down the equilibrium constant expression:
Kc=[Ba2+][CO32-]/[BaCO3]
By looking at this expression we can see that the molar expression is 1 and considering that the
concentrations of these ions are small, we are able to say that the activity of the ions are equal to
their molar concentrations which equals 1. Now we can see that Ksp is equal to the product of
the ions.
Ksp= [Ba2+][CO32-] = 5.1 x 10-9
The value of this constant can be found in Table 18.1 of the General Chemistry Principles &
Modern Applications or in Appendix D of the same text book.
4. How does Ksp and solubility relate?
The relation between solubility and the solubility product constants is that one can be used to
find the other. In other words, there is a relationship between the solute's molarity and the
solubility of the ions because Ksp is literally the product of the solubility of each ion in moles per
liter.
For example , if we wanted to find the Ksp of a slightly soluble compound from its solubility, we
would first have to make sure to correctly convert the (mass/ some volume) to (mols/liters) to
find the molarity of each ion. After we find the molarity of each ion (molar solubility) we can
plug these numbers into the Ksp formula which is the product of the solubility of each ion.
Here is an example of what is explained above: Let's say that the aqeous solubility is given to us
for compound PbI2 is 0.54 grams/100 ml at 25 degrees celsius and we want to find the Ksp of
PbI2 at 250C.
PbI2(s)<=> Pb2+ + 2 Ia. Take 0.54 grams of PbI2 to mols
0.54 grams x [1mol of PbI2/461.0 grams (molecular weight)]
=0.001167137 mols of PbI2
b. Convert ml to L 100ml/1000L= .100 L
c. Find Molarity
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0.001167137 mols/.100 L = 0.011714M of PbI2
d. Now find the molarity of each ion by using the stoichiometric ratio (remember there are two I)
[Pb2+]= 0.011714M/ 1L x 1 mol of Pb/ 1 mol PbI2 = 0.011714M
[I-] = 0.011714M/1L x 2 mol of I-/1 mol of PbI2 = 0.023427 M
e. Finally plug in your molarity to find Ksp
Ksp= [Pb2+][I-]2=(0.011714 M)(0.023427 M) 2= 6.4 x 10-6
*Because of this relation we can also solve for the molar solubility of the ionic compounds when
the Ksp is given to us. The process will be like working backwards from Ksp to the molarity of the
ionic compound.*
For example if we are given a Ksp at 25 0C that is 8.5 x 10-17 for the compound AgI how would
we find the molar solubility?
AgI (s) <=> Ag+(aq) + I- (aq)
a. Lets look at our goal, and substitute the letter "g" for number of mols
Ksp= [Ag2+][I-]
Ksp= g2= 8.5 x 10-17
b. Solve for "g"
g2= 8.5 x 10-17
g= (8.5 x 10-17)1/2
g= 9.0 x 10-9
The molar solubility of AgI is 9.0 x 10-9
References
1. General Chemistry, Principles & Modern Applications Ninth Edition: Petrucci, Harwood,
Herring, Madura
Outside Links
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

http://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Writing-Ksp-expression.html
http://en.wikipedia.org/wiki/Solubility_equilibrium
Contributors

Gianna Navarro
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