Solubility Equilibrium

Solubility is the ability of a substance to dissolve in water. The solubility is measured in

terms of concentration of an ion that is present in a smaller ratio in solution. On the other

hand, solubility equilibrium refers to the equilibrium between the dissolved salt (ions)

and undissolved salt that usually exists in a saturated solution or a solution of a sparingly

soluble salt. The word sparingly soluble salt refers to a salt that is partially (not

completely) soluble in water, as results of which, the equilibrium between dissolved ions

and undissolved salt is possible.

Solubility Product

To understand the concept of solubility product, consider the saturated solution of silver

chloride (AgCl), where the equilibrium exists between dissolved ions and undissolved

silver chloride according to the following equation.

Since this is an equilibrium reaction, we can write the equilibrium constant expression as

Ag Cl K

The concentrations of solids are either unknown or assumed to be constant. Hence we

combine [AgCl(s)] with K and label this constant as Ksp. Thus

[ ( )] [ ][ Kx AgCl s K Ag Cl sp ] + ? = =

The Ksp is called the solubility product constant or simply solubility product.

In general, the solubility product of a compound is the product of molar concentrations of

ions raised to the power of their respective stoichiometric coefficients in the equilibrium

reaction.

In the above example, writing the Ksp expression for AgCl is very simple because only

one mole of Ag+

are formed, then the things get more complex. Consider the following examples:

+ ?

[ ][

=

[ ()

AgCl s

]

]

ions and one mole of Cl-

ions are formed. If more than one mole of ions

• MgI2

2

• Ag2SO4

2

• Al(OH)3

2 MgI s Mg aq I aq () ( ) 2 ( ) ??? + ?

??? + 2 2 [ ][ K Mg I sp

2 4 4 Ag SO s Ag aq SO () 2 ( ) ??? + ?

??? + 2 2

3 Al OH s Al aq OH aq ( )() ( ) 3 ( ??? + ?

??? + 3 3 [ ][ K Al OH sp

3

• Ca3(PO4)2

2 3

Important

Before you make an attempt to write Ksp for a given compound, few things you

should know:

3 42 4 Ca PO s Ca aq PO aq ( )() 3 ( ) 2 ( ??? + ?

??? +

• How to break the compound into ions– identify the monatomic and

polyatomic ions

• Number of moles of each ion formed

• Charge on each ion

Write the equilibrium reaction with double arrows going in opposite directions with

proper phases first (s for solid, aq for solution) and then write Ksp.

The following table lists few Ksp values for low solubility hydroxides. You may refer to

any general chemistry textbook for more salts.

Compound Ksp Compound Ksp

[Al(OH)3] 1.8×10-33 Iron(II) hydroxide [Fe(II)(OH)2] 1.6×10-14

Aluminum hydroxide

[Ca(OH)2] 8.0×10-6 Iron (III) hydroxide [Fe(OH)3] 1.1×10-36

Calcium hydroxide

[Cr(OH)3] 3.0×10-29 Magnesium hydroxide

Chromium(III) hydroxide

[Cu(OH)2] 2.2×10-20 Zinc hydroxide [Zn(OH)2] 1.8×10-14

Copper(II) hydroxide

What do the Ksp values tell us?

Ksp values are derived from the concentrations of ions in equilibrium reactions; higher the

concentrations of ions, greater the Ksp. Higher the concentrations of ions means greater

the solubility of the salt. Therefore, the magnitude of Ksp directly indicates the solubility

of the salt in water. For example, comparing the Ksp of aluminum hydroxide and calcium

hydroxide, it is evident that calcium hydroxide is more soluble in water than aluminum

hydroxide.

Example

Arrange the following salts in order of increasing solubility.

(a) Aluminum hydroxide [Al (OH)3] (Ksp=1.8×10-33)

(b) Calcium hydroxide [Ca (OH)2] (Ksp=8.0×10-6)

(c) Chromium (III) hydroxide [Cr (OH)3] (Ksp = 3.0×10-29)

(d) Copper (II) hydroxide [Cu (OH)2] (Ksp=2.2×10-20)

Answer

Higher the Ksp means greater the solubility. Arranging the above salts from low Ksp to

high Ksp, we have

Al(OH)3 < Cr(OH)3 < Cu(OH)2 < Ca(OH)2

Solubility

In a simple term, solubility is the dissolving ability of a salt in a fixed amount of water. It

is measured by the concentration of an ion that is present in a smaller proportion (smaller

number of moles). The above discussed Ksp values are useful in comparing the

solubilities of group of salts, but not in assessing their actual solubilities. When we need

to calculate the actual solubility of the salt, we use either the molar solubility, which is

defined as the number moles of solute in 1 L of saturated solution (moles per liter), or

solubility that is defined as the number of grams of solute in 1 L saturated solution

(grams per liter). Thus you can see there are two ways to define solubility. However,

both are related through the molar mass.

Suppose, you want to convert solubility (g/L) to molar solubility (mol/L), you divide

solubility by molar mass (g/mol):

mol g g mol mol x x

L L molar mass g mol L g L = = =

Suppose, you want to convert molar solubility (mol/L) to solubility(g/L), you multiply

molar solubility by molar mass (g/mol):

g mol molar mass g mol mol g g x x

L L L = = mol L =

1

(/ )

(/ )

1

Relationship between Ksp and solubility

Solubility is the molar concentration and Ksp is the product of molar concentrations, and

hence both are related to one another. In the following, we develop relationships between

these two for various salts.

Let us illustrate the concept using a simple example of AgCl that dissociates into one

cation (Ag+

AgCl s Ag aq Cl aq () ( ) ( ) ??? + ?

The Ksp expression for this is

[ ][ K Ag Cl sp

Let s be the molar solubility(mol/L) of Ag+

ion because number of Ag+

Cl-

above equilibrium reaction. That is,

s = [Ag+

Substituting these into Ksp equation, we get

2 [ ][ ] ( ) ( ) K Ag Cl s x s sp

or sp s K =

Thus, the molar solubility is just square root of the Ksp.

We can extend similar procedure to develop the relationship between Ksp and s for more

complicated compounds:

Barium chloride (BaCl2)

2

s 2s

22 2 [ ][ ] ( )(2 ) 4 K Ba Cl s s sp

or s = (Ksp/4)1/3

) and one anion (Cl-

??? +

) at equilibrium:

+ ? = ]

ion. This will also be the molar solubility of

] and s = [Cl-

ions is equal to number of Cl-

]

+ ? = =

= s

2 BaCl s Ba aq Cl aq () ( ) 2 ( ) ??? + ?

??? +

+ ? = = 3 = s

Silver sulfide (Ag2S)

2

2s s

22 2 3 [ ] [ ] (2 ) ( ) 4 K Ag S s s sp

or s = (Ksp/4)1/3

Aluminum hydroxide (Al(OH)3)

2 Ag S s Ag aq S aq () 2 ( ) ( ) ??? + ?

??? +

+ ? = = = s

3 Al OH s Al aq OH aq ( )() ( ) 3 ( ??? + ?

s 3s

33 3 [ ][ ] ( )(3 ) 27 K Al OH s s sp

+ ? = = 4 = s

??? +

3

or s =(Ksp/27)1/4

Calcium phosphate (Ca3(PO4)2)

2 3

3s 2s

3 42 4 Ca PO s Ca aq PO aq ( )() 3 ( ) 2 ( ??? + ?

??? +

4 [ ] [ ] (3 ) (2 ) 108 K Ca PO s s sp

+ ? = =

23 32 3 2 5

or s=(Ksp/108)1/5

In each of the above salts, s is the concentration of the ion that is present in a smaller

mole ratio.

Calculating Ksp from solubility

Example

Calculate the Ksp of calcium sulfate (CaSO4) if the solubility is 0.56 g/L.

Answer

Solubility is the concentration of the ion present in a smaller proportion. Here both Ca2+

ion and SO4

2- ion are present in an equal proportion according to the following equation.

4 4 CaSO s Ca aq SO aq () ( ) ( ) ??? + ?

The Ksp is written as

4 [ ][ K Ca SO sp

In order to evaluate Ksp, we need the molar concentrations (mol/L), but the solubility is

given g/L. First, we need to convert g/L to mol/L:

0.56 1 4.1 10

Thus we have

[Ca2+] = [SO4

Now we can determine Ksp:

4 [ ][ ] (4.1 10 )(4.1 10 ) K Ca SO x x sp

= 1.7 x 10-5

2 2

??? +

2 2

+ ? = ]

g CaSO mol CaSO x x

4 4 3

1 ln 136.2

L so g CaSO

? =

4

2-] = 4.1×10-3

22 3

+? ? = = ?3

Calculating solubility from Ksp

Example

Calculate the molar solubility of Cu(OH)2 if Ksp = 2.2 x10-20.

Answer

First we write the Ksp expression for Cu(OH)2 as

2 Cu OH s Cu aq OH aq ( )() ( ) 2 ( ??? + ?

2 2 [ ][ K Cu OH sp

Let s be the molar solubility of Cu2+, i.e., [Cu2+]= s. Then 2s will be the concentration of

, i.e., [OH-

OH-

Ksp = 2.2x 10-20 = (s)(2s)2

= 2.2 x 10-20 / 4

3

s

Hence s = (2.2 x 10-20/4)1/3 =1.8 x 10-7 mol/L = 1.8 x 10-7 M

2

??? +

+ ? = ]

] = 2s. Then Ksp becomes

= 4s3

Predicting the Formation of Precipitate

Precipitate is the solid formed as a result of chemical reaction between two or more

solutions. Let say that you mix two solutions, how do you know whether the precipitate

forms or not? Before you can answer this question, you got to know what kind of

precipitate is going to form using the solubility rules. From the information given in the

problem, first calculate the molar concentrations of the ions responsible for the

precipitate. Next calculate the reaction quotient Q (this is exactly the same as Ksp except

it is based on the initial (given) concentrations) and compare it with Ksp.

If Q > Ksp precipitate will form otherwise no.

Example

Exactly 100 ml of 0.05 M AgNO3 are added to exactly 500 ml of 0.05 M HCl. Will a

precipitate form?

Answer

The ions present in solution are Ag+

only AgCl precipitate is possible. From the given information, first we calculate the

molar concentrations of Ag+

ml Ag xM x ml

100 3 [ ] 0.05 8.3 10 600

ml Cl x M x M

500 2 [ ] 0.05 4.2 10 600

Note the 600ml is the total volume of the solution. Then we calculate Q

Q = [Ag+

= 3.5×10-4

Therefore

Q ( = 3.5 x 10-4) > Ksp (=1.6×10-10))

The precipitate of AgCl will form.

and Cl-

, H+, and Cl-

, NO3

–

ions:

+ ? = = M

? ? = =

ml

][Cl-

] = (8.3 x 10-3)(4.2×10-2)

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