Ks and Chemistry


F.W. Boyle, Jr., Ph.D.
(C) - Copyright, 1991

The concept of a constant, designated as a K with a descriptive subscript, is widely used in chemistry. In all it's forms, the constant, K, is derived from the equilibrium reaction. The basic form of K expression is shown below:

                AB + CD <---> A + BC + D

                     [A] [BC] [D]
                K = ---------------
                       [AB] [CD]

As you can see the numerator (top of the fraction) is the mathematical product of the chemical concentrations of the products of the reaction. The denominator (bottom of the fraction) is the mathematical product of the chemical concentrations of the reactants of the reaction. Two other ways of remembering the proper positions are:


      Products               concentrations from the right side
K = ------------   and   K = ----------------------------------
      Reactants              concentrations from the left side

Additionally, each concentration must be raised to the power of its coefficient (the number in front of the chemical name). AND THOSE CHEMICALS WHICH APPEAR IN THE REACTION AS SOLIDS OR LIQUIDS, DO NOT APPEAR IN THE K EXPRESSION.

For example:

          Ca(OH)2(s) + 2H+(aq) <---> Ca2+(aq) + 2H2O(l)

                              [Ca2+]
                       Kc = ---------
                               [H+]2

The following K's will be discussed:

Kc - a CONCENTRATION constant. All terms are expressed in [ ] s.

Kp - a PRESSURE constant. All terms are expressed in pressures ( can be torr, atm, Pa)

Kw - a special concentration constant for WATER.

Kw = [H3O+] [OH-] = [H+] [OH-] = 1 x 10-14

Ka - an concentration constant for an ACID. All terms are expressed in [ ]s.

Kb - a concentration constant for a BASE. ALl terms are expressed in [ ]s.

Ksp - a concentration constant for the SOLUBILITY PRODUCT of a solid dissolved in pure water. All terms are expressed in [ ]s.

Kcomp - a concentration constant for complexes which can form with ions which are in solution.

MOLAR SOLUBILITY

The molar solubility is the solubility of the solid in moles per liter. The molar solubility can also be described as the amount of the solid in moles which dissolves in aqueous solution. A small "s" is used to indicate the molar solubility.

Calculating the molar solubility
Example 1:

A simple salt dissolving in water: PbCrO4(s) Ksp = 2.8 x 10-13 M2

The reaction is:

PbCrO4(s) <---> Pb2+(aq) + CrO42-(aq)

Ksp = [Pb2+] [CrO42-]

In pure water 1 mole of PbCrO4(s) dissolves and releases 1 mole of Pb2+(aq) and 1 mole of CrO42-(aq).

Let "s" be the molar solubility of PbCrO4(s). Since the dissolution is 1 to 1, "s" also equals the concentrations of Pb2+ and CrO42-.

Substituting:


                  Ksp = (s)(s) = 2.8 x 10-13 M2

s2 = 2.8 x 10-13 M2

s = 5.3 x 10-7 M

Therefore at equilibrium:


                 [Pb2+] = 5.3 x 10-7 M = [CrO42-]

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Note: For a compound (chemical) of the generic formula XY

                             Ksp = s2
                                         
                                      
                           s =  Ksp1/2
                                     
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Example 2:

A more complex chemical: Ca(OH)2(s) Ksp = 5.5 x 10-6 M3

1. Write the reaction.

               Ca(OH)2(s) <---> Ca2+(aq) + 2OH-(aq)

2. Write the Ksp expression.

                       Ksp = [Ca2+] [OH-]2

3. Let "s" equal the solubility of Ca(OH)2(s).

4. Therefore [Ca2+] = s and, from the stoichiometry, [OH-] = 2s. Substituting into the Ksp expression:

                 Ksp = (s) (2s)2 = 5.5 x 10-6 M3

                       (s) (4s2) = 5.5 x 10-6 M3

                             4s3 = 5.5 x 10-6 M3

                              s3 = 1.4 x 10-6 M3

                               s = 1.1 x 10-2 M
So [Ca2+] = 1.1 x 10-2 M

and [OH-] = 2(1.1 x 10-2 M) = 2.2 x 10-2 M

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Note: For a compound (chemical) of the generic formula XY2 or X2Y


                             Ksp = 4s3

                                Ksp  1/3
                           s = -----
                                 4  


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Example 3:

Use Al(OH)3(s) which has a Ksp = 1.3 x 10-33 M4

1. Write the chemical reaction.

Al(OH)3(s) <---> Al3+(aq) + 3OH-(aq)

2. Write the Ksp expression.

Ksp = [Al3+] [OH-]3

3. Let "s" equal the molar solubility of Al(OH)3(s).

From the reaction stoichiometry:

[Al3+] = s and [OH-] = 3s

Substitute into the Ksp expression:


                 Ksp = (s)(3s)3 = 1.3 x 10-33 M4

                      (s)(27s3) = 1.3 x 10-33 M4

                          27s4  = 1.3 x 10-33 M4

                            s4 = 4.8 x 10-35 M4

                             s = 2.6 x 10-9 M

So   [Al3+] = 2.6 x 10-9 M

and

[OH-] = 3(2.6 x 10-9 M) = 7.8 x 10-9 M

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Note: For a compound (chemical) of the generic formula XY3 or X3Y

                             Ksp = 27s4

                               
                                Ksp  1/4
                           s = -----
                                 27 


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If you know the molar solubility, you can calculate the Ksp of the solid.

Example 1:

The molar solubility, s, of PbSO4(s) is 1.3 x 10-4 M.

Since PbSO4(s) is a compound of the generic formula, XY, we would use the Ksp relationship shown in the note for the first example.


                       Ksp = s2

                       Ksp = (1.3 x 10-4 M)2

                           = 1.6 x 10-8 M2
Example 2:

The molar solubility, s, of CaF2(s) is 1.1 x 10-3 M.

CaF2(s) is of the generic formula XY2 so we would use the Ksp expression shown in the note section of Example 2.


                      Ksp = 4s3

                      Ksp = 4(1.1 x 10-3 M)3

                          = 5.3 x 10-9 M3

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An additional note:

For a compound of the generic formula X2Y3 or X3Y2:

                           Ksp = 108s5 

                                Ksp  1/5
                           s = -----
                                108 



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