Kb : Base ionisation constant


When Ammonia dissolves in water,the resulting solution has a pH greater than 7.00 , indicating that the solution is basic .In other words,the concentration of hydrogxide ions in aqueous ammonia is greater than a pure water.Ammonia molecules accept protons from water molecules,generating cations and hydroxide anions;


NH3 (aq) + H2O (aq)  <------------>  NH4+(aq)   +  OH- (aq)


Ammonia is an example of a weak base.A weak base generates hydroxide ions by accepting protons from water but reaches equilibrium when only a fraction of its molecules have done so .The equilibrium constant for this type of equilibrium is designated Kb :


                  Kb = [NH4+] eq [OH-]eq / [NH3]eq

Lets see a solved example:

Sample probelm : Ammonia has (Kb= 1.8x 10^-5) .What is the pH of 0.25M aqueous solution of ammonia?


Step 1 :Determine pH for which we need to know the equilibrium concentration of either H3O+ or OH-.The major species present in aqueous ammonia are molecules of NH3 and H2O.Both of these compounds produce hydroxide ions as minor species in solution:

                        NH3 (aq) + H2O (l) <-------------> NH4(aq)  + OH- (aq)   Kb = 1.8 x 10-5

                 2H2O (l) <---------> H3O+ (aq) + OH- (aq)      Kw = 1.0 x 10-14

Step 2: These two equilibria are linked by the fact that the solution can have only one hydroxide ion concetration.Both equilibria must be satisfied,but which reaction sholud we use to calculate [OH-]eq? Notice that both equilibrium constants are much less than 1,but Kb for ammonia is more than a billion times larges than Kw.This indicates that almost all the hydroxide ions in solution come from the ammonia equilibrium.Thus, ammonia reaction is appropriate choice for the calculation.

Step 3 : Determine [OH-]eq by setting up a concentration table , solving the equilibrium expression for the unknown and finding [OH-]eq .We know the initial concentrations but must identify x as the change in concentrations needed to reach equilibrium.

Reaction          NH3 (aq) + H2O (l) <-------------> NH4(aq)  + OH- (aq) 

I                        0.25                                                    0                      0

C                      - x                                                       +x                    +x

E                   0.25 -x                                                  x                        x


                   Kb = 1.8 x 10^-5  = x2 / 0.25 - x = x2 / .25

                      x 2 = 4.5 x 10 -6   so  x =  2.1 x 10-3


The table shows that [OH-] eq  = x  so [OH-] eq = 2.1 x 10 -3 M .

Step 4 : complete the the problem using equation pOH = -log [OH-] = -log (2.1 x 10-3) = 2.68

pH + pOH = 14.00 so pH =14.00 - 2.68 = 11.32