Ka : Acid ionisation constant

Most acids and bases  are weak . A solution of a weak acid contains the acid and water as major species and a solution of weak base contains the base and water as major species.

In acid ionization equilibria : An acid reacts with water to produce hyrdronium ion (hydrogen ion) and the conjugate base ion.This process is called acid ionisation or acid dissociation.

For example:

H2O(l)  + HF (aq) <------------------>  H3O+ (Aq)  + F- (aq)

For a strong acid ,which ionizes completely in solution,the concentrations of ions are determined by the stoichiometry of the reaction from initial concentrations of acid.However for a weak acid such as acetic acid, the concentrations of  ions in solution are determined from the acid-ionisation constant also called acid dissociation constant, which is the equilibrium constant for the ionization of a weak acid.   

To find the acid ionization constant , write HA for the general  formula of a weak,monoprotic acid.The acid-ionization equilibrium in aqueous solution is

               HA  (aq)  + H2O(l) <-----------------> H3O+ (aq) +  A- (aq)

The corresponding equilibrium constant is:

    Ka = [H3O+] [A-] / [HA] 


Lets see a solved example problem:

Sample problem 1 : The pH of a 0.25 M aqueous solution is 1.92 . Calculate Ka for this weak acid.

Solution :

Step 1: The problem states that hydrogen fluoride is a weak acid,so the major species in the solution are H2O and HF molecules.In aqueous solution ,HF trasfers protons to H2O:

H2O(l)  + HF (aq) <------------------>  H3O(Aq)  + F(aq)

From the pH, calculate the concentration of H3O+ ions at equilibrium:

 [H3O+] = 10 -pH = 10 -1.92 = 1.2 x 10 -2 M

Step 2 : Now construct a concentration table.The key is to use the intial and equilibrium concentrations of H3O+ to complete the change row.

                 H2O(l)  + HF (aq) <------------------>  H3O(Aq)  + F(aq)

I                                  .25                                         0                     0

C                             -1.2 x 10 -2                     +  1.2 x 10-2       +1.2 x 10-2

E                                     .24                            1.2 x 10-2            1.2 x 10-2

Step3 : Now substitute numerical  values of the conetration and solve for Ka:

                             Ka = (1.2 x 10-2 ) / .24 = 6.0 x 10-4     


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