Acertain weak acid, ha, has a ka value of 8.4×10−7. calculate the percent ionization of ha in a 0.10 m solution

To determine the percent ionization of the acid given, we make use of the acid equilibrium constant (Ka) given. It is the ratio of the equilibrium concentrations of the dissociated ions and the acid. The dissociation reaction of the HF acid would be as follows:

HA = H+ + A-

The acid equilibrum constant would be expressed as follows:

Ka = [H+][A-] / [HA] = 8.4 x 10^-7

To determine the equilibrium concentrations we use the ICE table,
         HF             H+              A-
I      0.10           0                 0
C      -x              +x               +x

E    0.10-x        x                   x 

8.4 x 10^-7= [H+][A-] / [HA] 
8.4 x 10^-7 = [x][x] / [0.10-x] 

Solving for x,

x = 0.0002894 = [H+] = [A-]

percent ionization = 0.0002894 / 0.10 x 100  = 0.289%

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.