Question

What is the percent ionization of a monoprotic weak acid solution that is 0.180 M? The acid-dissociation (or ionization) constant, Ka, of this acid is 8.93 × 10-8.

Answer 1

let the monoprotic acid be represented as HA

Its ionization reaction can be writen as HA+ H2O-------> A-+ H3O+

Ka= [A-] [H3O+]/[HA]

preparing the ICE table

component                                   initial concentration                                  change                  equilibrium

HA                                                       0.18                                                        -x                         0.18-x

A-                                                          0                                                            x                             x

H3O+                                                    0                                                            x                             x

Ka= x²/(0.18-x)= 8.93*10^-8, when solved for x using excel, x= 12.7*10-5

so % ionization =100*x/initial concentratino of HA= 100*12.7*10^-5/0.18=0.071%