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%

it have destructive character

the correct option is c.

isotopes refers to different forms of the same element with the same atomic number but different mass number as a result of difference in their neutron numbers. isotopes have different relative atomic mass but share the same chemical properties.