# A certain weak acid, ha, has a ka value of 8.4×10−7. Calculate the percent ionization of ha in a 0.10 m solution

## Question

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

## Answer

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%

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