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Find the percent ionization of a 0. 337 m hf solution. The ka for hf is 3. 5 × 10^-4.

Sagot :

Percent ionization = 3.17%

We utilize the supplied acid equilibrium constant (Ka) to calculate the percentage of the acid that is ionized. It is the ratio of the acid and dissociated ion equilibrium concentrations. The HF acid would dissociate in the manner described below:

HF = H+ + F-

The following is how the acid equilibrium constant might be written:

Ka = [H+][F-] / [HF] = 3.5 x 10-4

We utilize the ICE table to determine the equilibrium concentrations.

        HF             H+              F-

I      0.337           0                 0

C      -x              +x               +x

---------------------------------------------

E    0.337-x        x                   x

3.5 x 10-4 = [H+][F-] / [HF]

3.5 x 10-4 = [x][x] / [0.337-x]

Solving for x,

x = 0.01069 = [H+] = [F-]

percent ionization = 0.01069 / 0.337 x 100  = 3.17%

percent ionization = 3.17%

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