Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

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%

To learn more about percent ionization visit:

https://brainly.com/question/14225136

#SPJ4