Answered

Find trusted answers to your questions with the help of IDNLearn.com's knowledgeable community. Get accurate answers to your questions from our community of experts who are always ready to provide timely and relevant solutions.

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

We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. For dependable and accurate answers, visit IDNLearn.com. Thanks for visiting, and see you next time for more helpful information.