Percent Dissociation

[SunChip]

---

Members do not see this ad.

How do you calculate the % dissocation of HF (ka = 3.5 x 10-4) in 0.050 M HF? I can't get the right answer (it's 8.0 %)!! I get 2%. What am I doing wrong?

Thanks in advance! 😀

 

[Akarat]

Bad formatting is bad.

R: HF <---> H+ + F-
I: 0.05 0 0
C: -x +x +x
E: 0.05 - x +x +x

Since Ka < 1.0 x 10^-3, I can ignore the "-x" from "0.05 - x".

Ka = [H+][F-]/[HF]
3.5 x 10^-4 = (x^2)/(0.05)
x^2 = 1.75 x 10^-5, approximately 0.16 x 10^-4

(0.4)^2 = 0.16

=> x is approximately 0.4 x 10^-2 or 4 x 10^-3

[(4 x 10^-3)/(0.05)] x 100% = 8%

 

[SunChip]

Thanks, Akarat!