S SunChip Full Member 10+ Year Member Joined Jan 22, 2010 Messages 49 Reaction score 0 Jan 22, 2010 #1 Members don't 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! 😀
Members don't 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! 😀
A Akarat Full Member 10+ Year Member Joined Dec 31, 2009 Messages 69 Reaction score 19 Jan 23, 2010 #2 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% Upvote 0 Downvote
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%
S SunChip Full Member 10+ Year Member Joined Jan 22, 2010 Messages 49 Reaction score 0 Jan 23, 2010 #3 Thanks, Akarat! Upvote 0 Downvote