1 / 4

% IONIZATION

% IONIZATION. 1.) What is the % ionization of a 0.010 M HCOOH, formic acid, solution? (K a = 1.9 x 10 -4 ) HCOOH  H + + HCOO - I 0.010 0 0 C -x +x +x E 0.010 – x x x Ka = x 2 = 1.9 x 10 -4 0.010 - x. -x. x 2 = 1.9 x 10 -6 x = 1.4 x 10 -3

ishi
Download Presentation

% IONIZATION

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. % IONIZATION • 1.) What is the % ionization of a 0.010 M HCOOH, formic acid, solution? (Ka = 1.9 x 10-4) HCOOH  H+ + HCOO- I 0.010 0 0 C -x +x +x E 0.010 – x x x Ka = x2 = 1.9 x 10-4 0.010 - x -x

  2. x2 = 1.9 x 10-6 • x = 1.4 x 10-3 • %Ion = x/starting acid x 100 • %Ion = 1.4 x 10-3 / 0.010 x 100 = 14% • 2.) What is the pH of #1? pH = -log [H+]

  3. (.010)(.184)=.00184 • pH = -log (1.4 x 10-3) • pH = 2.85 • 3.) What is the Ka of a 0.010 M solution of HF which is 18.4% ionized? • HF  H+ + F- I 0.010 0 0 C -.00184 +.00184 +.00184 E .00816 0.00184 0.00184

  4. Ka = [H+][F-] [HF] Ka = (0.00184)2 = 4.15 x 10-4 0.00816 4.) What is the pH of #3? pH = -log [H+] = -log (0.00184) = 2.74

More Related