1 / 18

Ch. 16: Equilibrium in Acid-Base Systems

Ch. 16: Equilibrium in Acid-Base Systems. 16.3a: Acid-Base strength and equilibrium law. Definitions. Arrhenius A: produce H + in aqueous solution B: produces OH - in aqueous solution very limited Bronsted -Lowry A: H + donor B: H + acceptor more general. Acid ionization constant.

eliot
Download Presentation

Ch. 16: Equilibrium in Acid-Base Systems

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. Ch. 16: Equilibrium in Acid-Base Systems 16.3a: Acid-Base strength and equilibrium law

  2. Definitions Arrhenius A: produce H+ in aqueous solution B: produces OH- in aqueous solution very limited Bronsted-Lowry A: H+ donor B: H+ acceptor more general

  3. Acid ionization constant equilibrium expression where H+ is removed to form conjugate base so for: HA + H2O <--> H3O+ + A-

  4. Strength determined by equilibrium position of dissociation reaction strong acid: lies far to right, almost all HA is dissociated large Ka values creates weak conjugate base weak acid: lies far to left, almost all HA is stays as HA small Ka values creates strong conjugate base

  5. Water is a stronger base than the CB of a strong acid but a weaker base than the CB of a weak acid Water is a stronger acid than the CA of a strong base but a weaker acid than the CA of a weak base

  6. [H2O], pH and Kw conc. of liquid water is omitted from the Ka expression we assume that this conc. will remain constant in aqueous sol’n that are not highly concentrated pH= -log[H+] pOH = -log[OH-] 14.00= pH + pOH

  7. Example 1 The [OH-] of a solution at 25oC is 1.0x10-5 M. Determine the [H+], pH and pOH. Kw = 1.0x10-14 = [OH-] x [H+] [H+] = 1.0x10-9 pH= -log(1.0x10-9) = 9.00 pOH = -log(1.0x10-5) = 5.00 acidic or basic? basic

  8. Approximations If K is very small, we can assume that the change (x) is going to be negligible “rule of thumb” is if initial conc. of the acid is >1000 times its Ka value then cancel x this makes the answer true to +/- 5% and why Ka values are given to 2 sig. digs 0

  9. Calculating Weak Acids Write major species Decide on which can provide H+ ions Make ICE table Put equilibrium values in Ka expression Check validity of assumption (x must be less than 5% of initial conc) Find pH

  10. Example 2 Calculate the pH of 1.00 M solution of HF (Ka = 7.2 x 10-4) HF, H2O HF  H+ + F- Ka = 7.2x10-4 H2O  H+ + OH- Kw = 1.0 x 10-14 HF will provide much more H+ than H2O – ignore H2O

  11. Example 2

  12. Example 2 Check assumption: • pH = -log(0.027) = 1.57

  13. Example 3 Find pH of 0.100 M solution of HOCl (Ka = 3.5x10-8) HOCl, H2O HOCl will provide much more H+ than H2O, so we ignore H2O

  14. Example 3 • Check assumption: • pH = -log(5.9x10-5) = 4.23

  15. Example4 Find Ka for propanoic acid given the following information [C2H5COOH] = 0.10M and pH = 2.96 [H3O+] = 1.1 x 10-3 M Sol’n 1) C2H5COOH + H2O  C2H5COO- + H3O+ 2) Calculate [H3O+] using pH [H3O+] = 10-pH [H3O+] = 10-2.96 [H3O+] = 1.1 x 10-3 M 3) Pure water is not included as it does not change

  16. Example 4 con’t 4) Solve for Ka • % ionization: < 5% indicates a weak acid

  17. Homework • Textbook p743 #2a,c,e 5,7,9 • LSM 16.3A and 16.3D

More Related