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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.

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Ch. 16: Equilibrium in Acid-Base Systems

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  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

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