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

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

  17. Base Strength and Kb • follows same standard rules as for calculating Ka for acids • Kb is used with weak bases that react only partially with water (<50%)

  18. Bases Kb base ionization constant refers to reaction of base with water to make conjugate acid and OH- liquid water is again ignored like in Ka B(aq) + H2O(l) BH+ (aq) + OH- (aq)

  19. Example 4 Find the pH for 15.0 M solution of NH3 (Kb = 1.8x10-5) NH3 will create more OH- than water so self- ionization (H2O) can be ignored Sol’n NH3 + H2O <--> NH4+ + OH-

  20. Example 4 con’t % ion.

  21. Example 5 Codeine (C18H21NO3) is a weak organic base. A 5.0x10-3 M solution of codeine has a pH of 9.95. Calculate the Kb for this substance. Sol’n What is chemical reaction? C18H21NO3 + H2O <--> HC18H21NO3+ + OH- Find [OH-] using given pH pOH= 14.00-9.95 = 4.05 [OH-] = 10-4.05 = 8.9x10-5

  22. Example 5 con’t x = [OH-] = 8.9x10-5

  23. Ka - Kb relationship for conjugate pairs • Acid-Base strength tables do not give Kb values • we use the Ka - Kb relationship to solve this problem • this can be used with any conjugate acid-base pair Kw = Ka x Kb Kw = 1.0 x 10-14 or Kb = Kw / Ka

  24. Example 6 Sol’n • What is the Kb value for the weak base present when sodium cyanide dissociates into an aqueous solution? • NaCN --> Na+ +CN- (complete dissociation) • the CN- is the weak base so we write the equilibrium equation with water to determine its conjugate acid • CN- + H2O <--> HCN + OH- • the conjugate acid is found to be HCN and its Ka is 6.2 x 10-10 • Using Kb = Kw / Ka find Kb for CN- • Kb = 1 x 10-14 / 6.2 x 10-10 = 1.61 x 10-5

  25. Homework • Textbook p746 #11,12 • Textbook p750 #2,6,9 • LSM 16.3 A,B & D

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