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

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

3. Acid ionization constant equilibrium expression where H + is removed to form conjugate base equilibrium expression where H + is removed to form conjugate base so for: HA + H 2 O H 3 O + + A - so for: HA + H 2 O H 3 O + + A -

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

6. 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

7. [H 2 O], pH and K w conc. of liquid water is omitted from the K a expression conc. of liquid water is omitted from the K a expression we assume that this conc. will remain constant in aqueous sol’n that are not highly concentrated we assume that this conc. will remain constant in aqueous sol’n that are not highly concentrated pH= -log[H + ] pH= -log[H + ] pOH = -log[OH - ] pOH = -log[OH - ] 14.00= pH + pOH 14.00= pH + pOH

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

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

10. Calculating Weak Acids 1. Write major species 2. Decide on which can provide H + ions 3. Make ICE table 4. Put equilibrium values in K a expression 5. Check validity of assumption (x must be less than 5% of initial conc) 6. Find pH

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

12. Example 2 HF  H + + F - HF  H + + F - I 1.00 M 00 C-x+x+x E 1.00 -x xx

13. Example 2 Check assumption: Check assumption:  pH = -log(0.027) = 1.57

14. Example 3 Find pH of 0.100 M solution of HOCl (K a = 3.5x10 -8 ) Find pH of 0.100 M solution of HOCl (K a = 3.5x10 -8 ) HOCl, H 2 O HOCl, H 2 O HOCl will provide much more H + than H 2 O, so we ignore H 2 O HOCl will provide much more H + than H 2 O, so we ignore H 2 O HOCl  H + + OCl - HOCl  H + + OCl - I 0.100 M 00 C-x+x+x E 0.100 -x xx

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

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

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

18. Base Strength and K b follows same standard rules as for calculating K a for acids K b is used with weak bases that react only partially with water (<50%)

19. Bases K b K b base ionization constant base ionization constant refers to reaction of base with water to make conjugate acid and OH - refers to reaction of base with water to make conjugate acid and OH - liquid water is again ignored like in K a liquid water is again ignored like in K a B (aq) + H 2 O (l)  BH + (aq) + OH - (aq) B (aq) + H 2 O (l)  BH + (aq) + OH - (aq)

20. Example 4 Find the pH for 15.0 M solution of NH 3 (K b = 1.8x10 -5 ) Find the pH for 15.0 M solution of NH 3 (K b = 1.8x10 -5 ) NH 3 will create more OH - than water so self- ionization (H 2 O) can be ignored NH 3 will create more OH - than water so self- ionization (H 2 O) can be ignored NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - I15.000 C-x+x+x E15.0-xxx NH 3 + H 2 O NH 4 + + OH - NH 3 + H 2 O NH 4 + + OH - Sol’n

21. Example 4 con’t % ion.

22. Example 5 Codeine (C 18 H 21 NO 3 ) is a weak organic base. A 5.0x10 -3 M solution of codeine has a pH of 9.95. Codeine (C 18 H 21 NO 3 ) is a weak organic base. A 5.0x10 -3 M solution of codeine has a pH of 9.95. Calculate the K b for this substance. Calculate the K b for this substance. Sol’n Sol’n What is chemical reaction? What is chemical reaction? C 18 H 21 NO 3 + H 2 O HC 18 H 21 NO 3 + + OH - C 18 H 21 NO 3 + H 2 O HC 18 H 21 NO 3 + + OH - Find [OH - ] using given pH Find [OH - ] using given pH pOH= 14.00-9.95 = 4.05 pOH= 14.00-9.95 = 4.05 [OH - ] = 10 -4.05 = 8.9x10 -5 [OH - ] = 10 -4.05 = 8.9x10 -5

23. Example 5 con’t C 18 H 21 NO 3 + H 2 O HC 18 H 21 NO 3 + + OH - C 18 H 21 NO 3 + H 2 O HC 18 H 21 NO 3 + + OH - I 5.0x10 -3 00 C-x+x+x E 5.0x10 -3 - x xx x = [OH - ] = 8.9x10 -5 x = [OH - ] = 8.9x10 -5

24. K a - K b relationship for conjugate pairs Acid-Base strength tables do not give K b values we use the K a - K b relationship to solve this problem this can be used with any conjugate acid-base pair K w = K a x K b or K b = K w / K a K w = 1.0 x 10 -14

25. Example 6 What is the K b value for the weak base present when sodium cyanide dissociates into an aqueous solution? Sol’n 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 - + H 2 O HCN + OH -  the conjugate acid is found to be HCN and its K a is 6.2 x 10 -10  Using K b = K w / K a find K b for CN -  K b = 1 x 10 -14 / 6.2 x 10 -10 = 1.61 x 10 -5

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