1. SCH 4U

2.  Ionize only partially in water, exist primarily in molecule form  Dynamic equilibrium established between unreacted molecules and ions formed from rxn with water  Like all equilibria, can be shifted by removal/ addition of reactants or products

3.  Have a weak attraction for protons  Recall: the conjugate base of a strong acid is a weak base HA + H 2 O A – + H 3 O +  Usually non-hydroxide bases (Recall Arrhenius vs. Bronsted-Lowry definitions)

4. For weak acids: p = [H 3 O + ] x 100% [HA] For weak bases: p = [OH - ] x 100% [B] Ex. The pH of a 0.10 mol/L methanoic acid solution is 2.38. What is the percent ionization of methanoic acid?

5. HCO 2 H (aq) H + (aq) + HCO 2 - (aq) [H + (aq) ] = 10 -pH = 10 -2.38 = 4.2 x 10 -3 mol/L

6. p = conc. of acid ionized x 100% conc. of acid solute p = 4.2 x 10 -3 mol/L x 100% 0.10 mol/L p = 4.2 % Therefore methanoic acid ionizes 4.2% in a 0.10 mol/L solution i.e. HCO 2 H (aq) H + (aq) + HCO 2 – (aq) 4.2 %

7.  Equilibrium constant found as before, called “acid ionization constant,” K a  E.g. for acetic acid: HC 2 H 3 O 2(aq) H + (aq) + C 2 H 3 O 2 - (aq) K a = [H+][C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ]

8. Calculate the acid ionization constant, K a, of acetic acid if a 0.1000 M solution at equilibrium at SATP has a percent ionization of 1.3% HC 2 H 3 O 2(aq) H + (aq) + C 2 H 3 O 2 - (aq) K a = [H+][C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ]

9.  A percent ionization of 1.3% means initial [HC 2 H 3 O 2 ] is diminished by 1.3% by time system reaches equilibrium  Note in this example molar ratio is 1:1:1  Use an ICE table:  Solve for x: x = 0.1 mol/L X 0.013 x = 0.0013 mol/L HC 2 H 3 O 2(aq) H + (aq) + C 2 H 3 O 2 - (aq) Initial0.100000 Change- x+ x Equilibrium0.1000 - x0 + x HC 2 H 3 O 2(aq) H + (aq) + C 2 H 3 O 2 - (aq) Initial0.100000 Change- 0.0013+ 0.0013 Equilibrium 0.0987 0.0013

10.  Now we can calculate K a : K a = [H+][C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] = (0.0013) (0.0013) (0.0987) K a = 1.7 x 10 -5 HC 2 H 3 O 2(aq) H + (aq) + C 2 H 3 O 2 - (aq) Initial0.100000 Change- 0.0013+ 0.0013 Equilibrium 0.0987 0.0013

11.  K a values provide a means of comparing relative strengths of acids  Can also compare % ionization values, but only when acids are equal in initial conc. Data for acetic acid

12.  More dilute the solution, greater the degree of ionization  Can explain using Le Chatelier’s Principle: HA (aq) A – (aq) + H + (aq)

13.  Equilibrium constant called “base ionization constant,” K b  E.g. for ammonia: NH 3(aq) + H 2 O (l) OH - (aq) + NH 4 + (aq) K b = [OH - ][NH 4 + ] [NH 3 ]  Note: many weak bases contain one or more N atom, others are conjugate bases of strong acids

14.  Consider an acetic acid solution at equilibrium: HC 2 H 3 O 2(aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 - (aq)  As a base, the acetate ion also reacts with water, establishing an equilibrium: C 2 H 3 O 2 - (aq) + H 2 O (l) OH - (aq) + HC 2 H 3 O 2(aq) K a = [H 3 O + ][C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] K b = [OH - ][HC 2 H 3 O 2 ] [C 2 H 3 O 2 - ]

15. K a x K b = K w = K a x K b [H 3 O + ] [C 2 H 3 O 2 - ] x [OH - ] [HC 2 H 3 O 2 ] [HC 2 H 3 O 2 ] [C 2 H 3 O 2 - ] = [H 3 O + ] [OH - ] K a = K w K b K b = K w K a

16.  The conjugate base of a strong acid is a ______________ base  The conjugate base of a weak acid is a ______________ base  The conjugate base of a very weak acid is a ______________ base * Go over page 562 together very weak weak strong

17.  The conjugate acid of a strong base is a ______________ acid  The conjugate acid of a weak base is a ______________ acid  The conjugate acid of a very weak base is a ______________ acid very weak weak strong

18. Calculate the hydrogen ion conc. and pH of a 0.10 mol/L acetic acid solution. K a for acetic acid is 1.8 x 10 -5. *First need to compare K a ’s of all equilibria that may contribute H + to the system... K a = 1.8 x 10 -5 K w = 1.0 x 10 -14 HC 2 H 3 O 2(aq) H + (aq) + C 2 H 3 O 2 - (aq) I 0.10 0 0 C - x +x +x E0.10 – x x x

19. K a = [H+][C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] 1.8 x 10 -5 = x 2 0.10 –x Could solve quadratic or could make it simpler... FINISH... on p.563!

20. For diethylamine (CH 3 CH 2 ) 2 NH, the pH of a 2.6 x 10 -2 M solution is 11.56. What is K b for diethylamine? B (aq) + H 2 O (l) OH - (aq) + HB + (aq) Can use pH to det. [OH]... pH = 11.56 therefore pOH = 14.00 – 11.56 = 2.44 pOH = -log[OH - ] [OH - ] = 10 -2.44 [OH - ] = 3.6 x 10 -3 M

21. B (aq) + H 2 O (l) OH - (aq) + HB + (aq) I 2.6 x 10 -2 0 0 C -x +x +x E 2.6 x 10 -2 – xx x x = 3.6 x 10 -3 K b = [OH - ][HB + ] [B] K b = (3.6 x 10 -3 ) 2 2.6 x 10 -2 – 3.6 x 10 -3 = 5.8 x 10 -4 See summary of problem-solving steps p.574

22.  E.g. Sulfuric acid H 2 SO 4, boric acid H 3 BO 3  Different K a values (K a1, K a2, etc.)  In general, K a1 > K a2 > K a3...  See p. 574, 575 * Because K a1 is usually >> K a2, K a3, etc., typically use just Ka1 to determine pH

23.  p. 579 # 3-6, 11-13