1. Chapter 14 Acids and Bases AP*

2. Section 14.6 Bases  Arrhenius bases:  Brønsted–Lowry bases:  The pH of a basic solution:  Ionic compounds containing ____ are generally considered strong bases.  LiOH, NaOH, KOH, Ca(OH) 2 Also useful:  pOH = –log[OH – ]  pH = 14.00 – pOH  K a  K b = K w & K w = 1×10 -14 2 produce OH – ions. are proton acceptors >7 OH -

3. Section 14.6 Bases Calculate the pH of a 1.0 × 10 –3 M solution of sodium hydroxide. pOH = -log(1.0 × 10 –3 ) pOH = 3.00 pH = 14 – pOH pH = 14 - 3 pH = 11.00 Two significant figures!! Copyright © Cengage Learning. All rights reserved 3 CONCEPT CHECK!

4. Section 14.6 Bases Calculate the pH of a 1.0 × 10 –3 M solution of calcium hydroxide. Calcium hydroxide = Ca(OH) 2 1.0 × 10 –3 M Ca(OH) 2, so… [Ca 2+ ] = 1.0 × 10 –3 M [OH - ] = 2.0 × 10 –3 M pOH = -log(2.0 × 10 –3 ) pOH = 2.699 pH = 14 – pOH pH = 14 – 2.699 pH = 11.30 4

5. Section 14.6 Bases  Comparing K a to K b. HCN(aq) + H 2 O(l)   CN – (aq) + H 3 O + (aq); K a = 6.2 × 10 –10 CN – (aq) + H 2 O(l)   HCN(aq) + OH – (aq) ; K b = Kb = Kw ÷Ka = 1×10 -14 ÷ 6.2 × 10 –10 = 1.6 × 10 –5 Copyright © Cengage Learning. All rights reserved 5 x

6. Section 14.6 Bases Calculate the pH of a 2.0 M solution of ammonia (NH 3 ). (K b = 1.8 × 10 –5 ) pH = 11.78 Copyright © Cengage Learning. All rights reserved 6 CONCEPT CHECK!

7. Section 14.7 Polyprotic Acids  Acids that can furnish more than one proton.  Always dissociates in a stepwise manner, one proton at a time.  The conjugate base of the first dissociation equilibrium becomes the acid in the second step.  For a typical weak polyprotic acid: K a1 > K a2 > K a3  For a typical polyprotic acid in water, only the first dissociation step is important to pH. Copyright © Cengage Learning. All rights reserved 7

8. Section 14.7 Polyprotic Acids Calculate the pH of a 1.00 M solution of H 3 PO 4. K a1 = 7.5 × 10 -3 K a2 = 6.2 × 10 -8 K a3 = 4.8 × 10 -13 pH = 1.06 Copyright © Cengage Learning. All rights reserved 8 EXERCISE!

9. Section 14.7 Polyprotic Acids Calculate the equilibrium concentration of PO 4 3- in a 1.00 M solution of H 3 PO 4. K a1 = 7.5 × 10 -3 K a2 = 6.2 × 10 -8 K a3 = 4.8 × 10 -13 [PO 4 3- ] = 3.6 × 10 -19 M Copyright © Cengage Learning. All rights reserved 9