Chapter 14

Acids and Bases

AP*

Section 14.6Bases

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 KaKb = Kw & Kw = 1×10-14

2

produce OH– ions.are proton acceptors

>7OH-

Section 14.6Bases

Calculate the pH of a 1.0 × 10–3 M solution of sodium hydroxide.pOH = -log(1.0 × 10–3)pOH = 3.00pH = 14 – pOHpH = 14 - 3

pH = 11.00Two significant figures!!

Copyright © Cengage Learning. All rights reserved 3

CONCEPT CHECK!CONCEPT CHECK!

Section 14.6Bases

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…

[Ca2+] = 1.0 × 10–3 M[OH-] = 2.0 × 10–3 M pOH = -log(2.0 × 10–3)pOH = 2.699pH = 14 – pOHpH = 14 – 2.699

pH = 11.30

4

Section 14.6Bases

Comparing Ka to Kb.

HCN(aq) + H2O(l) CN–(aq) + H3O+(aq); Ka = 6.2 × 10–10

CN–(aq) + H2O(l) HCN(aq) + OH–(aq) ; Kb =

Kb = Kw ÷Ka = 1×10-14 ÷ 6.2 × 10–10 = 1.6 × 10–5

Copyright © Cengage Learning. All rights reserved 5

][]][[

CNOHHCNKb

][]][[ 3

HCNOHCNKa

x

Wba KOHHKK ]][[

Section 14.6Bases

Calculate the pH of a 2.0 M solution of ammonia (NH3).

(Kb = 1.8 × 10–5)

pH = 11.78

Copyright © Cengage Learning. All rights reserved 6

CONCEPT CHECK!CONCEPT CHECK!

Section 14.7Polyprotic 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:

Ka1 > Ka2 > Ka3

For a typical polyprotic acid in water, only the first dissociation step is important to pH.

Copyright © Cengage Learning. All rights reserved 7

Section 14.7Polyprotic Acids

Calculate the pH of a 1.00 M solution of H3PO4.

Ka1 = 7.5 × 10-3

Ka2 = 6.2 × 10-8

Ka3 = 4.8 × 10-13

pH = 1.06Copyright © Cengage Learning. All rights reserved 8

EXERCISE!EXERCISE!

Section 14.7Polyprotic Acids

Calculate the equilibrium concentration of PO43- in a

1.00 M solution of H3PO4.

Ka1 = 7.5 × 10-3

Ka2 = 6.2 × 10-8

Ka3 = 4.8 × 10-13

[PO43-] = 3.6 × 10-19 M

Copyright © Cengage Learning. All rights reserved 9