Download presentation
1
ACIDS AND BASES Dissociation Constants
2
Write the equilibrium expression (Ka or Kb) from a balanced chemical equation.
Use Ka or Kb to solve problems for pH, percent dissociation and concentration. Additional KEY Terms
3
Larger Ka : stronger acid : more product : more H+
HA(aq) H+(aq) A-(aq) Strong Acid HA(aq) + H2O(l) H3O+(aq) A-(aq) Weak Acid Ka - acid dissociation constant Larger Ka : stronger acid : more product : more H+
4
Larger Kb : stronger base : more product : more OH-
BOH (aq) B+(aq) OH-(aq) Strong Base B (aq) + H2O(l) BH+(aq) OH-(aq) Weak Base Kb - base dissociation constant Larger Kb : stronger base : more product : more OH-
5
Type III – all initial and one equilibrium concentration
Initially a 0.10 M solution of acetic acid, it reaches equilibrium with a [H3O+] = 1.3 x 10-3 M. What is the acid dissociation constant, Ka? CH3COOH(aq) + H2O(l) H3O+(aq) + C2H3O2-(aq) I C x x x 10-3 Ka = 1.7 x 10-5 Ignore the units for K. E x 10-3 1.3 x 10-3 Type III – all initial and one equilibrium concentration
6
Type IV– all initial and NO equilibrium
HA is a weak acid with a Ka of 7.3 x What are the equilibrium concentrations if the initial [HA] is 0.50 M? HA(aq) + H2O(l) H3O+(aq) + A-(aq) [I] [C] -x +x +x [E] 0.5-x x x Ka = [H3O][A-] [HA] Type IV– all initial and NO equilibrium
7
√ √ 7.3 x 10-8 = [x][x] 0.50 - x (7.3 x 10-8)(0.50) = x2
*Ka is small - assume that x is negligible compared to 0.50 - x (7.3 x 10-8)(0.50) = x2 √ √ 3.65 x = x2 1.9 x 10-4 = x [H3O+] = [A-] = 1.9 x 10-4 M [HA] = x = x 10-4 = M *Ka is small – OK to ignore it 0.50 M
8
Ka = [H3O+][HS-] [H2S] H2S (aq) + H2O (l) H3O+(aq) + HS-(aq)
Calculate the pH of a 0.10 mol/L hydrogen sulfide solution. (Ka=1.0 x 10-7) H2S (aq) + H2O (l) H3O+(aq) + HS-(aq) [I] [C] -x +x +x [E] x x x Ka = [H3O+][HS-] [H2S]
9
√ √ 1.0 x 10-7 = [x][x] 0.10 - x (1.0 x 10-7)(0.10) = x2
*Ka is small - x is negligible - x (1.0 x 10-7)(0.10) = x2 √ √ 1.0 x = x2 1.0 x = x [H3O+] = [HS-] = 1.0 x 10-4 M pH = - log [H3O+] = - log(1.0 x 10-4) pH = 4.00
10
Each acid/base has K associated with it
polyprotic acids lose their hydrogen one at a time - each ionization reaction has separate Ka Sulfuric acid H2SO4 H2SO4(aq) H+(aq) + HSO4-(aq) Ka1 HSO4- (aq) H+(aq) + SO4-2(aq) Ka2
11
Percent Dissociation Ka / Kb represent the degree of dissociation
(how much product has formed) Another way to describe dissociation is by percent dissociation
12
CH2O2H (aq) + H2O(l) H3O+(aq) + CH2O2¯(aq)
Calculate the percent dissociation of a solution of formic acid (CH2OOH) if the hydronium ion concentration is 0.100 M 4.21 x 10-3 M CH2O2H (aq) + H2O(l) H3O+(aq) + CH2O2¯(aq)
13
Use the %diss formula to find [OH-]
Calculate the Kb of hydrogen phosphate ion (HPO42-) if 0.25 M solution of hydrogen phosphate dissociates 0.080%. HPO42- + H2O H2PO4- + OH- Use the %diss formula to find [OH-]
14
HPO42- + H2O H2PO4- + OH- Kb = [H2PO4-][OH-] [HPO42-]
[I] [C] x x x [E] x x [OH-] = [H2PO4-] = x 10-4 M Kb = [H2PO4-][OH-] [HPO42-] Kb= [2.0 x 10-4][2.0 x 10-4] 0.25 Kb = 1.6 x 10-7
15
The smaller the Ka or Kb, the weaker the acid / base
percent dissociation describes the amount of acid/base dissociated
16
CAN YOU / HAVE YOU? Write the equilibrium expression (Ka or Kb) from a balanced chemical equation. Use Ka or Kb to solve problems for pH, percent dissociation and concentration. Additional KEY Terms
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.