ACIDS AND BASES Dissociation Constants.

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Presentation transcript:

ACIDS AND BASES Dissociation Constants

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

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+

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-

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 0.10 0 0 C -1.3 x 10-3 +1.3 x 10-3 +1.3 x 10-3 Ka = 1.7 x 10-5 Ignore the units for K. E 0. 0987 1.3 x 10-3 1.3 x 10-3 Type III – all initial and one equilibrium concentration

Type IV– all initial and NO equilibrium HA is a weak acid with a Ka of 7.3 x 10-8. What are the equilibrium concentrations if the initial [HA] is 0.50 M? HA(aq) + H2O(l) H3O+(aq) + A-(aq) [I] 0.50 0 0 [C] -x +x +x [E] 0.5-x +x +x Ka = [H3O][A-] [HA] Type IV– all initial and NO equilibrium

√ √ 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 10-8 = x2 1.9 x 10-4 = x [H3O+] = [A-] = 1.9 x 10-4 M [HA] = 0.50 - x = 0.50 - 1.9 x 10-4 = 0.49981 M *Ka is small – OK to ignore it 0.50 M

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] 0.10 0 0 [C] -x +x +x [E] 0.10 - x x x Ka = [H3O+][HS-] [H2S]

√ √ 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 10 -8 = x2 1.0 x 10 -4 = x [H3O+] = [HS-] = 1.0 x 10-4 M pH = - log [H3O+] = - log(1.0 x 10-4) pH = 4.00

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

Percent Dissociation Ka / Kb represent the degree of dissociation (how much product has formed) Another way to describe dissociation is by percent dissociation

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)

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-]

HPO42- + H2O H2PO4- + OH- Kb = [H2PO4-][OH-] [HPO42-] [I] 0.25 0 0 [C] -x +x +x [E] 0.25 +x +x [OH-] = [H2PO4-] = 2.0 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

The smaller the Ka or Kb, the weaker the acid / base percent dissociation describes the amount of acid/base dissociated

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