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

HA(aq) H+(aq) + A-

(aq)

Ka - acid dissociation constant

HA(aq) + H2O(l) H3O+(aq) + A-

(aq)

Strong Acid

Weak Acid

Larger Ka : stronger acid : more product : more H+

BOH (aq) B+

(aq) + OH-(aq)

Larger Kb : stronger base : more product : more OH-

Kb - base dissociation constant

Strong Base

B (aq) + H2O(l) BH+

(aq) + OH-(aq)

Weak Base

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

E 0. 0987 1.3 x 10-3

1.3 x 10-3

Type III – all initial and one equilibrium concentration

Ka = 1.7 x 10-5

Ignore the units for K.

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?

[I] 0.50 0 0[C] -x +x +x[E] 0.5-x +x +x

HA(aq) + H2O(l) H3O+(aq) + A-

(aq)

Ka = [H3O][A-] [HA]

Type IV– all initial and NO equilibrium

*Ka is small - assume that x is negligible compared to 0.50

7.3 x 10-8 = [x][x] 0.50 - x

(7.3 x 10-8)(0.50) = x2

3.65 x 10-8 = x2 √ √1.9 x 10-4 = x

[HA] = 0.50 - x= 0.50 - 1.9 x 10-4 = 0.49981 M

*Ka is small – OK to ignore it

[H3O+] = [A-] = 1.9 x 10-4 M

0.50 M

Calculate the pH of a 0.10 mol/L hydrogen sulfide solution. (Ka=1.0 x 10-7)

H2S (aq) + H2O (l)

[I] 0.10 0 0 [C] -x +x +x [E] 0.10 - x x x

Ka = [H3O+][HS-] [H2S]

H3O+(aq) + HS-

(aq)

*Ka is small - x is negligible1.0 x 10-7 = [x][x] 0.10 - 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)

HSO4-

(aq) H+(aq) + SO4

-2(aq)

Ka1

Ka2

Percent Dissociation

• Ka / Kb represent the degree of dissociation (how much product has formed)

• Another way to describe dissociation is by percent dissociation

Calculate the percent dissociation of a solution of formic acid (CH2OOH) if the hydronium ion concentration is .

CH2O2H (aq) + H2O(l) H3O+(aq) + CH2O2¯(aq)

0.100 M

4.21 x 10-3 M

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-]

[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

HPO42- + H2O H2PO4

- + OH-

[I] 0.25 0 0[C] -x +x +x[E] 0.25 +x +x

· 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