ACIDS AND BASES

Dissociation Constants

weaker the acid, the stronger its conjugate base

stronger the acid, the weaker its conjugate base

stronger the base, the weaker its conjugate acid

weaker the base, the stronger its conjugate acid

· Write equilibrium expression for an acid or base

· Calculate the acid/base dissociation constant

· Calculate the percent dissociation

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

(aq)

Ka - acid dissociation constant

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

(aq)

Strong Acid

Weak Acid

Larger Ka : strong acid: more product : more H+.

BOH (aq) B+

(aq) + OH-(aq)

Larger Kb : strong 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 reached 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 0.987 1.3 x 10-3

1.3 x 10-3

Ka = 1.7 x 10-5

There are no units for the Ka value

HA is a weak acid with a Ka of 7.3 x 10-8. What are the equilibrium concentrations (HA, H3O+ and A¯) if the initial concentration of HA is 0.50 mol/L?

I 0.50 0 0

C -x +x +x

E 0.5-x +x +x

HA(aq) + H2O(l) H3O+(aq) + A¯(aq)

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

[H3O+] = [A¯] = x = 1.9 x 10¯4 mol/L

[HA] = 0.50 - x= 0.50 - 1.9 x 10¯4 = 0.49981 mol/L

[HA] = 0.50 mol/L

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

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

[H3O+] = x = 1.0 x 10-4 mol/L

pH = -log [H3O+] = -log(1.0 x 10-4)

pH = 4.00

Each acid/base has K associated with it.

Diprotic/triprotic acids lose their hydrogens 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 DissociationThe dissociation constants represent the acid / base degree of dissociation.

Another way to describe the amount of dissociation is by percent dissociation.

Calculate the percent dissociation of a 0.100 M solution of formic acid (CH2OOH) if the hydronium ion concentration is 4.21 x 10-3 M.

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

Calculate the Kb of hydrogen phosphate ion (HPO42¯)

if a 0.25 mol/L solution of hydrogen phosphate is dissociated is 0.080%.

HPO42¯ + H2O H2PO4¯ + OH¯

[OH-] = [H2PO4-] = 2.0 x 10-4 mol/L

[OH-] = [H2PO4-] = 2.0 x 10-4 mol/L

HPO42¯ + H2O H2PO4¯ + OH¯

· The smaller the Ka or Kb, the weaker the acid / base

· The percent dissociation also describes the amount of acid/base dissociated

· The percent dissociated is calculated by