Ka, Kb and Kw
The values of Ka, and Kb relate to the equilibrium constants for acids and
bases at specific temperature and pressure
Large Ka 
Ka, or Kb = 10-4 – 10-5 
Ka, or Kb = 10-12 – 10-13 
Large Kb 
 Kw is the equilibrium constant for water its value is 1.00 x 10-14 @
25.0 0C
 This means that water exists in an equilibrium
H2O(l) + H2O(l) ↔ H3O+(aq) + OH-(aq)
 relationship between Ka, Kb and Kw
Kw = KaKb
Example 1
Find the Kb for Acetic Acids conjugate base, Ka = 1.8 x 10
–5.
The values for Ka, Kb and Kw can be used to determine the equilibrium
concentration for weak acids and bases, as well as, the values for pH and
pOH.
pH = -log [H3O+]
pOH = -log [OH-]
Example 2
Acetic Acid is a weak acid with a Ka = 1.8 x 10 –5 @ 25 0 C. If given a
solution of 1.00M determine the pH, pOH and the % disassociation.
% disassociation
=
# of H+ or OH- moles disassociated
# of moles initially present
Example 3
Calculate the Ka of hydrofluoric acid, HF(aq), if a 0.100 mol/L solution at
equilibrium at SATP has a percent ionization of 7.8%.
Example 4
Hydrazine, a component of Rocket Fuel, is a weak base that has a Kb of
9.6 x 10-7. What is the pH of a 4.56M solution of hydrazine?
pH of Salt Solutions
Salt
- a complex composed of anions and cations
- the force of attraction between the positive (cation) and
negative (anion) charges allow the complex to exist.
Acids
- have the general form HA
- when placed into water they form H30+(aq)/H+ ions and Aions
Example 1
HCl 
HBr 
H2SO4 
Bases
- have the general formula:
1
BOH
2
BNH
- when placed in water they form:
1
H2O(l) + BOH(aq)  B+(aq) + OH-(aq)
2
H2O(l) + BNH(aq)  BNH +(aq) + OH-(aq)
Example 2
KOH 
H2O(l) + NH3(aq) 
The anions from the acids and the cations from the bases can be
used/found in salts
Example of Acid Anion Salts (Conjugate Bases)