Strong and Weak Bases – Notes
Strong base – 100% of base dissociates in water
Example: NaOH Na+ + OHAll of NaOH dissociates into Na+ and OH- ions
These are like strong acid, and denoted by a single arrow (only the forward
reaction)
In a strong acid the concentration of OH will equal the original concentration of the
base
[OH] = [NaOH]
Examples of strong bases:
LiOH
NaOH
KOH
RbOH
CsOH
Mg(OH)2
Ba(OH)2
Sr(OH)2
Weak Bases – less than 100% dissociates in water
NH3 + H20
NH4+ + OH-
Not all of NH3 will dissociate into NH4 and OH. There will be a mixture of both the
product (NH4+ and OH-) and the original base (NH3)
These are like weak acids, and are deonted by a double arrow, meaning the forward
and reverse reaction can occur.
All other acids are considered weak (besides those in strong list)
Examples:
IO3
PO43NH3 (ammonia)
Kb – equilibrium constant for bases
Kb = [products]
[reactants]
Recall that we omit water because liquids concentrations are relatively stable so we
do not include them in the equation
Example:
NH3 + H20
Kb= [NH4+] [OH-]
[NH3]
NH4+ + OH-
Calculations using Kb
**(all very similar to Ka so don’t stress that there are MORE calculations)
Example 1: what is pH of a 0.1 solution of ammonia (NH3). The Kb=1.77x10-5
Step 1: write ionization for ammonia
NH3
+
H20
NH4+
Step 2: make an ICE chart and fill in with information you know
NH3
+
H20
NH4+
+
OH+ OH-
I
0.1
0
0
C -(insignificant)
+x
+x
E
0.1
x
x
Recall that x will be small in comparison to 0.1 so we can just omit it in the
equilibrium calculation for NH3.
Step 3: Find OH- using Kb
Kb= [NH4+] [OH-]
[NH3]
1.77x10-5 = (x)(x)
0.1
1.77x10-5(0.1) = x2
1.77x10-6 = x2
√1.77x10-6 = x
1.33x10-3 = x
Step 4: find pOH using [OH]
pOH = -log[OH-]
pOH = -log[1.33x10-3)
pOH= 2.89
Step 5: find pH using pOH
pH + pOH = 14
14-2.89 = pH
11.11 = pH
Example 2: A 0.2 M solution of generic weak base (call it B) has a a pH of
10.25. Find Kb.
Step 1: write ionization for ammonia
B
+
H20
BH+
+
OHStep 2: make an ICE chart and fill in with information you know
B
+
H20
BH+
+ OH-
I
0.2
0
0
C (insignificant)
E
0.2
Recall that BH and OH concentration will be small in comparison to 0.2 so we can
just omit it in the equilibrium calculation for B
Step 3: find pOH using pH
14-pH = pOH
14-10.25 = pOH
3.75 = pOH
Step 4: find OH using pOH
pOH = -log[OH-]
3.75 = -log[OH-]
1.78x10-4 = [OH]
(can fill this in ice chart )
B
+
I
C
E
H20
0.2
- (insignificant)
0.2
BH+
0
+1.78x10-4
1.78x10-4
Step 5: Find Kb
Kb= [BH+] [OH-]
[B]
Kb = (1.78x10-4) (1.78x10-4)
(0.2)
Kb= 1.6x10-7
Complete questions on Worksheet #9
+ OH0
+1.78x10-4
1.78x10-4