Previously in Chem104:
Today in Chem104:
• plant pigments do
acid/base chemistry
• pH scale
•it’s just equilibrium
•How Ka relates to
Kb and pKa to pKb
• new names: Ka, Kb for
same K expressions
• the concept of Kw
•More ways to use
the Kw circle
• the concept of the Kw
circle
• p-functions (pH, pKa,
pKw)
•Group worksheet
on The Most
Important
Equilibrium on the
Planet (Part 1)
P-Function simplifies a large range of numbers:
graphically
10
1 10-1
10-3 10-5
10-7
10-9
10-11
10-13 10-14
[H3O+], M
converts to a simpler scale
-1
0
1
3
5
7
9
11
13
pH
Note that on a p-scale,
the smaller the p-number, the larger the actual number
Working in P-Functions can simplify problems
Recall Kw = [H3O+][OH-] = 10-14
Apply the P-function to each side
p of Kw = p of [H3O+][OH-] = p of 10-14
-log Kw = -log ( [H3O+][OH-] )= -log 10-14
-log Kw = -log [H3O+] + ( -log [OH-] ) = -log 10-14
pKw = pH + pOH = 14
Recall how we used this picture and this relationship:
Kw = [H+] x [OH-] = 10-14
[H3O+]
[OH-]
Kw =
-14
10
Now apply this equation: pKw = pH + pOH = 14
to this picture
pH
pOH
pKw = 14
When the solution is acidic
[H3O+] > 10-7 M, pH < 7 : pH is a small number
pH < 7
pOH > 7
pKw
Because pKw = pH + pOH must be 14
Fill in the
blanks!
pKw
pH is
_______
When the solution is ____________
[H3O+] __10-7 M, pH ___ 7
pOH is
_______
Let’s do some problems !!
Example problems to be used with reaction:
[Fe-OH2]2+ + H2O
[Fe-OH]+ + H3O+
Keq = 10-10
When is the conjugate base (or acid) important in
acid / base equilibria?
Here?
HCl
acid
+ H2O
base
Cl-
+
H3O+
conjugate conjugate
base
acid
Write the Ka expression for AH and
the Kb expression for A- .
AH
acid
+ H2O
base
A-
+
H3O+
conjugate conjugate
base
acid
Alright, now we can understand why Cl- isn’t basic:
We proved Kw = Ka x Kb
Use the Kw circle!
Ka
Kb
Kw =
-14
10
If AH has a larger Ka, like 10-4
then A- must have a smaller Kb like 10-10
Kb
Ka
Kw
Because Kw = Kax Kb must = 10-14
The stronger the acid (Ka large), the
weaker the conjugate base, (Kb small)
If A- has a larger Kb, like 10-3
then AH must have a smaller Ka like 10-11
Kb
Ka
Kw
Because Kw = Kax Kb must = 10-14
The stronger the base (Kb large), the
weaker the conjugate acid, (Ka small)
Let’s apply P-Functions
We already did this one:
Kw = [H3O+][OH-] = 10-14  pKw = pH + pOH = 14
Now do the same with
Kw = Ka x Kb = 10-14
p of Kw = p of [Ka x Kb ] = p of 10-14
-log Kw = -log (Ka x Kb )= -log 10-14
-log Kw = -log Ka+ ( -log Kb ) = -log 10-14
pKw = pKa + pKb = 14
Now apply this equation: pKw = pKa + pKb = 14
to this picture
pKa
pKb
pKw