Previously in Chem104:
• plant pigments do acid/base chemistry
•it’s just equilibrium
• new names: Ka, Kb for same K expressions
Today in Chem104:
• the concept of Kw
• the concept of the Kw
circle
• p-functions (pH, pKa, pKw)
• pH scale
• P-functions in calcs
Dual Personality of Water: both an acid and a baseAmphoteric (or amphiprotic)
The Autoionization Reaction of water.
2 H20 H3O+ + OH-
Keq =
Kw
Because Kw = [H3O+][OH-] = 10-14 and Kw is a constant: then
the solution is neutral when [H3O+] = [OH-] = 10-7 M
[H3O+] [OH-]
Kw
But, if [H3O+] > [OH-], [H3O+] > 10-7 M,the solution is acidic
[OH-] < 10-7 M
[H3O+] > 10-7 M
Kw
Or if [OH-] > [H3O+], [OH-] > 10-7 M, the solution is basic
Because Kw = [H3O+] x [OH-] must be 10-14
[OH-] > 10-7 M
[H3O+] < 10-7 M
Because all these exponential numbers are a pain:i.e.Kw = [H3O+][OH-] = 10-14
[H3O+] = [OH-] = 10-7 M (in neutral water)Ka = 10-10
we will use: The P-Function
p of (a number) = -log10(a number)
so “pH” means p of [H3O+], or -log10 [H3O+]
so the pH of neutral water, where [H3O+] = 10-7 M is pH 7
The P-Function and K’s
p of (Keq) = -log10Keq = pKeq
p of (Ka) = -log10Ka = pKa
p of (Kw) = -log10Kw = pKw
Examples: Keq = 10-10 the pKeq = 10
Ka = 1.20 x 10-5 the pKeq = 4.92
Kw = 10-14 the pKw = 14
The P-Function simplifies exponential numbers
Ka = 1.20 x 10-5 , pKeq = 4.92
[H3O+] = 4.20 x 10-5 M, pH = 4.376
[OH-] = 6.66 x 10-6 M, pOH = 5.176(note number of sig figs)
Examples converting from p-function: if pH = 7.47,
[H3O+] = 10-7.47 M= 3.39 x 10-8 M,
if pKa = 4.92, Ka = 10-4.92 = 1.20 x 10-5
P-Function simplifies a large range of numbers: graphically
10 1 10-1 10-3 10-5 10-7 10-9 10-1110-13 10-14
[H3O+], M
-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
converts to a simpler scale
Apply the P-function to each side
p of Kw = p of [H3O+][OH-] = p of 10-14
Working in P-Functions can simplify problems
Recall Kw = [H3O+][OH-] = 10-14
-log Kw = -log ( [H3O+][OH-] )= -log 10-14
pKw = pH + pOH = 14
-log Kw = -log [H3O+] + ( -log [OH-] ) = -log 10-14
Kw = 10-14
Now apply this equation: pKw = pH + pOH = 14
to this picture
[H3O+] [OH-]
pKw = 14
Now apply this equation: pKw = pH + pOH = 14
to this picture
pH pOH
pKw
When the solution is acidic[H3O+] > 10-7 M, pH < 7 : pH is small
Because pKw = pH + pOH must be 14
pH < 7 pOH > 7
pKw
When the solution is ____________[H3O+] __10-7 M, pH ___ 7 : pH is ________
pOH is_______
pH is_______
Fill in the blanks!
Let’s do some problems !!