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 !!