Previously in Chem104:

• plant pigments do acid/base chemistry

•it’s just equilibrium

• new names: Ka, Kb for same K expressions• the concept of Kw

• the concept of the Kw

circle

• p-functions (pH, pKa, pKw)

Today in Chem104:

• pH scale

•How Ka relates to Kb and pKa to pKb

•More ways to use the Kw circle

•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

-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

Recall how we used this picture and this relationship: Kw = [H+] x [OH-] = 10-14

[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 a small number

Because pKw = pH + pOH must be 14

pH < 7 pOH > 7

pKw

When the solution is ____________[H3O+] __10-7 M, pH ___ 7

pOH is_______

pH is_______

Fill in the blanks!

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?

HCl + H2O Cl- + H3O+

acid conjugate base

base conjugate acid

Here?

AH + H2O A- + H3O+

acid conjugate base

base conjugate acid

Write the Ka expression for AH andthe Kb expression for A- .

Kw = 10-14

Alright, now we can understand why Cl- isn’t basic:We proved Kw = Ka x Kb

Use the Kw circle!

Ka Kb

Kw

If AH has a larger Ka, like 10-4

then A- must have a smaller Kb like 10-10

KaKb

The stronger the acid (Ka large), the weaker the conjugate base, (Kb small)

Because Kw = Kax Kb must = 10-14

Kw

KaKb

If A- has a larger Kb, like 10-3

then AH must have a smaller Ka like 10-11

The stronger the base (Kb large), the weaker the conjugate acid, (Ka small)

Because Kw = Kax Kb must = 10-14

Now do the same withKw = Ka x Kb = 10-14

p of Kw = p of [Ka x Kb ] = p of 10-14

Let’s apply P-Functions

We already did this one:Kw = [H3O+][OH-] = 10-14 pKw = pH + pOH = 14

-log Kw = -log (Ka x Kb )= -log 10-14

pKw = pKa + pKb = 14

-log Kw = -log Ka+ ( -log Kb ) = -log 10-14

pKw

Now apply this equation: pKw = pKa + pKb = 14

to this picture

pKa pKb