Expressions similar to the equilibrium constant expression can also be written for acids, bases, water, & things that dissolve (solubility product).

Weak acids exist in equilibrium. An ionization constant, Ka, can be written to show

how much the weak acids are ionized. EX: acetic acid HC2H3O2(aq) + H2O(l) H3O

+(aq)+C2H3O2

- (aq)

Ka= [H3O+][C2H3O2

- ] (pure H2O left out)

[HC2H3O2]

Ka

Write the acid equilibrium expression, Ka, for the acid HC3H5O3.

If the [HC3H5O3] is 0.750 M and the [C3H5O3-]

is the same as [H3O+] and the pH is 3.38, calculate the Ka.

You try it!

HC3H5O3 (aq) + H2O (l) C3H5O3- + H3O+

Ka = [C3H5O3-][H3O+]

[HC3H5O3 ]

10-pH = [H3O+] = 10-3.38 = 4.17 x 10-4

Ka =[4.17 x 10-4][4.17 x 10-4]= [0.750]

2.32 x 10-7

Weak bases have the same properties as weak acids.

A base dissociation constant can be written. B(aq) + H2O(l) BH+

(aq) + OH-(aq)

Kb = [BH+][OH- ] (pure H2O left out)

[B]

Kb

NH3 is a weak base with a Kb of 1.8 x 10-5. Write the base equilibrium expression Kb.

If [OH-] and [NH4+] are both 0.01, calculate the

[NH3] at equilibrium.

Calculate Kb

NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)

Kb = [NH4+] [OH-]

[NH3 ]

Kb = 1.8 x 10-5 = [0.01[0.01] [NH3 ]

[NH3 ] = 5.55 M

The weaker the acid, the smaller the Ka will be (the acid would rather stay together than separate).

The weaker the base, the smaller the Kb will be.

What does Ka or Kb tell you?

Buffered solutions – resist changes in pH

They are usually made of a weak acid/base and its conjugate ionic compound.

Ex: HF is acid, HNO2 is acid,

Buffered Solutions

F- is conjugate

NO2- is conjugate

HF (aq) + H2O (l) F- (aq) + H3O+(aq)

F- (aq) + H2O (l) HF (aq) + OH- (aq)

This will resist changing in pH by shifting equilibrium.

If OH- is added, it will react with the HF If H+ is added, it will react with the F-

Either way, pH remains relatively stable.

Example of a Buffer: HF/F-

Solubility product constants follow the same idea, but require more thinking.

If a substance is slightly soluble, it can still be in an equilibrium, but not much of the chunk of solid has dissolved. EX: AgCl (s) Ag+

(aq) +Cl-

(aq) K= [Ag+][ Cl- ] Ksp= [Ag+][ Cl- ] AgCl is left out because it is a pure solid &

also because its concentration is basically unchanging.

If numbers are known, Ksp can be used for concentration calculations.

Solubility Product

A copper(I) bromide solution has a solubility of 2.0 x 10-4 M at 298 K. Calculate the Ksp.

Calculate Ksp

CuBr (s) Cu+ (aq) + Br- (aq)

CuBr will break into two parts, so the [Cu+] will be the same as the [Br-]; 2.0 x 10-4 M

Ksp = [Cu+ ][Br-] = (2.0 x 10-4 M)(2.0 x 10-4 M) =

4.0 x 10-8

The pH of a 0.400 M solution of iodic acid, HIO3 is 0.726 at 25 C. What is the Ka at this temperature?

HIO3(aq) + H2O(l) H3O+(aq)

+ IO3-1

(aq)

pH = -log[H3O+ ] [H3O+ ] = 10–0.726 = 0.188 MKa = [H3O+][IO3

–1] = [0.188][0.188] =0.088

[HIO3] [0.400]Ka = 0.088

Your Turn!

The pH of a 0.150 M solution of hypochlorous acid, HClO, is found to be 4.55 at 25 C. Calculate the ka for HClO at this temperature.

HClO(aq) + H2O(l) H3O+(aq)

+ ClO-1(aq)

pH = -log[H3O+ ] [H3O+ ] = 10–4.55= 2.82 x 10-5 MKa = [H3O+][ClO–1] = [2.82 x 10-5 ][2.82 x 10-

5 ] [HClO] [0.150]

Ka = 5.3 x 10-9 M

Work it Out!

The compound propylamine, CH3CH2CH2NH2, is a weak base. At equilibrium, a 0.039 M solution of propylamine has an OH- concentration of 3.74 x 10-3 M. Calculate the pH of this solution and Kb for propylamine.

CH3CH2CH2NH2 + H2O(l) CH3CH2CH2NH3+ + OH-

Kb = [CH3CH2CH2NH3+][OH–1] = [3.74 x 10-3]2=

CH3CH2CH2NH2 [0.039]

Kb = 3.59 x 10-4 M

pOH = -log 3.74 x 10-3 M pOH = 2.42

pH = 14-2.42 pH = 11.58

Last One!

What are the only two letter that do not appear in any postal abbreviations of states in the US?

B+Q