8.2 Strong and Weak Acids and Bases

Learning Goals ……determine [H30+] and [OH-] using Kw, pH or pOH…determine Ka or Kb using Kw…find pH and pOH of a substance

Water as an Acid and a Base

•Pure water dissociates according to the following reaction:

2 H2O(l) H3O+(aq) + OH-

(aq)

• There is an equal amount of H+ and OH- ions in solution (neutral, pH = 7)

• at 25°C [H3O+] = [OH-] = 1x10-7 mol/L

• equilibrium constant for the dissociation of water: Kw

Kw = [H3O+][OH-]

= (1x10-7)(1x10-7)= 1x10-14

* small k, reactants are favoured(does not go to completion)

@ 25°C acids: [H3O+] > [OH-] [H3O+] > 1x10-7

[OH-] < 1x10-7

bases: [OH-] > [H3O+] [H3O+] < 1x10-7

[OH-] > 1x10-7

We can use Kw to calculate [H3O+] and [OH-] in solutions [H3O+][OH-]= 1x10-14

The Relationship Between Kw, Ka and Kb

HA (aq) + H2O (l) H3O+ (aq) + A-

(aq)

Ka = [H3O+][A-][HA]

A- (aq) + H2O (l) HA (aq) + OH- (aq)

Kb = [HA][OH-] [A-]

Ka x Kb = [H3O+][A-] [HA][OH-] [HA] [A-]

= [H3O+][OH-] = Kw

Kw = Ka x Kb

Ex) What is the Kb value for F- if the Ka value of HF is 6.6x10-4?HF(aq) + H2O(l) H3O+

(aq) + F-(aq)

acid conjugate base

Kw = Ka Kb Kb = Kw / Ka = (1.0x10-14)/(6.6x10-4) = 1.5x10-11

• A strong acid or base (large Ka or Kb) will have a very weak conjugate (very small Kb or Ka)

• A weak acid or base (small Ka or Kb) has a weak conjugate (small Ka or Kb)

Ex) H2CO3 + H2O H3O+ + HCO3- Ka1 = 4.4x10-7

HCO3- + H2O H3O+ + CO3

2- Ka2 = 4.7x10-11

HCO3- is amphoteric. It can act as an acid or a base.

HCO3- + H2O H2CO3

+ OH- Kb = ?

Kb = Kw / Ka1

= (1.0x10-14)/(4.4x10-7) = 2.3x10-8

POLYPROTIC ACIDS• acids that have more than one proton• only one proton is transferred at a time

Since Kb > Ka, HCO3- is more likely to act as a base

pH = -log [H3O+] pOH = -log [OH-]

In neutral water, pH = -log [H3O+] = -(log(1 x 10-7) = 7pOH = -log [OH-] = -(log(1 x 10-7) = 7

Note: pH + pOH = 14, always, regardless of solution!

Another way to calculate [H3O+] & [OH-] in solution:

[H3O+] = 10-pH [OH-] = 10-pOH

pH and pOH

Ex) A liquid shampoo has a [OH-] of 6.8x10-5 mol/L(a) Is the shampoo acid, basic or neutral?(b) What is [H3O+]?

(c) What is the pH and pOH of the shampoo?

(a) [OH-] = 6.8x10-5 > 1.0x10-7, basic

(b) [H3O+] = Kw / [OH-] = (1.0x10-14)/(6.8x10-5)

= 1.5x10-10 mol/L

(c) pOH = -log [OH-] = -log [6.8x10-5] = 4.17

pH = 14 – 4.17 = 9.83

HOMEWORKp509 #1, 2, 3ab, 4ab, 5ab, 6ab, 7-9

Self CheckHow prepared am I to start my homework? Can I ……determine [H+] and [OH-] using Kw, pH or pOH…determine Ka or Kb using Kw…find pH and pOH of a substance