# 4.8-4.9: Ka, Kb and the Conjugate Pair - PowerPoint PPT Presentation

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4.8-4.9: Ka, Kb and the Conjugate Pair. Chemistry 12. 4.8 – 4.9. Keq = Ksp = Kw = Ka = Kb =. equilibrium constant solubility product eqb exp for ionization of water acid ionization constant base ionization constant. Concept 1.

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4.8-4.9: Ka, Kb and the Conjugate Pair

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## 4.8-4.9: Ka, Kb and the Conjugate Pair

Chemistry 12

### 4.8 – 4.9

• Keq =

• Ksp =

• Kw =

• Ka =

• Kb =

equilibrium constant

solubility product

eqb exp for ionization of water

acid ionization constant

base ionization constant

### Concept 1

Using Kw to calculate unknown [H3O+] and [OH-] from a strong acid or a strong base.

Eg.) At 250C an HClsolution has a concentration of 0.0100 M. What is the [OH-]?

HCl + H20 ↔ H3O++ Cl-

[H3O+] = 0.0100 M

[H3O+][OH-] = Kw

0.0100 M [OH-] = 1.00 x 10-14

[OH-] = 1.00 x 10-12 M

### Concept 2: Equilibrium equations and Ka and Kb Expressions

What is the Ka expression for oxalic acid?

(Hint: use the table)

H2O(l) + H2C2O4(aq)↔ H3O+(aq) + HC2O4-(aq)

Ka = [H3O+][HC2O4-]

[H2C2O4]

What is the equilibrium equation for HC2O4- acting as a base and the corresponding Kb expression?

HC2O4-(aq) + H2O(l) ↔H2C2O4(aq) +OH-(aq)

Kb = [H2C2O4][OH-]

[HC2O4-]

### Concept 3: Finding Ka and calculating Kb

What is Ka value for the following?

(simply use the table)

H2C2O4 Ka = 5.9 x 10-2

HC2O4- Ka = 6.4 x 10 -5

…but solving for Kb is not that easy.

What is the Kb value of HPO42-?

We must use the Ka value of H2PO4-

to determine Kb of HPO42-

HPO42- is acting as a base

HPO42- is acting as an acid

Ka value of HPO42- so we cannot use this to calculate Kb

### Calculation

You should notice that this is the Ka value of the conjugate acid for HPO42-

Kb of HPO42- = Kw/ Ka(H2PO4-)

= 10-14/6.2 x 10-8

= 1.6 x10-7

### Now you try

Actually Ka of H2C2O4

Kb for HC2O4-?

Find HC2O4- onright side of table

Ka = 5.9 x 10 -2

Kw = Ka x Kb

1.0 x 10-14 = 5.9 x 10 -2 x Kb

Kb = 1.7 x 10-13

### Concept 4: Explaining the connection between Kw, Ka and Kb.

This can be explained using HC2O4- but other weak acids could also be used

Why do we use the formula Kw = Ka x Kb?

H2O(l) + HC2O4-(aq)↔ H3O+(aq) + C2O42-(aq)

Acid dissociation

Base dissociation

### 4.8 – 4.9

To get Kb we must figure out the base dissociation equation of C2O42-.

### 4.8 – 4.9

H2O(l) + HC2O4-(aq)↔ H3O+(aq) + C2O42-(aq)

Invert

H3O+(aq) + C2O42-(aq) ↔ H2O(l) + HC2O4-(aq)

+ OH- (aq)↔ OH-(aq)

2 H2O(l)

H2O(l) + C2O42-(aq) ↔ OH-(aq) + HC2O4-(aq)

### 4.8 – 4.9

H2O(l) + C2O42-(aq) ↔ OH-(aq) + HC2O4-(aq)

H2O(l) + HC2O4-(aq)↔ H3O+(aq) + C2O42-(aq)

Kb = [HC2O4-][OH-]

[C2O42-]

Ka = [H3O+][C2O42-]

[HC2O4-]

### 4.8 – 4.9

 if

Kb = [HC2O4-][OH-] Ka = [H3O+][C2O42-]

[C2O42-] [HC2O4-]

Then Ka x Kb

[HC2O4-][OH-] x [H3O+][C2O42-]

[C2O42-] [HC2O4-]

 Ka x Kb = [OH-][H3O+] = Kw

### Learning Check

Can you calculate [H3O+] and [OH-] of a strong acid or base using Kw?

Can you write equilibrium equations for weak acids and bases?

Can you write out Ka and Kb expressions?

Can you find Ka on the B-L table?

Can you calculate Kb using the B-L table?