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1

ANALYTICAL CHEMISTRY

 Salts

 Hydrolysis

 The Common-Ion Effect

 Buffer Solutions

 Acid-Base Titrations

 Solubility

AQUEOUS EQUILIBRIA

Brown et al ., Chapter 15, 569 - 606 2

nd Ed.

Brown et al ., Chapter 18 p 706-751 3

rd Ed.

EXAMPLE

What percentage of molecules have dissociated in a 0.100 mol dm

• 3 solution

of acetic acid?

K

a

(CH

3

COOH) = 1.75 × 10

• 5

SOLUTION

CH

3

COOH( aq ) ⇌ CH 3

COO

• ( aq ) + H

( aq )

Initial (M) 0.100 0 0

Change (M) - x + x + x

Equilibrium (M) (0.100 – x ) x x

K a

=

H

+

CH 3

COO

−

CH 3

COOH

x x

0.100 − x

= 1.75 × 10

− 5

% dissociated = (1.32 × 10

• 3 /0.1) × 100% = 1.32%

Assume that x  0.

x

2 = 1.75 × 10

• 6  x = 1.32 × 10 - 3

SALTS CAN REACT WITH WATER TO BE EITHER ACIDIC, BASIC OR

NEUTRAL.

SALTS OF STRONG ACIDS AND STRONG BASES

NaOH + HCl  NaCl + H 2

O

NaCl( aq )  Na

• Cl

Na

• H 2

O  no reaction and Cl

• • H 2

O  no reaction

a solution of NaCl is neutral.

SALTS OF WEAK ACIDS AND STRONG BASES

CH

3

COOH + NaOH  CH 3

COONa + H 2

O

CH

3

COONa  CH 3

COO¯ + Na

Na

does not react with water, but CH 3

COO

 undergoes hydrolysis

CH

3

COO



• H 2

O  CH

3

COOH + OH



 pH of sodium acetate solution > 7 (basic).

acidic if K a

(cation) > K b

(anion)

neutral if K a

(cation) = K b

(anion)

basic if K a

(cation) < K b

(anion)

Calculate the pH of a 0. 0924 mol dm

• 3 solution of NH 4

Cl at 25

o C, K a

× 10

• 10 .

NH

4

( aq ) + H 2

O( l ) ⇌ NH 3

( aq ) + H 3

O

( aq )

EXAMPLE

SOLUTION

Initial (M) 0.0924 0 0

Change (M) - x + x + x

Equilibrium (M) (0.0924 – x ) x x

From, pH = - log [H 3

O

], pH = 5.

x = 7.26 × 10

• 6 K a

=

NH 3

H 3

O

+

NH

4

+

Calculate the pH of a 0. 154 mol dm

• 3 solution of sodium acetate, K b

× 10

• 10 .

pOH = 5.03, therefore pH = 8.

EXAMPLE

SOLUTION

CH

3

COO

• ( aq ) + H 2

O( l ) ⇌ CH 3

COO

• ( aq ) + OH - ( aq )

Initial (M) 0.154 0 0

Change (M) - x + x + x

Equilibrium (M) (0.154 – x ) x x

x = 9.38 × 10

• 6

EXAMPLE

What is the pH of a solution made by adding 0. 30 mol acetic acid

(HC

2

H

3

O

2

) and 0. 15 mol of sodium acetate (NaC 2

H

3

O

2

) to enough water

to make 1. 0 dm

3 of solution? K a

= 1. 8 x 10

• 5

SOLUTION

CH

3

COOH ⇌ H

• CH 3

COO

Initial (M) 0.30 0 0.

Change (M) - x + x + x

Equilibrium (M) (0.30 – x ) x (0.15 + x )

K a

=

H

+ CH 3

COO

−

CH 3

COOH

= 1.8 × 10

− 5

x = 3.6 × 10

• 5 M = [H

]

pH = - log(3.6 × 10

• 5 ) = 4.

EXERCISE FOR THE IDLE MIND

Practice Exercises: p 571, p 572 and Exercise 15.10 p 602 2

nd Ed.

Practice Exercises: p 709, p 710 and Exercise 18.10 p 746 3

rd Ed.

pH < 7.00: two components - a weak acid and the salt of a weak acid.

CH

3

COOH( aq ) + OH

• ( aq ) ⇌ CH 3

COO

• ( aq ) + H 2

O( l )

WEAK ACID + SALT OF A WEAK ACID

CH

3

COOH + CH

3

COO

• Na

acetic acid sodium acetate

The weak acid is there to react with any OH

• ions which might be introduced into

the system.

RESULT - to convert a strong base (OH

• ) into a weak base (CH 3

COO

• ). Although

the pH will increase due to more base being present, the increase will be

relatively small as the acetate ion is a weak base ( K b

= 5. 7 × 10

• 10 ).

CH

3

COO

• ( aq ) + H 3

O

( aq ) ⇌ CH 3

COOH( aq ) + H 2

O( l )

The salt of the weak acid is there to react with any H 3

O

ions which might be

introduced into the system.

Result - a strong acid (H 3

O

) has been converted into a weak acid ( K a

(CH

3

COOH) = 1. 75 × 10

• 5 ). The pH of the solution will decrease but only by a

small amount.

CH

3

COOH( aq ) + H 2

O( l ) ⇌ CH 3

COO

• ( aq ) + H 3

O

( aq )

This buffer solution can be described by the following equilibrium:

K a

=

H 3

O

+ CH 3

COO

−

CH 3

COOH

EXAMPLE

Calculate the pH of a solution which contains 0. 1046 mol dm

• 3 of CH 3

COOH

and 0. 1247 mol dm

• 3 of sodium acetate.

SOLUTION

We first determine the pH using a method described before.

CH

3

COOH( aq ) ⇌ H

( aq ) + CH 3

COO

• ( aq )

Initial (M) 0.1046 0 0.

Change (M) - x + x + x

Equilibrium (M) (0.1046 – x ) x (0.1247 + x )

EXAMPLE

Calculate the pH of the solution prepared from adding 50. 00 mL of 0. 1011

mol dm

• 3 NaOH to 100. 0 mL of a 0. 1122 mol dm - 3 CH 3

CH

2

COOH solution.

(p K a

CH

3

CH

2

COOH = 4. 88 ).

SOLUTION

CH

3

CH

2

COOH + OH

• → CH 3

CH

2

COO

• • H 2

O

Initial (M) 0.0411 0.

Change (M) - x + x + x

Equilibrium (M) (0.0411 – x ) (0.0337 + x ) x

Before (mol) 0.01122 0.005055 0

After (mol) 0.006165 0 0.

CH

3

CH

2

COOH + H

2

O ⇌ CH

3

CH

2

COO

• • H 3

O

mol dm

• x = 1.61 ×
  • • •  pH = - log(1.61 ×
        • • ) = 4.
• pH = 4.88 + log(0.0337/0.0411) = 4. OR

Then, can write:

K b

=

NH 4

+

OH

−

NH 3

EXAMPLE

Calculate the pH of a solution prepared from 100. 0 mL of 0. 02041 mol NH 3

and 100. 0 mL of 0. 03267 mol NH 4

Cl ( K a

(NH

4

) = 5. 71 × 10

• 10 ).

SOLUTION

NH

4

( aq ) + H 2

O( l ) ⇌ NH 3

( aq ) + H 3

O

( aq )

Initial (mol) 0.03267 0.

Change (mol) - x + x + x

Equilibrium (mol) (0.03267 – x ) (0.02041 + x ) x

Equilibrium (M) (0.1634 – x ) (0.1021 + x ) x

EXERCISE FOR THE IDLE MIND

Practice Exercise p 575and Exercise15.14, 15.16 p 603 2

nd Ed.

x = 9.14 × 10

• 10 mol dm - 3  pH = - log(9.14 × 10 - 10 ) = 9.

OR

pH = 9.24 + log(0.1021/0.1634) = 9.

Practice Exercise p 713 and Exercise 18.14, 18.16 p 746 3

rd Ed.