PHYSICAL CHEMISTRY IN BRIEF

CHAP. 11: ELECTROCHEMISTRY [CONTENTS] 370

Example

Calculate the ionic strength of

a) a solution of FeCl 3 of the molalitym= 0.1 mol kg−^1 ,

b) a solution containing KCl of the molality 0.1 mol kg−^1 and CuCl 2 of the molality 0.2 mol kg−^1.

Solution

a) From the material balance it follows that the molality of the ferric ions is 0.1 mol kg−^1 , the

molality of the chloride ions is 3×0.1 = 0.3 mol kg−^1. From equation (11.41) we obtain

I=

1

2

(

0. 1 × 32 + 0. 3 × 12

)

= 0. 6 mol kg−^1.

b) From the material balance it follows that the molality of the potassium ions is 0.1 mol kg−^1 ,

the molality of the cupric ions is 0.2 mol kg−^1 , and the molality of the chloride ions is

0 .1 + 2× 0 .2 = 0. 5 mol kg−^1.

From equation (11.41) we obtain

I=

1

2

(

0. 1 × 12 + 0. 2 × 22 + 0. 5 × 12

)

= 0. 7 mol kg−^1.

11.4.4 Debye-H ̈uckel limiting law

Based on a theoretical model of a dilute solution of a strong electrolyte, Debye and H ̈uckel
derived a relation for the calculation of the activity coefficient of an ioni

lnγi=−A z^2 i

√

I , (11.43)

whereziis the ion charge number andIis the ionic strength of the solution. For the parameter
Awe have

A=

e^3 NA^2

√

2 /ρ 1

8 π( 0 rRT)^3 /^2

, (11.44)