PHYSICAL CHEMISTRY IN BRIEF

CHAP. 11: ELECTROCHEMISTRY [CONTENTS] 376

Equation (11.53) can then be written in the form

K=

α^2 c γ±^2

(1−α)γHA

1

cst

, (11.54)

wherecis the initial concentration of the acid.

Example

The dissociation degree of acetic acid of the concentrationc= 0.0025 mol dm−^3 isα= 0.0825.

Calculate the dissociation constant of the acid

a) given thatγHA=γ±= 1,

b) given thatγHA= 1 andγ±= 0.983, which is the estimate from relation (11.45).

c) Compare the two results.

Solution

Substituting into relation (11.54) yields

a)

K=

0. 08252 × 0. 0025 × 12

(1− 0 .0825)× 1

= 1. 85 × 10 −^5 ,

b)

K=

0. 08252 × 0. 0025 × 0. 9832

(1− 0 .0825)× 1

= 1. 79 × 10 −^5.

c) If we neglect the effect of the activity coefficients, the error in the dissociation constant will

be about 3 per cent in this case.