PHYSICAL CHEMISTRY IN BRIEF

CHAP. 14: DISPERSION SYSTEMS [CONTENTS] 464

the dissociation of water pass to the other part of the system together with the K+ions. The
solution in the first subsystem becomes acidic and that in the other subsystem becomes alkaline.
The equilibrium condition may be written in the form

(aK+aOH−)I= (aK+aOH−)II. (14.9)

Example

A subsystem I contains a high-molecular electrolyte NaR of the molar concentration c 1 =

0. 1 mmol dm−^3. A subsystem II, separated from the subsystem I by a membrane, contains pure

water at the beginning. Determine the equilibrium concentration of hydroxyl ions in both sub-

systems at 25◦C (Kw= 1× 10 −^14 , γi= 1).

Solution

Substituting into (14.9) yields

(c 1 −x)

Kw

x

=x·x.

For the first estimate ofxwe may use the relation

x=^3

√

c 1 Kw=^3

√

1 × 10 −^4 × 10 −^14 = 1× 10 −^6 mol dm−^3.

The exact solution of the equationx^3 +Kwx−c 1 Kw= 0would yieldx= 0. 9966 × 10 −^6 mol cm−^3.

In the second subsystem,(cOH−)II =x= 0. 9966 × 10 −^6 mol dm−^3. In the first subsystem the

concentration of the OH−ions will be(cOH−)II=Kxw= 1. 0034 × 10 −^8 mol dm−^3.

For a high-molecular electrolyte RB, which dissociates into R+and B−, equation (14.9)
rearranges to
(aR+aB−)I= (aR+aB−)II. (14.10)