Page 206 - Modern physical chemistry

198 Equilibria in Condensed Phases

gives us
VI
v2
RTlna=RTlnal +RTlna2 , [8.115]
whence
[8.116]
The activity of a completely ionized electrolyte equals the activity of the cation raised to
a power equal to the number of cations times the activity of the anion raised to a power
equal to the number of anions in the formula of the electrolyte.

Example 8.12

A membrane is permeable to sodium ions and to chloride ions but not to an organic
ion R- . Calculate the equilibrium concentrations if initially 100 rnl of a solution 0.100 M
in NaCI and 0.0100 M in NaR were separated by the membrane from 1.00 1 of water.
Neglect any movement of water and consider the molar activity coefficients to equal 1.
Let [Na+h and [Cl-lel) represent the sodium and chloride ion concentrations in phase
(1), [Na+l(2) and [Cq(2) the sodium and chloride ion concentrations in phase (2). Initially
we have

[Na+ ](1) = 0.110 M, [CI-](1) = 0.100 M,

[Na+ ](2) = 0 M, [CI-](2) = 0 M,

To preserve electrical neutrality in both phases, the amount of Na+ migrating must
equal the amount of Cl" migrating. Let x equal this amount in moles when equilibrium
has been established. If VI is the volume of phase (l) and V 2 the volume of phase (2), we
then have
[Na+]() =0.110-"£= 0.110-~ =0.110-10x,
1 ~ Q1

[ CI-]( ) = 0.100-"£ = 0.100-~ = 0.100-10x,
1 ~ Q1

[Na+l2)=~ =~=x, [CI-](2)=~ =~=x.

For electrolyte NaCI, the cation and the anion are singly charged,
v+=v_=l.
So formulas (8.107) and (8.116) yield

Setting the activities equal to the corresponding concentrations,

and introducing the equilibrium expressions already constructed leads to
(0.110 -lOx)( 0.100 -lOx) = xx,