Physical Chemistry Third Edition

272 6 The Thermodynamics of Solutions

Solution

I

1

2

[

m

(

Na+

)

(+ 1 )^2 +m(Ca^2 +)(+2)^2 +m(Cl−)(−1)^2

]



1

2

(

0 .100 mol kg−^1

)

+

(

0 .200 mol kg−^1

)(

22

)

+

(

0 .500 mol kg−^1

)

 0 .700 mol kg−^1

Notice the large contribution of a multiply charged ion such as Ca^2 +.

Exercise 6.17

Calculate the ionic strength of a solution that is 0.150 mol kg−^1 in K 2 SO 4 and 0.050 mol kg−^1

in Na 2 SO 4. Assume complete dissociation.

Exercise 6.18

Show that for water at 298.15 K, the value ofαis

α 1 .171 kg^1 /^2 mol−^1 /^2 (in water, 298.15 K) (6.4-25)

The density of water at 298.15 K is equal to 997.14 kg m−^3.

Exercise 6.19

Show that for water at 298.15 K,

β 3. 281 × 109 kg^1 /^2 mol−^1 /^2 m−^1 (in water, 298.15 K) (6.4-26)

The density of water at 298.15 K is equal to 997.14 kg m−^3.

We must convert Eq. (6.4-21) to an equation for the activity coefficient of a neutral

electrolyte. Using the definition of the mean ionic activity coefficient and the require-

ment of electrical neutrality, we write Eq. (6.4-21) once for the cation and once for the

anion and combine them using Eq. (6.4-6) to obtain

ln(γ±)−z+|z−|

αI^1 /^2

1 +βaI^1 /^2

(6.4-27)

Exercise 6.20

Carry out the algebraic steps to obtain Eq. (6.4-27).