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
2 2
+
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 of α is
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 × 10 9 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 requirement 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).