Question
Consider a solution of 8.5 g NaCl and 1.7 g glucose per liter. What would be (a) the
freezing point of the solution and (b) the ionic strength of the freeze-concentrated
solution at 128 C? You may assume that the glucose does not crystallize. Consult
also Section 2.3.
Answer
(a) According to Eq. (2.16); the freezing point depression of a solution is given by
103 lnxW(in K), wherexWis the mole fraction of water. A kg of solution contains
8.5/58.5¼0.145 mole NaCl, 17/180¼0.0944 mole glucose, and 974.5/18.015¼54.1
mole water. Assuming that NaCl is fully dissociated, we obtainxW¼0.993, which
yields a freezing point of0.72 8 C. (b) Assuming that glycerol will act like glucose,
and taking into account that the molar mass of glucose is twice that of glycerol, the
curve noted 1 in Figure 16.14 would apply to the present solution. This then implies
that at 128 C the concentration factor equals 11.4. A kg of concentrated solution
then contains 709 g of water and 69.9 g¼1.66 mole of NaCl. The salt molality then is
1.66/0.709¼2.34. This is not the ionic strength of the solution, since substantial
association of Naþand Cloccurs at high concentration. Rewriting Eq. (2.21) for
the association constantKA(¼1/KD) yields
KA¼
½NaCl
g^2 +½Naþ½Classuming the activity coefficient for undissociated NaCl to equal unity. According to
Table 2.4, the association constant for NaCl equals about 1 L?mol^1 , hence also
about 1 kg?mol^1 , and according to Figure 2.11, the ion activity coefficientg+for
NaCl will equal about 0.7. Solution of the equation then yields [Naþ]¼[Cl]¼1.4
mole per kg water. This is also the ionic strength.
The reader will by now be aware that calculation of the data asked is not
simple and could only be done because essential basic quantities are given in the
book. It is therefore advisable to find an independent check of the result. This can be
done via the water activity as a function of temperature given in Figure 16.12b. It can
readily be calculated that the mole fraction of water at 128 C is 0.89 (you may check
this). For a not too concentrated and simple solution as considered, we have
xW&aW, and the figure gives indeedaW¼0.89 at 128 C.