Physical Chemistry , 1st ed.

Solutions containing ions that have larger absolute charges have greater

coulombic effects affecting their properties. One way to keep track of this is by

defining the ionic strength, I, of the solution:

number of

I

1

2

 

ions

i 1

mizi^2 (8.47)

where ziis the charge on the ith ion. Ionic strength was originally defined in

1921 by Gilbert N. Lewis. Recall that for ionic solutes that do not have a 1 1

ratio of cation and anion, the individual molalities miwill not be the same. The

following example illustrates.

Example 8.10

a.Calculate the ionic strengths of 0.100 mNaCl, Na 2 SO 4 , and Ca 3 (PO 4 ) 2.

b.What molality of Na 2 SO 4 is needed to have the same ionic strength as

0.100 mCa 3 (PO 4 ) 2?

Solution

a.Using equation 8.47, we can find that

INaCl^12 [(0.100m)( 1)^2 (0.100m)( 1)^2 ] 0.100 m

INa 2 SO 4  2 ^1 [( 2 0.100m)( 1)^2 (0.100m)( 2)^2 ] 0.300 m

↑

n  2

ICa 3 (PO 4 ) 2 ^12 [( 3 0.100m)( 2)^2 (20.100m)( 3)^2 ] 1.50 m

↑↑

n  3 n  2

Notice how high the ionic strength gets when the charges on the individual

ions increase.

b.This part asks what molality of Na 2 SO 4 is needed to get an ionic strength

the same as 0.100 mCa 3 (PO 4 ) 2 , which we found in part a to be 1.50 m.We

can set up the INa 2 SO 4 ionic strength expression, but use 1.50 mfor the value

and set the molality as the unknown. We have

INa 2 SO 4 1.50 m^12 [(2 m)( 1)^2 (m)( 2)^2 ]

1.50 m^12 (2m 4 m)  2 ^1  6 m  3 m

Therefore,

m0.500 m

So we need a solution of Na 2 SO 4 with five times the molality to have the same

ionic strength as Ca 3 (PO 4 ) 2. As an exercise, what molality of NaCl would be

needed to have this same ionic strength?

As with any other chemical species, solvated ions also have enthalpies and

free energies of formation, and entropies. From equation 8.23, we can see that

^12 H 2 (g) →H^ (aq) 1e^ E°0.000 V

This is (almost) the formation reaction of H^ (aq) from its elements, and using

the relationship between Eand G, we might suggest that fG[H^ (aq)] 0.

However, this argument presents a problem. First of all, the presence of the

electron as a product is problematic in terms of defining this equation as the

228 CHAPTER 8 Electrochemistry and Ionic Solutions