Page 191 - Modern physical chemistry

8.4 Key Thermodynamic Considerations 183

mHA =0,
Y2 = 1.00 x 10-6 m.
Substituting these into fonnula (8.44) yields

6
m H + = 1.00 x 10-6 - -0.90 x 10- (-0.046 x 10-6) = 1.00 x 10-6 - 0.040 x 10- 6 = 0.960 x 10- 6 m.
1.046 x 10-6
Employing this as the new X 2 , we obtain

Y2 = -0.002 x 10-6 m.
Using the same Xl and Yl as before, we now have

mw = 1.00 x 10-6 - -0.040 x 10-6 (-0.046 x 10-6) = 1.00 x 10-6 - 0.042 x 10- 6 = 0.958 x 10-6 m.
1.044 x 10-6
This checks the result found by the first method.

8.4 Key Thermodynamic Considerations
The standard state for a solid or liquid is the pure substance under 1 bar pressure at
the chosen temperature. The standard state for a solute here is the hypothetical 1 molal
solution obeying Henry's law at 1 bar total pressure and the chosen temperature. Thus,
we are employing an activity system based on the infinitely dilute solution as the refer-
ence state, with concentrations in the molality system.
A difficulty arises with ions, however. For, a cation cannot be fonned apart from the
fonnation of an anion or an electron. Nor can an anion be fonned apart from the for-
mation of a cation or reaction with an electron. So a person can arbitrarily assign ther-
modynamic properties to a chosen ion or to the electron. By convention, the hydrogen
ion is taken. One assigns to H+ (aq) zero SJ, !:J.lJO f , and I1Go f •
The results from many experiments with condensed phases are sununarized in table 8.3.

Example 8.8
From table 8.3, determine ~, !:J.lJO, log K rm , and Kym for the reaction

HCOOH (aq) ( ) H+ (aq) + HCOO- (aq).

With the numbers in the table, we find that

Also
MlO = Mlo + Mlo - MlP HCOOH = 0.00 - 425.55 + 425.43 = 0.12 J,
r H+ r HCOO-
and

10gKym = (IOgKr) +(IOgKr) -(logKr) +0.00+61.49-65.22=-3.73,
H+ HCOO- HCOOH
whence
Kym = 1.86 x 10-4 m.