Page 189 - Modern physical chemistry

8.3 Conditions Determining Equilibria 181

But in the aqueous solution at 25° C, we have

Consequently, we find that

2)
2 [mH+]( mw mow = 1.013 x 10 1.008 x 10
mH + = -- ) ( X -14 m = 1.021 x 10 -13 2
m,
mow
whence

ExampleB.7

Calculate m H+ in 1.00 x 10- 6 m acetic acid.
In this dilute solution we have to consider both equilibria

simultaneously. Here HA represents the acetic acid molecule, A the acetate ion.
The 1.00 x 10"" mol acetic acid per 1000 g H 20 appears as HA and A-; there is no other
contribution to m A -. So
1.00 x 10-6 m =mHA +m A -.
From the electrical neutrality condition, we have

And governing the equilibria, we have

mH+mA - = 1.754 x 10-5 m,
mHA

2
mw mow = 1.008 x 10- 14 m .
Let us rearrange these simultaneous equations in a set of type (8.37) - (8.40):