You have to consider the overall ionization scheme. You have two

sources of protons going on here, the water and the acetic acid.

Say you have one mole of CH3COONa and 1 mole of CH3COOH mixed in with

100 liters of water. Then you have in solution the following species:

a moles per liter of CH3COOH

b moles per liter of CH3COO(-)

c moles per liter of H(+)

d moles per liter of OH(-)

e moles per liter of Na(+)

The Na(+) from the sodium acetate does not react with anything, so e =

0.01. That leaves four unknowns governed by the following equations:

a + b = 0.02 to balance the acetate ions

c + 0.01 = b + d to balance charges

b*c = K_a*a from acetic acid dissociation

c*d = K_w from water dissociation

I won't go through the mathematical detailsof the solution to this

nonlinear system, but what happens is that because K_w is so small, c

and d are both much smaller than 0.01 and then you can assume that the

second equation is reducible to b = 0.01. Then a = 0.01 and the third

equation leads to the "standard" buffer pH.

But now suppose that instead of 100 liters of water there are 10^6 or

10^9 liters. Then 0.01 is correspondingly replaced by 10^(-6) or

10^(-9) above, and these become small enough so that c and d are no

longer negligible in the second equation even with their product K_w as

small as 10^(-14). In other words, dissociation of the water is

becoming important, competing with dissociation of the acetic acid for

generation of the protons. That's when the standard buffering

approximation breaks down and the pH begins to reflect that of pure

water.

Real lakes have other buffers acting from their impurities (natural and

pollution). But that is another story.

--OL