Concentrations appearing in the Ka of buffer solutions

I've just finished studying buffer solutions and I think I understood more or less how they work, but there's something I can't make sense of:

In the $K_a$ of equilibrium of a buffer solution made from Sodium acetate and acetic acid, we just have the concentrations of acid, $\ce{H+}$ and acetate ion. So, why isn't the concentration of $\ce{Na+}$ taken into consideration in the $K_a$? Why just the others?

Moreover as $[\ce{Na+}] = [\ce{CH3COO-}]$ (the acetate ion in solution is almost only the one dissociated from the salt) shouldn't in the Henderson-Hasselbach solution occur a $a^2$ factor? So in the numerator having $[\ce{CH3COO-}]^2$ and no just a single one?

Thank you

Because Na+ is a spectator ion in this situation.

$$\ce{Na+ + CH3COOH <=> Na+ + CH3OO- + H+}$$

In other words, Na+ is on both sides of the chemical equation, so [Na+] would appear in both the numerator and denominator of the equilibrium constant if Na+ is included.

$[\ce{Na+}][\ce{CH3COO-}][\ce{H+}]/[\ce{Na+}][\ce{CH3COOH}] = [\ce{CH3COO-}][\ce{H+}]/[\ce{CH3COOH}]$

• I don't have any objections to make here, but I'd step away if people deem this answer a bit low quality. I mean, short answers are generally just not acceptable. Sep 22 '16 at 16:55
• @Rubisco my female teacher would say "your answers should be like a skirt: long enough to cover the subject, but short enough to be interesting". Sep 22 '16 at 17:01
• I agree with the sentiment, but something gets triggered when people see that they don't need to scroll down to see the whole answer. And I don't have the rep to vote to undelete if anything happens. Sep 22 '16 at 17:12
• I might go a little deeper and explain why sodium is a spectator ion - e.g. what's the critical reaction under discussion and why sodium is not involved. (At the very least, the questioner would be helped by a short discussion of how to identify spectator ions in similar chemical scenarios.)
– R.M.
Sep 22 '16 at 18:57
• So the answer below about "not changing much" is wrong? Also I can't understand why is Na a spectator ion while others aren't in this case as there are no major reactions going on...I see things Na does also done by other ions? Thank you Sep 24 '16 at 9:38

Chemistry is based on the good-enough as opposed to theoretical perfection. Although $\ce{Na+}$ doesn't affect the pH directly, it does change the ionic strength of the solution. The pKa of acetic acid would be affected slightly by the ionic strength of the solution as a secondary correction. Textbook problems ignore such factors because it leads to calculation complications that require extra equations that in turn require iterative computer programs to solve.