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. – M.A.R. ಠ_ಠ 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". – DavePhD 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. – M.A.R. ಠ_ಠ 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 – Nur Bedeir 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.