Concentrations appearing in the Ka of buffer solutions
Because Na+ is a spectator ion in this situation. 

$$ \ce{Na+ + CH3COOH &lt;=&gt; Na+ + CH3OO- + H+}$$

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.  

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