What does the equilibrium equation look like when calculating the solubility from Ksp?

For example, given a reaction


$\ce{NaCl -> Na+ + Cl-}$


and its $Ksp$ value, I know that you can set up
$$Ksp\; =\; \left[ \ce{Na^{+}} \right]\left[ \ce{Cl^{-}} \right]$$ and solve from there.


My question is why do you not put $\ce{[NaCl]}$ in the denominator, as isn't that part of the equilibrium expression also?




The reason why $K\_{sp}$ does not include $[\ce{NaCl}]$ is because it is solid:


$$\ce{NaCl(s) -> Na+ (aq) + Cl- (aq)}$$


The equilibrium constant $K$ would include $[\ce{NaCl}]$:


$$K=\dfrac{[\ce{Na+}][\ce{Cl-}]}{[\ce{NaCl}]}$$


However, the concentration of sodium chloride in solid sodium chloride is constant, thus we can write an equation for the observed rate constant that is based only on the variable concentrations. $K\_{sp}$ is this observed rated constant:


$$K\_{sp}=K\_{obs}=K[\ce{NaCl}]=[\ce{Na+}][\ce{Cl-}]$$


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