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:

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

• But since $\ce{NaCl}$ is soluble shouldn't you write it as $Na\mbox{C}l_{\left( aq \right)}$? – 1110101001 Mar 13 '14 at 3:45
• But then we're not talking about sodium chloride's solubility product constant, we're talking about its dissociation constant. Also, there is a limit to sodium chloride's solubility in water (359 g/L according to Wikipedia) – Ben Norris Mar 13 '14 at 10:16