Writing Ksp when the ions are on the left of the equilibrium equation? For the equation:
$\ce{Ca^{2+} {(aq)} + ^-OOC-COO- {(aq)} &lt;=&gt; Ca(OOC-COO) {(s)}}$
My book gives $K_{sp} = \pu{3.0e-9 M} = [\ce{Ca^{2+}}] \times [\ce{^-OOC-COO-}]$.
However, as per Chem.Libretexts, for any  $K_c$, the right hand side substances go in the numerator and the left hand side substances go in the denominator.

&quot;It is really important to write down the equilibrium reaction whenever you talk about an equilibrium constant. That is the only way that you can be sure that you have got the expression the right way up - with the right-hand substances on the top and the left-hand ones at the bottom.&quot;

So for $K_{sp}$, why is it not the same?
I guess intuitively for a $K_{sp}$ I would probably put the $K_{sp}$ = product of the ions. But is there any other exceptions to when the above rule from Chem.Libretexts is not followed (in regards to K values)?
By convention, the solid is on the left of the side dissolving reaction. If you do it the $K_{sp}$ has the correct form.

The value of the $K_{sp}$ given above suggests that it conforms to this convention.