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For the equation:

$\ce{Ca^{2+} {(aq)} + ^-OOC-COO- {(aq)} <=> Ca(OOC-COO) {(s)}}$

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.

"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."

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)?

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For the equation:

$\ce{Ca^{2+}_{(aq)} + ^-OOC-COO-_{(aq)} <=> Ca(OOC-COO)_{(s)}}$

$K_{sp} = \pu{3.0e-9 M} = \ce{[Ca^{2+}] \cdot [\ce{^-OOC-COO-}]}$

The equation for the solubility product, Ksp, is correct.

The reason $\ce{Ca(OOC-COO)}$ is not in the equilibrium equation is because it is a solid and in another phase as indicted by the (s) subscript. The calcium oxalate is in the solid phase not the liquid phase of the solution. The gist is it doesn't matter how much solid calcium oxalate is in contact with the solution. It could be 1 milligram, or 53.6 kilograms, and the solubility constant would have the same value.

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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.

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  • $\begingroup$ Thanks for the response. How about if my exam question gives a certain Ksp value but has the equation with the solids on the right side and the aqueous products on the left side? Would I need to write 1/the given Ksp=[product][product]? They did not do that in the example question in the book. $\endgroup$ – K-Feldspar Sep 10 '16 at 9:52
  • $\begingroup$ But based on chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/…, if it was Kc value I would have to do that (if I interpreted it correctly) as it says.. $\endgroup$ – K-Feldspar Sep 10 '16 at 9:53
  • $\begingroup$ "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." $\endgroup$ – K-Feldspar Sep 10 '16 at 9:53
  • $\begingroup$ The inverse of $\pu{3e-9 M}$ would not be a reasonable value. $\endgroup$ – aventurin Sep 10 '16 at 10:05

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