1
$\begingroup$

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

If $\ce{Ca^{2+}(aq)}$ is removed, the reaction shifts to the left to produce more $\ce{Ca^{2+}(aq)}$ to compensate.

But when the reaction shifts to the left, isn't a side effect that more $\ce{^-OOC-COO^-}$ is also produced (as when $\ce{Ca(OOC-COO)(s)}$ dissociates it produces BOTH $\ce{Ca^{2+}(aq)}$ and $\ce{ ^-OOC-COO^- }$, and not $\ce{Ca^{2+}(aq)}$ by itself?

So if there is more $\ce{^-OOC-COO^-}$, won't the reaction want to shift to the right?

So wouldn't these two (i.e. wanting to shift left and shift right) counteract and the reaction not shift anywhere? I think there is something about Le Chateliers's that I have misunderstood, so please point it out if you see it.

$\endgroup$
2
$\begingroup$

The thing to be careful about with Le Chatelier's principle is that it relates to equilibria :

Le Châtelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium.

Le Châtelier's Principle

Now when you say in the above "But when the reaction shifts to the left, isn't a side effect ..." note you can't use Le Chatelier's principle here again so fast because the system has not reached an equilibrium yet, it is just moving towards one.

It is likely that the next thing you will study is how to actually calculate the new equilibrium conditions. This gives usually a deeper understanding of how changes in various conditions make the reaction move forward/backward establishing new equilibrium conditions.

$\endgroup$
  • $\begingroup$ It's quite likely that it will overshoot as the OP suggests. However, it will gradually reach equilibrium. $\endgroup$ – DHMO Jan 22 '17 at 11:52
1
$\begingroup$

According to Wikipedia, regarding the Law of Mass Action:

In chemistry, the law of mass action is the proposition that the rate of a chemical reaction is directly proportional to the product of the activities or concentrations of the reactants. It explains and predicts behaviors of solutions in dynamic equilibrium. Specifically, it implies that for a chemical reaction mixture that is in equilibrium, the ratio between the concentration of reactants and products is constant.

So the key with respect to your question is that the equilibrium will shift in a manner that is proportional to the product of the reactants.

However, I think your specific example needs a bit of a clarification. As written, you are suggesting removing only positively charged species, plucking out just the $\ce{Ca^2+}$ ions without adding more cations or also removing anions. If you did this through some sort of ion-exchange technique, you would have to consider whether the cation replacing $\ce{Ca^2+}$ would also react with your other reactant or if it would purely be a spectator ion. The discussion of the Law of Mass Action above assumes the latter.

$\endgroup$
1
$\begingroup$

Calcium oxalate is relatively unsolvable. So there will be a $\text{K}_{\text{sp}}$ such that

$\text{K}_{\text{sp}}$ = $\ce{[Ca^{2+}] [^-OOC-COO^-]}$

So if the calcium oxalate is dropped into a sodium chloride solution, then

$\ce{[Ca^{2+}] \approx [^-OOC-COO^-]}$

It isn't quite true because a bit of $\ce{HOOC-COO^-}$ will form. Let's ignore this factor which will be small. Let's define $x$ as:

$x = \sqrt{\text{K}_{\text{sp}}} = \ce{[Ca^{2+}]_i = [^-OOC-COO^-]_i}$

where "i" stands for initial. Now we take out our magic tweezers and remove half the calcium ions and twice that number of chloride ions. After all the magic tweezers aren't magic enough to leave the solution charged.

But now more $\ce{Ca^{2+}}$ must dissolve as well as more $\ce{^-OOC-COO^-}$. But we know that

$0.5x = \ce{[^-OOC-COO^-]_2 - [Ca^{2+}]_2}$

and that

$\text{K}_{\text{sp}}$ = $\ce{[Ca^{2+}]_2 [^-OOC-COO^-]_2}$

So we have two equations with two unknowns and the new equilbrium conditions can be solved.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.