Common-ion example 1 in notes

[Zellmer, Robert]

Students often ask many questions about common-ion problems,
mainly how we can do them by treating them as if we add the substances
simultaneously to the beaker (when in reality that's not what is actually
happening).  I've attached an explanation as to why we can do these problems
as I did in class.  I did the problem first as it would be done following what
is actually physically occurring (adding the common ion to an existing
equilibrium causing it to shift back to the left).  I showed this page in lecture
and told everyone not to write down what was on that page.  Then I did it
assuming we add everything simultaneously to the beaker.  You will see
we get the same results, but the 2nd way is a whole lot easier.

The easiest way to think of what I told you to do in the NH3/NH4^+ example
is the following.  Just picture starting the reaction with the reactant (NH3)
and one of the products (NH4^+ in my example) rather than just starting
with the reactant.  So, we started the reaction with NH3 and NH4^+ and no
OH-.  Since one of the products is still missing the reaction must go to
the right (in the forward direction).  Also, since we started with one of the
products the reaction will not go as far in the forward direction as it would
if we started w/o any products.  That's why less OH- is produced from the
reaction of NH3 with H2O in the presence of the common ion, NH4^+.

By the way, we could have done a common-ion problem by adding OH- to
an NH3 solution.  That means we would have started the reaction (and ICE
table) with NH3 and OH- and no NH4^+.  The reaction would still then go
to the right (in the forward direction), but again not as much as it would if
we started with no products.

Therefore, a common-ion problem boils down to starting a reaction with
not just a reactant but also one of the products.


Dr. Zellmer
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