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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">Students often ask many questions about common-ion problems,<br>
mainly how we can do them by treating them as if we add the substances<br>
simultaneously to the beaker (when in reality that's not what is actually<br>
happening). I've attached an explanation as to why we can do these problems<br>
as I did in class. I did the problem first as it would be done following what<br>
is actually physically occurring (adding the common ion to an existing<br>
equilibrium causing it to shift back to the left). I quickly showed this page<br>
and actually told everyone to not write down what was on that page. Then<br>
I did it assuming we add everything simultaneously to the beaker. You will<br>
see we get the same results but the 2nd way is a whole lot easier.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">The easiest way to think of what I told you to do in the NH3/NH4^+ example<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">is the following. Just picture starting the reaction with the reactant (NH3)<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">and one of the products (NH4^+ in my example) rather than just starting<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">with the reactant. So we started the reaction with NH3 and NH4^+ and no<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">OH-. Since one of the products is still missing the reaction has to go to<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">the right (in the forward direction). Also, since we started with one of the<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">products the reaction will not go as far in the forward direction as it would<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">if we started w/o any products. That’s why less OH- is produced from the<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">reaction of NH3 with H2O in the presence of the common ion, NH4^+.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">By the way, we could have done a common-ion problem by adding OH- to<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">an NH3 solution. That means we would have started the reaction (and ICE<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">table) with NH3 and OH- and no NH4^+. The reaction would still then go<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">to the right (in the forward direction), but again not as much as it would if<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">we started with no products.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">Therefore, a common-ion problem boils down to starting a reaction with<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">not just a reactant but also one of the products.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-family:"Arial",sans-serif"><br>
<br>
Dr. Zellmer <o:p></o:p></span></p>
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