Common Ion Effect

[danny89]

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PbI2--->Pb + 2I
When solid KI is then added to the solution, why is there an increase concentration of I ions and decrease
concentraion of Pb ions in solution and not a decrease in both I and Pb ions?
This is what I thought: When KI is dissolved, we are left with more I ions (common ion) and K ions (spectator ion). To decrease the concentration of I ions, by Le Chatelier's Principle, there is a leftward shift when I ions combine with Pb ions; thus, decreasing both I and Pb ions.

 

[Sammy1024]

Did the answer say that? Maybe it meant to say that there is twice as much increase in I because for every 2 moles I, 1 mole of Pb forms?

 

[Sammy1024]

It could be that Pb does not dissolve, so there would be an increase in I in the solution but not Pb. It is a solubility rule. "Most Ag+, Pb2+ and Hg22+ salts are insoluble (unless paired with something from group 1 of periodic table"

 

[Entadus]

It's La Chatelier's principle. When you throw in the KI solid, it will dissolve in the water (dissociate) to K+ and I- (the aqueous ions, correct?)

K+ is merely a spectator ion.

However by adding I- we shift the equilibrium of the reaction: (....WHICH WAY???)
PbI2 <---> Pb2+ + 2I-

(PS I cried a little when you wrote a single direction arrow)

This clearly drives [Pb2+] down. It TRIES to bring [I-] down to compensate, but realize that the amount of I- from the KI we added far surpasses any shift in the equilibrium of the lead iodide (KI is soluble, that is, it dissociates completely, whereas the PbI2 is pretty insoluble comparatively. Check the Ksp )