1
$\begingroup$

I was wondering why

$$\ce{Pb2+ + 2I- <=> PbI2}$$

occurs in low $[\ce{I-}]$, however when $[\ce{I-}]$ is in excess, $\ce{PbI4^2-}$ is produced with

$$\ce{Pb2+ + 4I- <=> [PbI4]^2-}$$

Why is the formation of complex ion "favoured"? Furthermore, since the formation of insoluble precipitate will reduce entropy, wouldn't the formation of insoluble precipitate be favoured?

$\endgroup$
  • $\begingroup$ @ Ong What do you mean by "Why is there no formation of complex ion "favoured"?" $\endgroup$ – Mockingbird Aug 4 '17 at 8:11
  • $\begingroup$ @Mockingbird Hi there Mockingbird, my auto correct screwed up my sentence a bit. To correct myself, I wanted to know why formation of soluble tetraiodo plumbate(ii) ion complex is more observed than the formation of insoluble lead (ii) iodide (ie why the overall effect of added $\ce{I-}$ is the dissolving of ppt) $\endgroup$ – Ong Hai Xiang Aug 4 '17 at 9:21
  • $\begingroup$ When $\ce{Cl^-}$ is in greater concentration, there are just enough ions to favor the 2nd reaction. $\endgroup$ – Mockingbird Aug 4 '17 at 10:07
  • $\begingroup$ @Mockingbird is there any principle behind this? $\endgroup$ – Ong Hai Xiang Aug 4 '17 at 11:48
  • $\begingroup$ Well, the rule of thumb is when you need to have grater number reactants for some reaction the greater concentration will work better. $\endgroup$ – Mockingbird Aug 4 '17 at 14:40

Your Answer

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

Browse other questions tagged or ask your own question.