How do we formulate #"iodide ion"# in aqueous solution?

1 Answer
Apr 20, 2017

Answer:

Iodide anion is treated as #I^-# in aqueous solution..........

Explanation:

Elemental iodine is binuclear, i.e. we treat it as #I_2#, but this species can be reduced to give water-soluble ions, #I^-#........ This species can react with (some) heavy metal ions to give insoluble precipitates.

And thus we interrogate the equilibrium:

#AgI(s)rightleftharpoonsAg^(+) + I^(-)#

This is an equilibrium reaction, and typically we write the #"solubility product"#, where #K_(sp)=[Ag^+][I^-]=8.1xx10^-17# under standard conditions..........

And so, let us call the solubility of #AgI#, a yellow salt, #S#, and thus............

#K_"sp"=S^2#, and thus we can make a quantitative measurement of solubility, #S=sqrt(K_"sp")=sqrt(8.1xx10^-17)=9xx10^-9*mol*L^-1#.

So, what do you have to learn? Iodide, like other halides, exists in aqueous solution as the ion, #I^-#, #Cl^-#, #Br^-#, #F^-# etc. The addition of solid iodine to solutions of iodide CAN produce triiodide species such as #I_3^(-)#?