The Ksp for AgI is 8.5 x 10-17. The Ksp for TlI (thallium iodide) is 5.5 x 10-8. A one litre solution contains 0.035 mol L-1 Ag+ ion and Tl+ ion. When TlI begins to precipitate, what can be said about the extent of precipitation of AgI?

If you want to know about the EXTENT of reaction, you need #Q#.

And you should find that #"AgI"# completely precipitates.

We have that at #25^@ "C"# and #"1 atm"#,

• #K_(sp)("AgI") = 8.5 xx 10^(-17)#
• #K_(sp)("TlI") = 5.5 xx 10^(-8)#

We have #"0.035 mol/L Ag"^(+)# and #"0.035 mol/L Tl"^(+)#, and we want to first find out what concentration of #"I"^(-)# causes precipitation of #"TlI"#, i.e.

For what #[I^(-)]# is #Q_(sp)(TlI) = 5.5 xx 10^(-8) = K_(sp)(TlI)#?

#"TlI"(s) rightleftharpoons "Tl"^(+)(aq) + "I"^(-)(aq)#

#5.5 xx 10^(-8) = ["Tl"^(+)]["I"^(-)]#

#= ("0.035 M")["I"^(-)]#

Therefore,

#["I"^(-)]_(eq) = (5.5 xx 10^(-8))/("0.035 M") = 1.6 xx 10^(-6) "M"#

causes #Q_(sp) > K_(sp)# and leads #"TlI"# to precipitate. So where is #"AgI"# at? Well, we know that:

#Q_(sp)("AgI") = ["Ag"^(+)]["I"^(-)]#

#= ("0.035 M")(1.6 xx 10^(-6) "M")#

#= 5.5 xx 10^(-8)# #">>"# #8.5 xx 10^(-17)#

When #Q > K#, the reaction is skewed towards the products, and so, Le Chatelier's principle states that equilibrium shifts towards the reactants (i.e. the solid!).

This is much larger than #K_(sp)# for #"AgI"#, so #"AgI"# will certainly precipitate.