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 percentage of Ag+ remains in the solution?

Since #"AgI"# has a much lower #K_(sp)#, I would expect it to precipitate first. So, the question seems to be asking about the common ion effect due to excess #"I"^(-)# beyond the amount required to precipitate #"AgI"#.

It was found that about #1.25 xx 10^(-7) %# is left of #"I"^(-)#, and #99.995%# is left of #"Ag"^(+)#. But I think the intent of the question is to ask for #"I"^(-)#, not #"Ag"^(+)#.

As far as I can tell,

• the #4.6 xx 10^(-3) %# is the percent of silver cation lost from solution (not the amount remaining in solution).
• the #1.5 xx 10^(-7)%# is closest to the percent of iodide anion remaining in solution, while the #1.5 xx 10^(-9)%# is a distractor for if you tack on a #%# without multiplying by #100#.
• it couldn't make sense to have #0%# of #"Ag"^(+)# in solution because its concentration is so much higher than the #"I"^(-)# introduced to force both precipitations, that the added #"I"^(-)# should hardly have any effect on suppressing #["Ag"^(+)]#.

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"# beyond precipitation of #"AgI"#.

First,

#8.5 xx 10^(-17) = ["Ag"^(+)]["I"^(-)]#

#=> ["I"^(-)]_1 = (8.5 xx 10^(-17))/(0.035) = 2.43 xx 10^(-15) "M"#

Since this is so small, we assume that #["I"^(-)]_2# required to precipitate #"TlI"# is approximately the same #["I"^(-)]_1# plus the #["I"^(-)]_2# required to precipitate #"AgI"#.

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

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

#= ("0.035 M")(2.43 xx 10^(-15) "M" + ["I"^(-)]_2)#

#~~ ("0.035 M")["I"^(-)]_2#

Therefore, to precipitate #"TlI"#, we need:

#["I"^(-)]_2 = (5.5 xx 10^(-8))/("0.035 M") = 1.6 xx 10^(-6) "M"#

At this state, we want to find what concentration #"I"^(-)# would have been formed by the time we re-establish the #"AgI"# equilibrium.

#"AgI"(s) rightleftharpoons "Ag"^(+)(aq) + "I"^(-)(aq)#

#"I"" "-" "" "" "0.035" "" "1.6 xx 10^(-6)#
#"C"" "-" "" "-s" "" "" "-s#
#"E"" "-" "" "0.035-s" "1.6 xx 10^(-6)-s#

Since #["Ag"^(+)]# is much larger than would be at equilibrium in a saturated #"AgI"(aq)# solution in water, we expect nearly all the #"Ag"^(+)# to be leftover, but nearly all the #"I"^(-)# to be gone.

Thus, if we solve the appropriate mass action expression for #s#:

#8.5 xx 10^(-17) = (0.035 - s)(1.6 xx 10^(-6) - s)#

we would get that #s ~~ 1.6 xx 10^(-6) "M"#. Wolfram Alpha would agree. Using that knowledge, we plug in an approximation for #s# only for the #0.035 - s# term to solve for the true value.

#8.5 xx 10^(-17) = (0.035 - 1.6 xx 10^(-6))(1.6 xx 10^(-6) - s)#

#s = {1.6 xx 10^(-6) - (8.5 xx 10^(-17))/(0.035 - 1.6 xx 10^(-6)) }"M"#

#= 1.599999998 xx 10^(-6)#

Yes indeed, the approximation works... We would have had that #["I"^(-)]_(eq) ~~ 2 xx 10^(-15) "M"#. [As a check, we would get #K_(sp) ~~ 7 xx 10^(-17)# for #"AgI"#, which is still within spitting range of the original.]

So now, we find that the percent of iodide anion leftover is:

#color(blue)((1.6 xx 10^(-6) - s)/(["I"^(-)]_i) xx 100%)#

#= (2 xx 10^(-15))/(1.6 xx 10^(-6)) xx 100%#

#= color(blue)(1.25 xx 10^(-7)%)#