Question

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?

Select one:
a. 1.5 x 10-9 %
b. 0%
c. 4.6 x 10-3 %
d. 1.5 x 10-7 %

I am not sure with the final result

Answer 1

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)%)`