Question

Calculate the molar solubility of `\sf{AgI}` in a 3.0 M `\sf{NH_3}` solution, if the `\tt{K_(sp)}` is `\tt{1.5xx10^-16}` and the `\tt{K_f}` for `\tt{[Ag(NH_3)_2]^(+)}` is `\tt{1.5xx10^7}`?

I did everything pretty much the way the picture shows, but I got `\approx1.4xx10^-4` and not `xx10^-5`...

Answer 1

`["I"^(-)]_(eq) = 1.42 xx 10^(-4) "M"`

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Well, what I would do first is find the new equilibrium reaction that now occurs.

`"AgI"(s) rightleftharpoons cancel("Ag"^(+)(aq)) + "I"^(-)(aq)`, `K_(sp) = 1.5 xx 10^(-16)`
`ul(cancel("Ag"^(+)(aq)) + 2"NH"_3(aq) -> "Ag"("NH"_3)_2^(+)(aq))`, `K_f = 1.5 xx 10^7`
`"AgI"(s) + 2"NH"_3(aq) -> "Ag"("NH"_3)_2^(+) + "I"^(-)(aq)`

For this, the composite equilibrium constant is the product:

`beta = K_(sp)K_f`

`= 2.25 xx 10^(-9) = (["Ag"("NH"_3)_2^(+)]["I"^(-)])/(["NH"_3]^2)`

The `"3.0 M"` `"NH"_3` then serves as the initial concentration.

`"AgI"(s) + 2"NH"_3(aq) -> "Ag"("NH"_3)_2^(+) + "I"^(-)(aq)`
`"I"" "-" "" "3.0" "" "" "" "0" "" "" "" "" "0`
`"C"" "-" "-2x" "" "" "+x" "" "" "" "+x`
`"E"" "-" "3.0-2x" "" "" "x" "" "" "" "" "x`

Thus,

`2.25 xx 10^(-9) = x^2/(3.0 - 2x)^2`

And this is easily solvable without the quadratic formula.

`4.74 xx 10^(-5) = x/(3.0 - 2x)`

We can see `beta` is small, so the small x approximation applies:

`4.74 xx 10^(-5) ~~ x/3.0`

Therefore,

`color(blue)(x ~~ 1.42 xx 10^(-4) "M" -= ["I"^(-)]_(eq))`

That is then the molar solubility of `"I"^(-)`, and hence `"AgI"`, as they are `1:1`.