# 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 $\setminus \approx 1.4 \times {10}^{-} 4$ and not $\times {10}^{-} 5$...

Aug 9, 2018

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

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

${\text{AgI"(s) rightleftharpoons cancel("Ag"^(+)(aq)) + "I}}^{-} \left(a q\right)$, ${K}_{s p} = 1.5 \times {10}^{- 16}$
$\underline{{\cancel{{\text{Ag"^(+)(aq)) + 2"NH"_3(aq) -> "Ag"("NH}}_{3}}}_{2}^{+} \left(a q\right)}$, ${K}_{f} = 1.5 \times {10}^{7}$
${\text{AgI"(s) + 2"NH"_3(aq) -> "Ag"("NH"_3)_2^(+) + "I}}^{-} \left(a q\right)$

For this, the composite equilibrium constant is the product:

$\beta = {K}_{s p} {K}_{f}$

$= 2.25 \times {10}^{- 9} = \left({\left[{\text{Ag"("NH"_3)_2^(+)]["I"^(-)])/(["NH}}_{3}\right]}^{2}\right)$

The $\text{3.0 M}$ ${\text{NH}}_{3}$ then serves as the initial concentration.

${\text{AgI"(s) + 2"NH"_3(aq) -> "Ag"("NH"_3)_2^(+) + "I}}^{-} \left(a q\right)$
$\text{I"" "-" "" "3.0" "" "" "" "0" "" "" "" "" } 0$
$\text{C"" "-" "-2x" "" "" "+x" "" "" "" } + x$
$\text{E"" "-" "3.0-2x" "" "" "x" "" "" "" "" } x$

Thus,

$2.25 \times {10}^{- 9} = {x}^{2} / {\left(3.0 - 2 x\right)}^{2}$

And this is easily solvable without the quadratic formula.

$4.74 \times {10}^{- 5} = \frac{x}{3.0 - 2 x}$

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

$4.74 \times {10}^{- 5} \approx \frac{x}{3.0}$

Therefore,

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

That is then the molar solubility of ${\text{I}}^{-}$, and hence $\text{AgI}$, as they are $1 : 1$.