# 0.10 mol of AgNO_3 is dissolved in "1.00 L" of "1.00 M "NH_3. If "0.010 mol" of NaCl is added, will "AgCl (s)" precipitate?

## Given: ${K}_{f} = 1.6 \times {10}^{7}$ ${K}_{s p} = 1.8 \times {10}^{-} 10$ Apparently the calculated ${Q}_{s p}$, $9.8 \times {10}^{-} 11$, is less than given ${K}_{s p}$. What does this mean for the matter of 'precipitates or not' ?

Aug 1, 2018

No. The reaction wants to consume the $\text{AgCl}$ before it can precipitate. Only if ${Q}_{s p}$ exceeds ${K}_{s p}$ will a precipitate form.

The complex formation reaction of "Ag"("NH"_3)_2^+ first occurs. In terms of $\text{M}$, we have:

"Ag"^(+)(aq) + 2"NH"_3(aq) rightleftharpoons "Ag"("NH"_3)_2^(+)(aq)

$\text{I"" "0.100" "" "" "1.00" "" "" "" } 0$
$\text{C"" "-x" "" "" "-2x" "" "" } + x$
$\text{E"" "0.100 - x" "1.00-2x" "" } x$

It just so happens that ${\text{NH}}_{3}$ will NOT easily get displaced by ${\text{Cl}}^{-}$ if $\text{NaCl}$ is added to the solution. ${\text{NH}}_{3}$ is a good Lewis base, but ${\text{Cl}}^{-}$ is not nearly as good (inorganic chemists call ${\text{Cl}}^{-}$ a "weak-field ligand").

So, all we care about here is $\left[{\text{Ag}}^{+}\right]$.

${K}_{f} = 1.6 \times {10}^{7} = \left({\left[{\text{Ag"("NH"_3)_2^+])/(["Ag"^(+)]["NH}}_{3}\right]}^{2}\right)$

$= \frac{x}{\left(0.100 - x\right) {\left(1.00 - 2 x\right)}^{2}}$

Since ${K}_{f}$ is huge, we make the approximation that $x \approx \text{0.100 M}$. We EXPECT that most of the ${\text{Ag}}^{+}$ will be gone, because the reaction is heavily PRODUCT-FAVORED. This is an important fundamental concept that you should note.

NOTE: In doing so, we must make sure we do NOT plug it in for $\left[A {g}^{+}\right]$. (Why? Well, just try it and evaluate the right-hand side.)

Hence:

$1.6 \times {10}^{7}$ "M"^(-2) ~~ ("0.100 M")/((0.100 - x)("1.00 M" - 2("0.100 M"))^2)

$= {\text{0.15625 M}}^{- 1} / \left(0.100 - x\right)$

So now, we can solve for $\left[{\text{Ag}}^{+}\right]$ at equilibrium, as well as $x$.

["Ag"^(+)]_(eq) = 0.100 - x = 9.77 xx 10^(-9) "M"

$\implies x \approx \text{0.100 M}$ as expected... just not EXACTLY $\text{0.100 M}$.

And as expected, the silver is almost all gone (but not quite). This silver cation can then react with the incoming ${\text{Cl}}^{-}$. The ${\text{0.010 mols Cl}}^{-}$ that came from $\text{NaCl}$ has a concentration of $\text{0.010 M}$ since it is in $\text{1.00 L}$.

Now, if a precipitate forms, it means the reaction given by

${\text{AgCl"(s) rightleftharpoons "Ag"^(+)(aq) + "Cl}}^{-} \left(a q\right)$

was previously PRODUCT-FAVORED, i.e ${Q}_{s p} > {K}_{s p}$.

We officially have $\left[{\text{Ag}}^{+}\right]$ and $\left[{\text{Cl}}^{-}\right]$ to use, so that we can determine the current state of the reaction.

• Is it at EQUILIBRIUM? If ${Q}_{s p} = {K}_{s p}$, then yes...
• Is it PRODUCT-FAVORED? If ${Q}_{s p} > {K}_{s p}$, then yes...
• Is it REACTANT-FAVORED? If ${Q}_{s p} < {K}_{s p}$, then yes...

${Q}_{s p} = {\left[{\text{Ag"^(+)]_i["Cl}}^{-}\right]}_{i}$

$= \left(9.77 \times {10}^{- 9} \text{M")("0.0100 M}\right)$

$= 9.77 \times {10}^{- 11}$

And since ${Q}_{s p} < 1.8 \times {10}^{- 10}$, the reaction will NOT form a precipitate.

The reaction favors the solid at the moment, so Le Chatelier's Principle suggests the reaction shifts to the right to dissolve more solid BEFORE precipitate is allowed to form, thus not allowing it to form.