# Calculate the final equilibrium concentrations (complex ion reaction)?

## If $\setminus \texttt{2.0 \text{ mol } C {u}^{2 +}}$ and $\setminus \texttt{1.0 \text{ mol } N {H}_{3}}$ are introduced into a solution that has a total volume of $\setminus \texttt{1.0 \text{ L}}$, calculate the final equilibrium concentrations of $\setminus \texttt{C {u}^{2 +} , N {H}_{3} , \text{ and } {\left[C u {\left(N {H}_{3}\right)}_{4}\right]}^{2 +}}$ if $\setminus \texttt{{K}_{f} = 1.03 \times {10}^{13}}$?

Aug 4, 2018

${\left[{\text{Cu"^(2+)]_(eq) = "2.0 M" - 1/4["NH}}_{3}\right]}_{i}$
["NH"_3]_(eq) = 3.43 xx 10^(-4) "M"
${\left[{\text{Cu"("NH"_3)_4^(2+)]_(eq) = 1/4["NH}}_{3}\right]}_{i}$

Remember that this is no different than any other equilibrium problem. This might as well be ${K}_{c}$ and it won't change the nature of the problem, except that ${K}_{f}$ is generally huge instead of sometimes small.

${K}_{f}$ is the formation constant of "Cu"("NH"_3)_4^(2+), the tetraamminecopper(II) ion, and describes the "equilibrium":

"Cu"^(2+)(aq) + 4"NH"_3(aq) -> "Cu"("NH"_3)_4^(2+)(aq)

You know what the starting concentrations are (even if you don't, it's in $\text{1.0 L}$ so the accidental guess that the $\text{mols}$ are concentrations would not numerically matter anyway), so let's just skip to the ICE table.

Remember the coefficients in the change in concentration and in the exponents.

"Cu"^(2+)(aq) + 4"NH"_3(aq) -> "Cu"("NH"_3)_4^(2+)(aq)

$\text{I"" "2.0" "" "" "" "1.0" "" "" "" } 0$
$\text{C"" "-x" "" "" "-4x" "" "" } + x$
$\text{E"" "2.0-x" "" "1.0-4x" "" } x$

The mass action expression is therefore:

${K}_{f} = \left({\left[{\text{Cu"("NH"_3)_4^(2+)])/(["Cu"^(2+)]["NH}}_{3}\right]}^{4}\right)$

$= \frac{x}{\left(2.0 - x\right) {\left(1.0 - 4 x\right)}^{4}}$

Since ${K}_{f}$ is huge, the reaction is very PRODUCT-FAVORED. Once we find the limiting reactant, we know what $x$ should be.

Ammonia is the limiting reactant, because there is less than $4$ times the ammonia as the copper cation. Thus, we expect it to run out first, and so, $x \approx \text{0.25 M}$.

We do NOT plug that in for ${\left[{\text{NH}}_{3}\right]}_{e q}$ (why?), but otherwise plug it in for ${\left[{\text{Cu}}^{2 +}\right]}_{e q}$ and ["Cu"("NH"_3)_4^(2+)]_(eq).

$1.03 \times {10}^{13} \approx \frac{0.25}{\left(2.0 - 0.25\right) {\left(1.0 - 4 x\right)}^{4}}$

$= \frac{1 / 7}{1.0 - 4 x} ^ 4$

From here,

$\textcolor{b l u e}{{\left[{\text{NH}}_{3}\right]}_{e q}} = 1.0 - 4 x = {\left(\frac{1 / 7}{1.03 \times {10}^{13}}\right)}^{1 / 4}$

$= \textcolor{b l u e}{3.43 \times {10}^{- 4} \text{M}}$

Now we can "re-solve" for $x$ to get ["Cu"("NH"_3)_4^(2+)]_(eq) and verify that it is still $\approx \text{0.250 M}$.

x = color(blue)(["Cu"("NH"_3)_4^(2+)]_(eq)) = (1.0 - 3.43 xx 10^(-4) "M")/4

$= {0.24}_{9914 \cdots}$ $\text{M}$

$\approx$ $\textcolor{b l u e}{\text{0.25 M}}$

And lastly, copper, which we already have from already knowing what value of $x$ is a good estimate.

color(blue)(["Cu"^(2+)]_(eq)) = 2.0 - 0.25 = color(blue)("1.75 M")

And now let's verify that ${K}_{f}$ is still satisfied.

${K}_{f} = \frac{x}{\left(2.0 - x\right) {\left(1.0 - 4 x\right)}^{4}}$

$= \frac{0.25}{\left(2.0 - 0.25\right) {\left(3.43 \times {10}^{- 4}\right)}^{4}}$

$= {1.03}_{21} \times {10}^{13}$ color(blue)(sqrt"")

Aug 7, 2018

$\setminus \texttt{{K}_{f} = \setminus \frac{\left[{\left[C u {\left(N {H}_{3}\right)}_{4}\right]}^{2 +}\right]}{\left[C {u}^{2 +}\right] {\left[N {H}_{3}\right]}^{4}} = 1.03 \times {10}^{13}}$
$\setminus \texttt{\frac{1}{1.03 \times {10}^{13}} = \setminus \frac{\left(1.75 + x\right) \left(4 {x}^{4}\right)}{0.25 - x}}$

$\setminus \textsf{1.03 \times {10}^{-} 13 \setminus \cong \frac{\left(1.75\right) {\left(4 x\right)}^{4}}{0.25}}$
$\setminus \textsf{{\left(4 x\right)}^{4} = \frac{\left(0.25\right) \left(1.03 \times {10}^{-} 13\right)}{1.75}}$
$\setminus \textsf{4 x = \setminus \sqrt[4]{\frac{\left(0.25\right) \left(1.03 \times {10}^{-} 13\right)}{1.75}} \text{ or } {\left(\frac{\left(0.25\right) \left(1.03 \times {10}^{-} 13\right)}{1.75}\right)}^{1 / 4}}$
\sf{" \approx0.000348}
$\setminus \textsf{x = \frac{0.000348}{4} = 0.0000950}$

\tt{[Cu^(2+)=1.785+0.0000950\approx1.75}
$\setminus \texttt{\left[N {H}_{3}\right] = 4 \left(0.0000950\right) = 0.000348}$
$\setminus \texttt{\left[{\left[C u {\left(N {H}_{3}\right)}_{4}\right]}^{2 +}\right] = 0.25 - 0.0000950 \setminus \approx 0.25}$

(please ignore anything under the red text)

## $\setminus \texttt{\setminus \textcolor{red}{\text{Old work that is miswritten.}}}$

My attempt to apply an ICF (ICE with all known values) table.

Since we usually follow an ICF with a normal (variable) ICE table, in our class lectures/notes, I attempted to do so:

Obviously, I messed up somewhere, or I'm not supposed to do this.