Calculate the final equilibrium concentrations (complex ion reaction)?

If \tt{2.0" mol "Cu^(2+)} and \tt{1.0" mol "NH_3} are introduced into a solution that has a total volume of \tt{1.0" L"}, calculate the final equilibrium concentrations of \tt{Cu^(2+),NH_3," and "[Cu(NH_3)_4]^(2+)} if \tt{K_f=1.03xx10^13}?

2 Answers
Aug 4, 2018

["Cu"^(2+)]_(eq) = "2.0 M" - 1/4["NH"_3]_i
["NH"_3]_(eq) = 3.43 xx 10^(-4) "M"
["Cu"("NH"_3)_4^(2+)]_(eq) = 1/4["NH"_3]_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 "1.0 L" so the accidental guess that the "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)

"I"" "2.0" "" "" "" "1.0" "" "" "" "0
"C"" "-x" "" "" "-4x" "" "" "+x
"E"" "2.0-x" "" "1.0-4x" "" "x

The mass action expression is therefore:

K_f = (["Cu"("NH"_3)_4^(2+)])/(["Cu"^(2+)]["NH"_3]^4)

= x/((2.0 - x)(1.0 - 4x)^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 ~~ "0.25 M".

We do NOT plug that in for ["NH"_3]_(eq) (why?), but otherwise plug it in for ["Cu"^(2+)]_(eq) and ["Cu"("NH"_3)_4^(2+)]_(eq).

1.03 xx 10^(13) ~~ (0.25)/((2.0 - 0.25)(1.0 - 4x)^4)

= (1//7)/(1.0 - 4x)^4

From here,

color(blue)(["NH"_3]_(eq)) = 1.0 - 4x = ((1//7)/(1.03 xx 10^(13)))^(1//4)

= color(blue)(3.43 xx 10^(-4) "M")

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

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

= 0.24_(9914cdots) "M"

~~ color(blue)("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 = x/((2.0 - x)(1.0 - 4x)^4)

= (0.25)/((2.0 - 0.25)(3.43 xx 10^(-4))^4)

= 1.03_21 xx 10^(13) color(blue)(sqrt"")

Aug 7, 2018

screenshot off Wordscreenshot off Word
\tt{K_f=\frac{[[Cu(NH_3)_4]^(2+)]}{[Cu^(2+)][NH_3]^4}=1.03xx10^13}
\tt{1/(1.03xx10^13)=\frac{(1.75+x)(4x^4)}{0.25-x}}

\sf{1.03xx10^-13\cong((1.75)(4x)^4)/0.25}
\sf{(4x)^4=((0.25)(1.03xx10^-13))/1.75}
\sf{4x=\root[4]{((0.25)(1.03xx10^-13))/1.75}" or "(((0.25)(1.03xx10^-13))/1.75)^(1//4)}
\sf{" \approx0.000348}
\sf{x=0.000348/4=0.0000950}

\tt{[Cu^(2+)=1.785+0.0000950\approx1.75}
\tt{[NH_3]=4(0.0000950)=0.000348}
\tt{[[Cu(NH_3)_4]^(2+)]=0.25-0.0000950\approx0.25}


(please ignore anything under the red text)


\tt{\color{red}{"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:

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