Question

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}`?

Answer 1

`["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`

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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")`

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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"")`

Answer 2

`\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}`

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(please ignore anything under the red text)

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`\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:

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