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

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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}# ?

If

##### 2 Answers

#["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 **generally huge** instead of sometimes small.

**formation constant** of

#"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 **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* *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**

We do NOT plug that in for

#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 = 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

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

And now let's verify that

#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"")#

#\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}#

(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:

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