Question

`"500.0 mL"` of solution contains `0.750` millimoles of `"Co"("NO"_3)_2`, `125.0` millimoles of `"NH"_3`, and `100.0` millimoles of ethylenediamine. What is the equilibrium concentration of `"Co"^(2+)` for this complexation reaction?

The `K_f` for the formation of `["Co"("NH"_3)_6]^(2+)` is `7.7 xx 10^4`. The `K_f` for the formation of `["Co"("en")_3]^(2+)` is `8.7 xx 10^13`.

Answer 1

Well, it ought to be close to zero...

I got `2.227 xx 10^(-10) "M"` by using natural logarithms. If you did it normally without logarithms, you may get `>10%` error.

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The initial concentrations are:

`["Co"^(2+)]_i = 0.750/500.0 "mmol"/"mL" = "0.00150 M"`

`["NH"_3]_i = 125.0/500.0 "mmol"/"mL" = "0.2500 M"`

`["en"]_i = 100.0/500.0 "mmol"/"mL" = "0.2000 M"`

This has two "equilibria", so let us form a composite "equilibrium" constant:

`beta = K_(f1)K_(f2) = 7.7 xx 10^4 cdot 8.7 xx 10^13 = 6.7 xx 10^18`

This is for the following two reactions combined:

`"Co"^(2+)(aq) + 6"NH"_3(aq) -> ["Co"("NH"_3)_6]^(2+)(aq)`
`"Co"^(2+)(aq) + 3"en"(aq) -> ["Co"("en")_3]^(2+)(aq)`
`"-----------------------------------------------------------"`
`2"Co"^(2+)(aq) + 6"NH"_3(aq) + 3"en"(aq) -> ["Co"("NH"_3)_6]^(2+)(aq) + ["Co"("en")_3]^(2+)(aq)`

And so, the composite formation constant is:

`beta = 6.7 xx 10^18 = ([["Co"("NH"_3)_6]^(2+)][["Co"("en")_3]^(2+)])/(["Co"^(2+)]^2["NH"_3]^6["en"]^3)`

The changes in concentration from an ICE table would then be:

`Delta["Co"^(2+)] = -2x`
`Delta["NH"_3] = -6x`
`Delta["en"] = -3x`
`Delta[["Co"("NH"_3)_6]^(2+)] = x`
`Delta[["Co"("en")_3]^(2+)] = x`

This gives:

`6.7 xx 10^18 = (x^2)/((0.00150 - 2x)^2(0.2500 - 6x)^6(0.2000 - 3x)^3)`

This is nearly impossible to solve without approximations, but using Wolfram Alpha we can confirm that `2x ~~ "0.00150 M"`, i.e. that all the cobalt is gone, but that we already knew that.

One way to solve this accurately while keeping it feasible in a calculator is to take the `ln` of both sides.

`ln(6.7 xx 10^18) = 2lnx - 2ln(0.00150 - 2x) - 6ln(0.2500 - 6x) - 3ln(0.2000 - 3x)`

Since `2x ~~ 0.00150`, use that info to evaluate the logarithms that do not go to `pmoo`.

`43.349 ~~ -14.391 - 2ln(0.00150 - 2x) - (-8.427) - (-4.862)`

`~~ -1.102 - 2ln(0.00150 - 2x)`

Therefore,

`0.00150 - 2x ~~ e^(-22.2255)`

And `color(blue)(["Co"^(2+)] ~~ 2.227 xx 10^(-10)"M")`.