Question

What is the equilibrium concentration of `"Al"^(3+)` for the complexation with six equivalents of `"F"^(-)` to form `"AlF"_6^(3-)` if the initial concentrations are `"0.013 M Al"^(3+)` and `"0.104 M F"^(-)`? `K_f = 7.0 xx 10^19`

Answer 1

Warning! Long Answer. `["Al"^"3+"] = 6.0 × 10^"-13"color(white)(l)"mol/L"`

Explanation:

Because the value of `K_text(f)` is so large, the reaction will go essentially to completion.

We are given the concentrations of two reactants, so this is really a limiting reactant problem.

Step 1. Identify the limiting reactant

Assume that we have 1 L of solution.

`color(white)(mmmmmmmm)"Al"^"3+" + "6F"^"-" ⇌ "AlF"_6^"3-"`
`"Moles":color(white)(mmmm)0.013color(white)(m)0.104`
`"Divide by": color(white)(lmmm)1color(white)(mmml)6`
`"Moles of rxn": color(white)(ll) 0.013color(white)(m)0.0173`

An easy way to identify the limiting reactant is to determine the "moles of reaction".

We divide the number of moles of each reactant by their corresponding coefficient in the balanced equation.

I did that for you in the table above.

We see that `"Al"^"3+"` is the limiting reactant because it gives the fewer moles of reaction.

Step 2. Calculate the moles of each substance at equilibrium

We can use an ICE table to solve this problem.

`color(white)(mmmmmm)"Al"^"3+" + "6F"^"-" ⇌ "AlF"_6^"3-"`
`"I/mol":color(white)(mm)0.013color(white)(m)0.104color(white)(mmll)0`
`"C/mol":color(white)(m)"-0.013"color(white)(m)"-0.078"color(white)(m)"+0.013"`
`"E/mol":color(white)(mmll)0color(white)(mm)0.026color(white)(mm)0.013`

Step 3. Calculate the equilibrium concentration of `"Al"^"3+"`

Obviously, the concentration of `"Al"^"3+"` is not zero, as implied above, but it will be very small.

The question is really, "What is the concentration of `"Al"^"3+"` in equilibrium
with 0.013 mol/L `"AlF"_6^"-"` and 0.026 mol/L `"F"^"-"`"?

It will be easier to write the equation in reverse.

`color(white)(mmmmmmll)"AlF"_6^"3-"⇌ "Al"^"3+" +color(white)(ml) "6F"^"-"`
`"I/mol·L"^"-1":color(white)(ml)0.013color(white)(mmll)0color(white)(mmml)0.026`
`"C/mol·L"^"-1":color(white)(mll)"-"xcolor(white)(mmm)"+"xcolor(white)(mmm)"+6"x`
`"E/mol·L"^"-1":color(white)(ll)"0.013-"xcolor(white)(mm)xcolor(white)(mm)"0.026+6"x`

`K = (["Al"^"3+"]["F"^"-"]^6)/(["AlF"_6^"3-"]) = (x("0.026+6"x)^6)/("0.013-"x) = 1/(7.0 × 10^19) = 1.43 × 10^"-20"`

Because `K` is so small, `6x≪0.026` and `x≪0.013`.

Then,

`(x×0.026^6)/0.013 =1.43 × 10^"-20"`

`3.09 ×10^"-10"x = 1.86 × 10^"-22"`

`x = (1.86 × 10^"-22")/(3.09 ×10^"-10") = 6.0 × 10^"-13"`

∴ `["Al"^"3+"] = 6.0 × 10^"-13"color(white)(l)"mol/L"`

Answer 2

`["Al"^(3+)] = 6.01 xx 10^(-13) "M"`

---

Ernest had a good way to think about it, but here's another way to do this.

Write out the reaction and its ICE table:

`"Al"^(3+)(aq)" "" "+" "" "6"F"^(-)(aq) " "->" " "AlF"_6^(3-)(aq)`

`"I"" ""0.013 M"" "" "" "" ""0.104 M"" "" "" "" "" ""0 M"`
`"C"" "-x" "" "" "" "" "-6x" "" "" "" "" "" "+x`
`"E"" "(0.013 - x)"M"" "(0.104 - 6x)"M"" "" "" "x " M"`

The mass action expression would then be:

`7.0 xx 10^(19) = x/((0.013 - x)(0.104 - 6x)^6)`

Since `K_f` is so large, we assume that pretty much all the `"Al"^(3+)` is completely gone. Therefore, and although it technically isn't all gone, `x ~~ "0.013 M"` is a good approximation.

This substitution gives, EXCEPT for the term `0.013 - x`:

`7.0 xx 10^(19) = 0.013/((0.013 - x)(0.104 - 6(0.013))^6)`

`=> 7.0 xx 10^(19) = 0.013/((0.013 - x)(0.026)^6)`

`=> ["Al"^(3+)] = 0.013 - x = 0.013/((0.026^6)(7.0 xx 10^(19)))`

NOTE: You could just solve this as it is, but sometimes you will get expressions that are impossible to solve (like if you have a 2nd and 6th order term in the denominator).

Instead, consider taking the `ln` of both sides.

This maximizes the accuracy, since usually when `K_f` is this large, it promotes rounding errors even when you keep many decimal places; logarithms help get rid of this rounding error, particularly if you're the kind of person who doesn't do an entire calculation in your calculator in one step.

`=> ln["Al"^(3+)] = ln[0.013/((0.026^6)(7.0 xx 10^(19)))]`

`= ln(0.013) - 6ln(0.026) - ln(7.0 xx 10^(19))`

Now we can evaluate the individual terms.

`=> ln["Al"^(3+)] = -4.343 - (-21.898) - 45.695`

Therefore,

`color(blue)(["Al"^(3+)]) = e^(ln(0.013 - x)) = 0.013 - x`

`~~ e^(-28.140)`

`~~ color(blue)(6.01 xx 10^(-13) "M")`

We can check whether this is right by plugging it back in as follows to find the actual value of `x`.

`0.013 - x = 6.01 xx 10^(-13)`

`=> x = 0.013 - 6.01 xx 10^(-13) ~~ 0.013`

Most calculators don't have enough precision to evaluate this, but using `x = 0.013` is fine, as long as you use `0.013 - x = 6.01 xx 10^(-13)`.

`=> K_f = x/((0.013 - x)(0.104 - 6x)^6)`

`= (0.013 - 6.01 xx 10^(-13))/((6.01 xx 10^(-13))(0.104 - 6*(0.013 - 6.01 xx 10^(-13))))`

`~~ (0.013)/((6.01 xx 10^(-13))(0.104 - 6*0.013))`

`= 7.0(0) xx 10^(19)` `color(blue)(sqrt"")`