Question

If given the information below, how do you calculate the `K_c` for the reaction `2HF(aq) + C_2O_4^(2-)(aq) rightleftharpoons 2F^(-)(aq) + H_2C_2O_4(aq)`?

Answer 1

Here's what I got.

Explanation:

You know that

`color(white)(aaaa)"HF"_ ((aq)) rightleftharpoons "H"_ ((aq))^(+) + "F"_ ((aq))^(-)" "K_(c1) = 6.8 * 10^(-4)`

`"H"_ 2"C"_ 2"O"_ (4(aq)) rightleftharpoons 2"H"_ ((aq))^(+) + "C" _ 2"O"_ (4(aq))^(2-)" "K_(c2) = 3.8 * 10^(-6)`

Before moving on, take the time to write the expressions of the two equilibrium constants given to you.

You will have

`K_(c1) = (["H"^(+)] * ["F"^(-)])/(["HF"])`

`K_(c2) = (["H"^(+)]^2 * ["C"_2"O"_4^(2-)])/(["H"_2"C"_2"O"_4])`

Now, notice that your target equilibrium

`color(red)(2)"HF" _ ((aq)) + "C" _ 2"O"_ (4(aq))^(2-) rightleftharpoons color(red)(2)"F"_ ((aq))^(-) + "H"_ 2"C"_ 2"O"_ (4(aq))`

has the oxalate anion on the products' side, so star by reversing the second equilibrium

`2"H"_ ((aq))^(+) + "C" _ 2"O"_ (4(aq))^(2-) rightleftharpoons "H"_ 2"C"_ 2"O"_ (4(aq))`

The new equilibrium constant for this reaction is

`K_(c2)^' = (["H"_2"C"_2"O"_4])/(["H"^(+)]^2 * ["C"_2"O"_4^(2-)])`

As you can see, you can write this as

`K_ (c2)^' = 1/K_(c2)`

`K_( c2)^' = 1/ (3.6 * 10^(-6))= 2.8 * 10^5`

Moreover, the target reaction uses `color(red)(2)` moles of hydrofluoric acid and `color(red)(2)` moles of fluoride anions, so multiply the first equilibrium by `color(red)(2)`.

This will get you

`color(red)(2)"HF"_ ((aq)) rightleftharpoons color(red)(2)"H"_ ((aq))^(+) + color(red)(2)"F"_ ((aq))^(-)`

The new equilibrium constant for this reaction will be

`K_(c1)^' = (["H"^(+)]^color(red)(2) * ["F"^(-)]^color(red)(2))/(["HF"]^color(red)(2))`

You can rewrite this as

`K_(c1)^' =((["H"^(+)] * ["F"^(-)])/(["HF"]))^color(red)(2)`

which means that you will have

`K_(c1)^' = K_(c1)^color(red)(2)`

`K_(c1) = (6.8 * 10^(-4))^color(red)(2) = 4.6 * 10^(-7)`

You can now add the two equilibrium reactions to get the target equilibrium

`2"H"_ ((aq))^(+) + "C" _ 2"O"_ (4(aq))^(2-) rightleftharpoons "H"_ 2"C"_ 2"O"_ (4(aq))" "K_(c2)^'`

`color(white)(aaaaaaaaa)color(red)(2)"HF"_ ((aq)) rightleftharpoons color(red)(2)"H"_ ((aq))^(+) + color(red)(2)"F"_ ((aq))^(-)" "K_(c1)^'`
`color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)`

`color(red)(cancel(color(black)(2"H"_ ((aq))^(+)))) + "C" _ 2"O"_ (4(aq))^(2-) + 2"HF"_ ((aq)) rightleftharpoons "H"_ 2"C"_ 2"O"_ (4(aq)) + color(red)(cancel(color(black)(2"H"_ ((aq))^(+)))) + 2"F"_ ((aq))^(-)`

This simplifies to

`2"HF" _ ((aq)) + "C" _ 2"O"_ (4(aq))^(2-) rightleftharpoons 2"F"_ ((aq))^(-) + "H"_ 2"C"_ 2"O"_ (4(aq))`

The equilibrium constant for this equilibrium will be

`K_"target" = K_(c1)^' * K_(c2)^'`

`K_"target" = 4.6 * 10^(-7) * 2.8 * 10^5 = color(darkgreen)(ul(color(black)(1.3 * 10^(-1))))`

The answer is rounded to two sig figs.