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)#?

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1 Answer
May 22, 2017

Answer:

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.