# 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)?

May 22, 2017

#### Answer:

Here's what I got.

#### Explanation:

You know that

$\textcolor{w h i t e}{a a a a} \text{HF"_ ((aq)) rightleftharpoons "H"_ ((aq))^(+) + "F"_ ((aq))^(-)" } {K}_{c 1} = 6.8 \cdot {10}^{- 4}$

$\text{H"_ 2"C"_ 2"O"_ (4(aq)) rightleftharpoons 2"H"_ ((aq))^(+) + "C" _ 2"O"_ (4(aq))^(2-)" } {K}_{c 2} = 3.8 \cdot {10}^{- 6}$

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

You will have

${K}_{c 1} = \left(\left[\text{H"^(+)] * ["F"^(-)])/(["HF}\right]\right)$

${K}_{c 2} = \left(\left[{\text{H"^(+)]^2 * ["C"_2"O"_4^(2-)])/(["H"_2"C"_2"O}}_{4}\right]\right)$

Now, notice that your target equilibrium

$\textcolor{red}{2} {\text{HF" _ ((aq)) + "C" _ 2"O"_ (4(aq))^(2-) rightleftharpoons color(red)(2)"F"_ ((aq))^(-) + "H"_ 2"C"_ 2"O}}_{4 \left(a q\right)}$

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

$2 {\text{H"_ ((aq))^(+) + "C" _ 2"O"_ (4(aq))^(2-) rightleftharpoons "H"_ 2"C"_ 2"O}}_{4 \left(a q\right)}$

The new equilibrium constant for this reaction is

${K}_{c 2}^{'} = \left(\left[{\text{H"_2"C"_2"O"_4])/(["H"^(+)]^2 * ["C"_2"O}}_{4}^{2 -}\right]\right)$

As you can see, you can write this as

${K}_{c 2}^{'} = \frac{1}{K} _ \left(c 2\right)$

${K}_{c 2}^{'} = \frac{1}{3.6 \cdot {10}^{- 6}} = 2.8 \cdot {10}^{5}$

Moreover, the target reaction uses $\textcolor{red}{2}$ moles of hydrofluoric acid and $\textcolor{red}{2}$ moles of fluoride anions, so multiply the first equilibrium by $\textcolor{red}{2}$.

This will get you

$\textcolor{red}{2} {\text{HF"_ ((aq)) rightleftharpoons color(red)(2)"H"_ ((aq))^(+) + color(red)(2)"F}}_{\left(a q\right)}^{-}$

The new equilibrium constant for this reaction will be

${K}_{c 1}^{'} = \left({\left[\text{H"^(+)]^color(red)(2) * ["F"^(-)]^color(red)(2))/(["HF}\right]}^{\textcolor{red}{2}}\right)$

You can rewrite this as

${K}_{c 1}^{'} = {\left(\left(\left[\text{H"^(+)] * ["F"^(-)])/(["HF}\right]\right)\right)}^{\textcolor{red}{2}}$

which means that you will have

${K}_{c 1}^{'} = {K}_{c 1}^{\textcolor{red}{2}}$

${K}_{c 1} = {\left(6.8 \cdot {10}^{- 4}\right)}^{\textcolor{red}{2}} = 4.6 \cdot {10}^{- 7}$

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

$2 \text{H"_ ((aq))^(+) + "C" _ 2"O"_ (4(aq))^(2-) rightleftharpoons "H"_ 2"C"_ 2"O"_ (4(aq))" } {K}_{c 2}^{'}$

$\textcolor{w h i t e}{a a a a a a a a a} \textcolor{red}{2} \text{HF"_ ((aq)) rightleftharpoons color(red)(2)"H"_ ((aq))^(+) + color(red)(2)"F"_ ((aq))^(-)" } {K}_{c 1}^{'}$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

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

$2 {\text{HF" _ ((aq)) + "C" _ 2"O"_ (4(aq))^(2-) rightleftharpoons 2"F"_ ((aq))^(-) + "H"_ 2"C"_ 2"O}}_{4 \left(a q\right)}$

The equilibrium constant for this equilibrium will be

${K}_{\text{target}} = {K}_{c 1}^{'} \cdot {K}_{c 2}^{'}$

${K}_{\text{target}} = 4.6 \cdot {10}^{- 7} \cdot 2.8 \cdot {10}^{5} = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{1.3 \cdot {10}^{- 1}}}}$

The answer is rounded to two sig figs.