equilibrium - How can you use ICE tables to solve multiple coupled equilibria?

If I have a problem involving multiple coupled equilibrium reactions, such as

Calcium fluoride, $\ce{CaF2}$, has a molar solubility of $\pu{2.1e−4 mol L−1}$ at pH = 7.00. By what factor does its molar solubility increase in a solution with pH = 3.00? The p$K_{\mathrm{a}}$ of $\ce{HF}$ is 3.17.

The relevant reactions are:

$$\ce{CaF2(s) <=> Ca^2+(aq) + 2 F-(aq)}$$ and $$\ce{HF(aq) <=> H+(aq) + F-(aq)}$$

They are coupled because fluoride occurs in both of them.

Is there a way to use ICE tables to organize the information (stoichiometry, initial concentrations, mass balance) as a way to solve the problem?

For example, I could try to set up one ICE table for each reaction (the column for $\ce{H+}$ is strange because in the problem, the pH is set to a value through an unspecified mechanism):

$$ \begin{array}{|c|c|c|} \hline &[\ce{Ca^2+}] & [\ce{F-}] \\ \hline I & \pu{2.1e−4} & \pu{4.2e−4} \\ \hline C & +x & +2x \\ \hline E & \pu{2.1e−4}+x & \pu{4.2e−4}+2x \\ \hline \end{array} $$

and

$$ \begin{array}{|c|c|c|} \hline &[\ce{HF}] & [\ce{H+}] & [\ce{F-}] \\ \hline I & 0 & \text{N/A} & \pu{4.2e−4} \\ \hline C & +x &\text{N/A} & -x \\ \hline E & +x & 10^{-3.00} & \pu{4.2e−4} - x\\ \hline \end{array} $$

However, how do the fluoride concentrations "talk to each other" in the two tables? Is the $x$ in one table the same as the $x$ in the other table?

Answer

The way to use ICE tables in this case would be to combine the ICE tables, and use "$x$" for the changes due to one reaction and "$y$" for the changes due to the other. For each reaction, you have one unknown (how far it reacted, i.e. $x$ and $y$), so you need two pieces of information to solve it, in this case the two equilibrium constants.

Here is the combined ICE table:

$$ \begin{array}{|c|c|c|c|c|} \hline &[\ce{Ca^2+}] & [\ce{F-}] & [\ce{H+}]&[\ce{HF}] \\ \hline I & \pu{2.1e−4} & \pu{4.2e−4} & \text{N/A} & 0 \\ \hline C & +x & +2x-y & \text{N/A} & +y \\ \hline E & \pu{2.1e−4}+x & \pu{4.2e−4}+2x-y & 10^{-3.00} & +y \\ \hline \end{array} $$

Now, you can use the equilibrium constants for the two reactions to solve for $x$ and $y$, giving you the equilibrium concentrations.

What the combined table shows is that the fluoride concentration depends on both reactions, so you can't first deal with one equilibrium and then with the other but have to solve a system of two equations with two unknowns.

Once you combine the tables, you could also have initial conditions where nothing is dissolved yet, to get easier expressions (I'm using $p$ and $q$ because they will be different from $x$ and $y$):

$$ \begin{array}{|c|c|c|c|c|} \hline &[\ce{Ca^2+}] & [\ce{F-}] & [\ce{H+}]&[\ce{HF}] \\ \hline I & 0 & 0 & \text{N/A} & 0 \\ \hline C & +p & +2p-q & \text{N/A} & +q \\ \hline E & +p & +2p-q & 10^{-3.00} & +q \\ \hline \end{array} $$