# 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?

• Your "initial state" seems to be pH 7 so you do know the [H+] at the start. – Tan Yong Boon Mar 16 '19 at 1:03
• Yes, but the pH is adjusted externally, so we can't take the change in [H+] to deduce how much [HF] and [F-] are changing. For that reason, I left that box as N/A. I could have left the entire column out instead. – Karsten Theis Mar 16 '19 at 1:10
• chemistry.stackexchange.com/questions/110607/… – Adnan AL-Amleh Mar 16 '19 at 22:20

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 $$y$$):

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

Alternatively, given the cross promotion of a related question, we can consider doing away with ICE tables all together.

$$K_{\mathrm{sp}} = \ce{[Ca^{2+}][F-]^{2}}=\pu{3.7e−11}$$ $$K_{a} = \frac{\ce{[H+][F-]}}{\ce{[HF]}}=10^{-3.17}$$

At this point, the unknowns are $$\ce{[Ca^{2+}]}$$, $$\ce{[F-]}$$, $$\ce{[HF]}$$. $$\ce{[H+]}$$ is known from the pH and it's $$10^{-3}$$.

We have two equations and 3 unknowns. The final piece that's missing is that because the source of all fluorine and calcium in this system is from calcium fluoride, we may impose the stoichiometry of the solid on the species present:

$$2\ce{[Ca^{2+}]} = \ce{[F-]} + \ce{[HF]}$$

Three equations and three unknowns. I cheated and used WolframAlpha:

$$\ce{[Ca^{2+}]} \approx 3.8\times 10^{-4}$$ This represents an approximately 1.8 fold increase in solubility.

(I sanity checked the calcium ion concentration for a neutral solution using this method and got back the molar solubility.)

• Solving it is not too bad. The ratio of fluoride to hydrofluoric acid is $10^{-0.17}$ (from pKa and pH), so you know which fraction of total soluble fluorine is fluoride, so you know the proportion of calcium ions and fluoride ions. You can plug this back into the solubility product to get the calcium concentration. – Karsten Theis Mar 18 '19 at 22:25
• @AdnanAL-Amleh The question asks about solubility, so that means all calcium and fluorine derive from dissolution of calcium fluoride. Given the composition of that compound, there must be twice as much fluorine as calcium at any point in time. – Zhe Mar 19 '19 at 1:02
• But you don't know what the starting concentration is. If you want to do it by looking at changes, use the other answer. – Zhe Mar 19 '19 at 1:43
• @AdnanAL-Amleh It's just division, and frankly, it's not ambiguous. At this point, it's probably more helpful if you try to work through these difficulties yourself than have everything spelled out. – Zhe Mar 19 '19 at 15:49
• Unfortunately, we have words because equations cannot always express what we mean. – Zhe Mar 19 '19 at 16:38