Plotting a mineral stability diagram

Question

I'm trying to figure out how to plot the stability of 4 minerals — $\ce{CaCO3}$, $\ce{CaSO4}$, $\ce{CaMg(CO3)2}$, and $\ce{CaMg3(CO3)4}$ — with axes of $\mathrm{ln}\,a_{\ce{Mg^{+2}}}$ vs $\mathrm{ln}\,a_{\ce{SO_4^{-2}}}$ where $\mathrm{ln}\,a$ is the natural log of the activities of each ionic species. The activity of carbonate is a given.

I found solubility $K$ for each dissolution, based on the Gibbs free energy, $R$, and $T = 25^\circ \mathrm{C}$, then tried to set up equations to plot the following:

1. $\mathrm{ln}\,K_{\ce{sp,\,CaCO3}} = \mathrm{ln}\,a_{\ce{Ca}} + \mathrm{ln}\,a_{\ce{CO_3}}$
2. $\mathrm{ln}\,K_{\ce{sp,\,CaSO4}} = \mathrm{ln}\,a_{\ce{Ca}} + \mathrm{ln}\,a_{\ce{SO_4}}$
3. $\mathrm{ln}\,K_{\ce{sp,\,CaMg(CO_3)_2}} = \mathrm{ln}\,a_{\ce{Ca}} + \mathrm{ln}\,a_{\ce{Mg}} + 2\,\mathrm{ln}\,a_{\ce{CO_3}}$
4. $\mathrm{ln}\,K_{\ce{sp,\,CaMg_3(CO_3)_4}} = \mathrm{ln}\,a_{\ce{Ca}} + 3\,\mathrm{ln}\,a_{\ce{Mg}} + 4\,\mathrm{ln}\,a_{\ce{CO_3}}$

Based on the thermo data each $\mathrm{ln}\,K$ is a constant (since $T$, $P$ are constant), since the activity of carbonate is given it seems like I should be able to solve equation 1 for the activity of calcium, but then each equation is a point not a line.

I'm confused as to how each of the equations is dependent on $\ce{SO_4}$ at all (which is supposed to be my y axis), and how to relate everything. I tried to treat activity $\ce{Ca}$ as variable, then solve for it in terms of activity $\ce{SO_4}$, to then make equations 3, 4 dependent on $\ce{SO_4}$, and $\ce{Mg}$, but then I had no idea how to treat equations 1 and 2. Also the values I got for that were really big, the lines were on $y=35+3x$ and $y=18+x$.

This question is based on a question for a textbook, which I'm trying to use to study for an upcoming exam (1 week), so any help would be greatly appreciated.

Answer 1

Remember that you are concerned about stability fields. The lines on your stability diagram are the places where two minerals are in equilibrium. One one side one mineral will be more stable, on the other side the other one will be.

Let's talk about $\ce{Ca}$ first. The reactions all have the same $\ce{Ca}$ so it's activity isn't a factor in their relative stability. If you write a combined reaction for any mineral to another, it will cancel out. Try it. Calcite stability has nothing to do with $\ce{Mg}$ or $\ce{SO4}$ it's stability will only depend on $\ce{CO3}$ and $\ce{Ca}$. But wait, you know $\mathrm{ln}\,a_{\ce{CO3}}$ so you can calculate $\mathrm{ln}\,a_\ce{Ca}$ in equilibrium with calcite. So you can use that to help figure out where calcite is the most stable.

You only have one equation where sulfate is included. What direction will that stability boundary take on your diagram? Say you want to divide your diagram space into areas of calcite vs gypsum stability, how could you combine the two solubility equations to come up with the conversion of one mineral to the other? What dissolved species activities enter into the equation (hint: you know one of the two). Given that you can figure out the other one and plot that on you diagram.

Can you follow a similar process for calcite and dolomite (hint: yes)? What about calcite and huntite? Is there any part of the diagram where huntite is more stable than dolomite (or vice versa)?

Now figure out the stability relationship between dolomite and gypsum. What angle will the line take on your plot? Figure out where it will occur by combining reactions. This is a bit trickier but the $\mathrm{ln}\,a$ values will depend on the $\mathrm{ln}\,K$ for the combined reaction.

Now erase any lines in a field where those two minerals are less stable than anything else.

Good luck and let us know if you get stuck. I didn't want to just give you the answer because I'd have to actually work it out.