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Suppose two dissolution reactions to a reference mineral $C$ have been normalized to the same amount of a conserved reference component. After division by the same positive energy factor, let their Gibbs energies of reaction be $y-{\ell }_1(x)$ and $y-{\ell }_2(x)$. Their relative thermodynamic potentials can then be written as displayed. The stable mineral has the smallest potential.

It follows that $C$ is stable below $y=\min ({\ell }_1,{\ell }_2)$, mineral 1 is stable where ${\ell }_1<{\ell }_2$ and $y>{\ell }_1$, and mineral 2 is stable where ${\ell }_2<{\ell }_1$ and $y>{\ell }_2$. The two nonreference minerals coexist stably where ${\ell }_1={\ell }_2<y$. Their equality below that height is a metastable continuation. This comparison proves the stable lower envelope and selects which portions of the pairwise mass-action lines belong on the mineral activity diagram.

An aluminium hydroxide mineral with composition $\mathrm{Al}(\mathrm{OH}{)}_3$. In a simplified mineral activity diagram, it is favored over kaolinite at sufficiently low dissolved silica thermodynamic activity.

Use decimal logarithms and write $x={\log }_{10}{a}_{{\mathrm{H}}_4\mathrm{Si}{\mathrm{O}}_4}$ and $y={\log }_{10}(a_{{\mathrm{Mg}}^{2+}}/a_{{\mathrm{H}}^{+}}^2)$. The question's concentration notation is interpreted as dimensionless thermodynamic activities, or concentrations divided by a fixed standard concentration with activity coefficients set to one.

For the chlorite–gibbsite reaction, unit solid activities giveHence $y=10.53-0.60x$. It is a descending straight mineral activity diagram boundary, passing through $(0,10.53)$ and $(-5,13.53)$. Its continuation alone does not establish which portions are stable against kaolinite.

Use base-ten logarithms and dimensionless thermodynamic activities at the stated temperature and pressure. Define $x={\log }_{10}{a}_{{\mathrm{H}}_4\mathrm{Si}{\mathrm{O}}_4}$ and $y={\log }_{10}(a_{\mathrm{N}{\mathrm{a}}^{+}}/a_{{\mathrm{H}}^{+}})$. The concentration notation is an ideal dilute-solution approximation with a consistent concentration reference.

With pure solids and unit water activity, the two reaction quotients are $10^{2y+2x}$ and $10^{2y+6x}$. Equating them to the corresponding equilibrium constants givesThe slopes on the mineral activity diagram are $-1$ and $-3$. The lines meet at $(x_{\ast },y_{\ast })=(-5.515,8.955)$, within the negative silica-logarithm range.

The three stable fields areThe bounding equalities give the stable two-mineral coexistences. In particular, the stable corundum region lies below the lower of the two oblique lines; treating every line extension as stable would mislabel parts of the mineral activity diagram.

Take base-ten logarithms and use dimensionless thermodynamic activities. The supplied concentration notation is interpreted in the ideal dilute-solution approximation, with a consistent reference concentration. PutWith pure solids and unit water activity, the reaction quotients are $Q_I=10^{2y+2x}$ and $Q_{II}=10^{2y+6x}$. At chemical equilibrium, $Q=K$, so the two lines on the mineral activity diagram areThey have slopes $-1$ and $-3$. Restricting the plotted silica activity to $x<0$ is compatible with their intersection,Their full straight extensions are equilibrium relations for the respective phase pairs; not every extension is a stable boundary when the third mineral is available.