 Solution

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.