 Solution

For the chlorite–kaolinite reaction, take water and both solids to have unit thermodynamic activity. Its reaction quotient is $Q=a_{{\mathrm{Mg}}^{2+}}^5{a}_{{\mathrm{H}}_4\mathrm{Si}{\mathrm{O}}_4}/a_{{\mathrm{H}}^{+}}^{10}$. ThereforeThis line has a shallower negative slope than the chlorite–gibbsite line, with points $(0,12.228)$ and $(-5,13.228)$. Plot both against predominantly negative $x$, because the dissolved silica thermodynamic activity is generally below one.