= Solution
For the <chlorite>–<kaolinite> reaction, take water and both solids to have unit <thermodynamic activity>. Its <reaction quotient> is $Q=a_{\mathrm{Mg}^{2+}}^5a_{\mathrm{H_4SiO_4}}/a_{\mathrm H^+}^{10}$. Therefore
$$
5y+x=61.14,\qquad \boxed{y=12.228-0.20x}.
$$
This 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.
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