= Solution
Use decimal <logarithms> and write $x=\log_{10}a_{\mathrm{H_4SiO_4}}$ 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 give
$$
K=\frac{a_{\mathrm{Mg}^{2+}}^5a_{\mathrm{H_4SiO_4}}^3}{a_{\mathrm H^+}^{10}},\qquad \log_{10}K=5y+3x=52.65.
$$
Hence \b[$y=10.53-0.60x$]. 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>.
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