= Solution
Subtract the <chlorite>–<gibbsite> reaction from the <chlorite>–<kaolinite> reaction. The common <chlorite>, protons and magnesium cancel, leaving
$$
2\,\mathrm{Al(OH)_3}+2\,\mathrm{H_4SiO_4}=\mathrm{Al_2Si_2O_5(OH)_4}+5\,\mathrm{H_2O}.
$$
This converts <gibbsite> and dissolved silica into <kaolinite>. <Gibbs energies of reaction> add with reaction stoichiometry, so decimal <logarithms> of <equilibrium constants> also subtract:
$$
\boxed{\log_{10}K=61.14-52.65=8.49}.
$$
With unit water and solid <thermodynamic activities>, $K=a_{\mathrm{H_4SiO_4}}^{-2}$; thus \b[$x=-8.49/2=-4.245$], a vertical boundary. Reversing the reaction would give $\log_{10}K=-8.49$ and the same boundary. Its intersection with either sloping line is \b[$(x,y)=(-4.245,13.077)$].
\Image[/past-exam-of-the-year-2-of-the-earth-sciences-geology-course-of-the-university-of-oxford/2023/paper-1-mineral-stability.png]
{title=Stable chlorite, gibbsite and kaolinite fields with metastable boundary continuations}
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