= Solution
The net <reaction quotient> is $a_{\mathrm{H_4SiO_4}}^2$ when the solid and water activities equal one. Therefore its equilibrium line is
$$
\boxed{x=\tfrac12\log_{10}K=-5.515,}
$$
a vertical line through the intersection of the other two. There is no dependence on $y$, because the net reaction contains neither hydrogen ions nor sodium ions.
\Image[/past-exam-of-the-year-2-of-the-earth-sciences-geology-course-of-the-university-of-oxford/2025/paper-3-q10-mineral-activity-diagram.png]
{title=Stable and metastable mineral boundaries for the common albite-nepheline-corundum equilibrium example}
{description=Original diagram computed from the supplied equilibrium constants. Solid lines show stable coexistence; dashed continuations are metastable. Logarithms are base ten.}
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