= Solution
For pure <albite>, pure <nepheline> and unit water activity, $Q_{III}=a_{\mathrm{H_4SiO_4}}^2$. Consequently
$$
\boxed{x=\tfrac12\log_{10}K_{III}=-5.515,}
$$
a vertical line through the intersection of (I) and (II). It does not depend on $y$ because the net reaction has no hydrogen or 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 boundaries for albite, nepheline and corundum}
{description=Original diagram derived from the two supplied equilibrium constants. Solid black segments delimit the stable mineral fields; dashed grey continuations are metastable pairwise equilibria. Logarithms are base ten and the plotted silica activities are below one.}
Back to article page
