 Solution

Use base-ten logarithms and dimensionless thermodynamic activities at the stated temperature and pressure. Define $x={\log }_{10}{a}_{{\mathrm{H}}_4\mathrm{Si}{\mathrm{O}}_4}$ and $y={\log }_{10}(a_{\mathrm{N}{\mathrm{a}}^{+}}/a_{{\mathrm{H}}^{+}})$. The concentration notation is an ideal dilute-solution approximation with a consistent concentration reference.

With pure solids and unit water activity, the two reaction quotients are $10^{2y+2x}$ and $10^{2y+6x}$. Equating them to the corresponding equilibrium constants givesThe slopes on the mineral activity diagram are $-1$ and $-3$. The lines meet at $(x_{\ast },y_{\ast })=(-5.515,8.955)$, within the negative silica-logarithm range.