 Solution

Set $x={\log }_{10}{a}_{{\mathrm{H}}_4\mathrm{Si}{\mathrm{O}}_4^0}$ and $y={\log }_{10}(a_{\mathrm{N}{\mathrm{a}}^{+}}/a_{{\mathrm{H}}^{+}})$, treating the supplied concentration notation as an ideal-dilute activity approximation. For pure solids and unit water activity, the first chemical equilibrium relation gives $\log K_a=2y+2x$ and the second gives $\log K_b=2y+6x$. Thus plotTheir slopes are $-1$ and $-3$ respectively. They intersect at $(x_{\ast },y_{\ast })=(-5.515,8.955)$. Both the relevant silica activities and the intersection lie at negative $x$, as expected for a dilute aqueous silica species.