= Solution
For the reaction direction in part ii, pure solids and unit water activity give $K_c=1/a_{\mathrm{H_4SiO_4^0}}^2$. Therefore
$$
11.03=-2x,\qquad\boxed{x=-5.515.}
$$
Plot a \b[vertical line] through the common intersection, independent of $y$. Increasing dissolved silica favors <albite>, consistently with its larger Si/Al ratio. The vertical line’s stable and metastable portions are selected by comparison with <corundum>, not by its own mass-action equation alone.
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