= Solution
Subtract reaction (I) from reaction (II), cancelling the common aqueous ions and <corundum>, and divide by two. Choose the albite-to-nepheline direction:
$$
\mathrm{NaAlSi_3O_8}+4\mathrm{H_2O}
\rightleftharpoons\mathrm{NaAlSiO_4}+2\mathrm{H_4SiO_4}.
$$
This is <albite> plus water converting to <nepheline> and dissolved silica. <Standard Gibbs energies of reaction> add with their <stoichiometric coefficients>, so logarithms of <equilibrium constants> do too. For this single-formula normalization,
$$
\boxed{\log_{10}K_{III}=\frac{-15.18-6.88}{2}=-11.03.}
$$
The doubled reaction has $\log_{10}K=-22.06$, while reversing either reaction changes the sign. Specifying the direction and normalization removes that apparent ambiguity.
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