Use the Henderson-Hasselbalch \( (\mathrm{H}-\mathrm{H}) \) equation, below, to find the ratio of deprotonated \( (+) \)-catechin (base) to protonated ( + )-catechin (acid) in the presence of a mild base ( \( \mathrm{pH}=12.2 \) ). ( + )-Catechin has a \( \mathrm{pK}_{\mathrm{a}} \) value of 10.1
\[
\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log (\text { base /acid })
\]
At this pH of 12.2, would you expect (+)-catechin to be more water soluble, less water soluble, or unchanged? Please explain using your answer from above (i.e. the \( \mathrm{H}-\mathrm{H} \) equation).