Hydrolysis of Radiolabeled Esters Hydrolysis of C$_6$H$_5$COO$^{18}$CH$_3$ with aqueous NaOH produces what final product.

(a) CH$_3$$^{18}$OH

(b) CH$_3$CO$^{18}$ONa

(c) CH$_3$C$^{18}$OONa

(d) CH$_3$$^{18}$ONa

(e) CH$_3$OH

The correct answer is (a), which is puzzling to me.  I think it should be (d), as when the radiolabeled methoxy leaving group is pushed out by the reformation of the carbonyl, it will be negatively charged.  The only cation that is floating around is Na$^+$. 

Where, oh were, did this magical proton appear in basic conditions?

EDIT:  Since I think my comments will get longer I'll just post my question here.  I agree that the methoxy is in equilibrium with water:

CH$_3$O$^-$ + H$_2$O $\rightleftharpoons$ CH$_3$OH + OH$^-$

But the pK$_a$ of methanol is 15.54, and the pK$_a$ of water is 15.74.  Thus this equilibrium will favor the methoxy/water side.  Thus, the slight majority of product should be a methoxy salt.
You are surely right in that the labeled oxygen will stay with the alcohol and not with the acid. Now is the time to recall what do you know about the comparative acidity of alcohol and... wait, what does the word "aqueous" stand for, really?