Is 4-nitrobenzaldehyde or 4-aminobenzaldehyde more likely to be hydrated? Of 4-nitrobenzaldehyde and 4-aminobenzaldehyde, which has the higher $\ce{K_{eq}}$ of hydration? 

 $\qquad \qquad$


Steric factors can likely be ignored in this problem since we are dealing with rather structurally similar compounds. 

I suspect the problem comes down to recognizing that a more polar compound is more favorably solvated by water. 

I know that the nitro group is always electron-withdrawing, and I drew a few resonance structures as well. The resonance structures for both, however, seem to suggest that both are rather polarized (given the distribution of formal charges). 

Resonance forms for 4-nitrobenzaldehyde


I expect the ones with the oxygen possessing the incomplete octet are very minor resonance contributors. 
Therefore it seems that the main thing about this molecule is that the nitro group is withdrawing electron density from the ring and the aldehyde group. 




Resonance forms for 4-aminobenzaldehyde


The resonance structures here are all somewhat reasonable.
I think you're asking about the following equilibrium



This equilibrium will shift to the left as the aldehyde becomes more stabilized.  So the question becomes, which R group, $\ce{NO2}$ or $\ce{NH2}$, will tend to stabilize the aldehyde the most?  A substituent that further delocalizes the positive charge shown in the dipolar resonance form would tend to be stabilizing.  Just like in electrophilic aromatic substitution, additional stabilizing resonance structures can be drawn involving the amino substituent, but similar structures involving the nitro group are not as significant due to electrostatics (2 adjacent positive charges). Therefore, since the amino will stabilize the aldehyde better than the nitro substituent, the equilibrium involving the amino substituent should lie further to the left, and the aldehyde with the nitro substituent should be more hydrated.