Acidity of aldehydes
Honestly, I don't see how anyone could have predicted this outcome before seeing the actual $\ce{pK_a}$ data.  If you look up the acidities of methanal and ethanal, you find $\ce{pK_a}$ = 13.27 and 13.57 respectively.  A pretty small difference to begin with. As the picture below illustrates, in the case of methanal we are removing the proton directly bound to the carbonyl carbon to produce the anion pictured below. 



Now if someone asked, "would pulling the same proton off from ethanal be easier or harder?", you could answer "harder" because that methyl group (being electron releasing) that is attached to the carbonyl carbon would act to slightly destabilize the anion with the negative charge on carbon.  But it turns out that when ethanal is treated with base a different proton is removed!  The proton that is on the methyl group (said to be alpha to the carbonyl carbon) is the proton that is removed.  Without knowing the actual $\ce{pK_{a}{'s}}$ beforehand, how could you know that the anion involved in the methanal case would be ever so slightly more stable than the resonance structures involved in the ethanal case?  

Edit: Thanks to Greg E and Nicolau Saker Neto for pointing out an error in the earlier version of the figure and the corresponding text respectively.