Reaction of Iron(III) with hydroxide ion The observations:

1) Treatment of 1 mL of 0.05 M $\ce{Fe(NH4)SO4}$ solution with excess 1 M $\ce{NaOH}$ results in the formation of a rust-colored red-brown iron-based solution with many suspended particles in the solution. These particles match the description of $\ce{Fe(OH2)3(OH)3}$. This is the vial on the right in the picture. 

2) Treatment of 1 mL of 0.05 M $\ce{Fe(NH4)SO4}$ solution with excess 6 M $\ce{NaOH}$ results in the formation of a clear but colored solution with no suspended particles in the solution or any sort of precipitate. The yellow-brown colored solution matches the description $\ce{Fe(OH2)4(OH)2+}$. This is the vial on the left. 



Why is it that treatment with more concentrated base results only two apparent deprotonations rather than three? 

More observations (after about 10 minutes): 

1) The 1 mL of 0.05 M $\ce{Fe(NH4)SO4}$ solution treated with excess 6 M $\ce{NaOH}$ has developed suspended particles. This is the vial on the left. 

2) The 1 mL of 0.05 M $\ce{Fe(NH4)SO4}$ solution treated with excess 1 M $\ce{NaOH}$ has started to become clear (the suspended particles are disappearing). It seems like much of the particles have settled on the bottom. This is the vial on the right. 



It seems that the 6 M $\ce{NaOH}$ is eventually "getting the job done" by going through with the third deprotonation of $\ce{Fe(OH2)6^{3+}}$. I don't think that the 1 M $\ce{NaOH}$ is deprotonating $\ce{Fe(OH2)3(OH)3}$. 

Still, why is it that treatment of iron(III) ions with concentrated base does not result in the formation of $\ce{Fe(OH2)3(OH)3}$ immediately but treatment with less concentrated base does? 

Even more observations:

1) Treatment of 10 mL of 0.05 M $\ce{Fe(NH4)SO4}$ with excess 6 M NaOH results in immediate and obvious $\ce{Fe(OH2)3(OH)3}$ formation. This seems to support the nucleation explanation; the higher solution volume means lower concentration and lower supersaturation and therefore fewer and larger crystals of $\ce{Fe(OH2)3(OH)3}$ being formed. 

Possible explanations I came up with:

1) I ran the results past a grad student and he said his best guess is that the excess $\ce{HO-}$ present stabilizes $\ce{Fe(OH2)4(OH)2+}$. He suggested a conductivity test to make sure that there are $\ce{Fe(OH2)4(OH)2+}$ ions in the clear solution. However, I did not have a conductivity meter handy. 

2) A problem in the lab manual from an unrelated section notes that strongly electrolytic solutions, such as saturated $\ce{H4NCl}$ solution, can cause dispersion rather than dissolution of $\ce{Al(OH2)3(OH)3}$ in solution. A 6 M $\ce{NaOH}$ solution is strongly electrolytic, and as a result, the ions may be keeping the $\ce{Fe(OH2)3(OH)3}$ from clumping together. 

What are your thoughts?
I wouldn't expect the reactions to yield different products but rather different particle size. The higher concentrated $\ce{NaOH}$ should hereby produce smaller particles†.

The "solution" in the left vial is probably rather a colloid of microscopically dispersed iron(III) hydroxide. It should show the Tyndall effect if you send a beam of light through it. However, the particles will aggregate over time to form larger particles that also precipitate, as seen in the second image.


† Higher concentration means higher supersaturation. Higher supersaturation leads to a higher nucleation rate, so more and thus smaller crystals are being formed.