Why does ligand substitution occur when OH- is a better ligand than NH3?

$$\ce{[Cr(H2O)3(OH)3](s) + 6NH3 -> [Cr(NH3)6]^3+ + 3H2O + 3OH-}$$

When excess $$\ce{NH3}$$ is added to triaquatrihydroxychromium(III), why does ligand substitution occur when $$\ce{OH-}$$ is a better ligand than $$\ce{NH3}$$?

Wouldn't the 3 $$\ce{H2O}$$ molecules be deprotonated to $$\ce{OH-}$$ and therefore form $$\ce{[Cr(OH)6]^3-}$$?

This also seems to occur when excess $$\ce{NH3}$$ is added to tetraaquadihydroxocobalt(II):

$$\ce{Co(H2O)4(OH)2(s) + 6NH3 -> [Co(NH3)6]^2+ + 4H2O + 2OH-}$$

• "Excess" may mean that this occurs simply because the equilibrium is pushed to the right to form the hexaammine complex due to overwhelming conc. of NH3 – Tan Yong Boon Jan 3 at 1:34
• Have a look at Jan's answer here: chemistry.stackexchange.com/q/68547/44877 – Tan Yong Boon Jan 3 at 1:36