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

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):

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

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    $\begingroup$ "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 $\endgroup$
    – Tan Yong Boon
    Commented Jan 3, 2019 at 1:34
  • $\begingroup$ Have a look at Jan's answer here: chemistry.stackexchange.com/q/68547/44877 $\endgroup$
    – Tan Yong Boon
    Commented Jan 3, 2019 at 1:36

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This is an equilibrium reaction. Using e.g. LeChatelier's principle, you can figure you that if you increase the concentration of $\ce{NH3}$ sufficiently, you shift the equilibrium so far to the right that you get the hexamine as the main product. At neutral pH in water, the concentration of hydroxide is quite low (100 nanomolar). You would expect this reaction to be pH-dependent, and of course depend on the concentration of ammonia in solution.

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