The Student Room Group

Reply 1

Velouria
Are these equations correct for the reaction of Cr3+ ions with NaOH solution?

[Cr(H2O)6]3+ + 3OH- ---> [Cr(H2O)3(OH)3] + 3H2O
green solution----------------light green ppt

And in excess:

[Cr(H2O)3(OH)3] + OH- ---> [Cr(H2O)2(OH)4]- + 4H2O
light green ppt--------------dark green solution


yes

Reply 2

Original post by charco
yes


Hi sorry for bringing up such an old post, but why does the [Cr(H2O)3(OH)3] react further with XS NaOH to produce [Cr(H2O)2(OH)4]?

Reply 3

The same reason the [Cr(H2O6)]3+ reacted in the first place. Although the effect is weaker, that positive metal ion still makes the water ligands bound to it acidic.

Reply 4

Original post by Blocker
The same reason the [Cr(H2O6)]3+ reacted in the first place. Although the effect is weaker, that positive metal ion still makes the water ligands bound to it acidic.


Thanks Blocker

Reply 5


So what is stopping it becoming the full blown hexahydroxy molecule? =/

Reply 6

In excess won't it go to [Cr(OH)6]3- ?

Reply 7

With each deprotonation, you essentially exchange a water ligand for an -OH. This has a negative charge, and hence is better able to balance the charge of the positive Cr3+ cation at the centre. The complex is acidic because this cation is withdrawing electron density from the surrounding waters. as you introduce more hydroxides into the system the cation becomes less electron withdrawing and the complex less acidic.

kapeesh? ;D

[EDIT] Also the process starts getting really unfavourable once you get to [Cr(OH)4(OH2)2]-. Youve now got an overall negatively charged complex, which to deprotonate further you need to bring into close proximity to -OH. This is also negatively charged, and the ions will repel, making the reaction harder.
(edited 14 years ago)

Reply 8

What happens when aqueous ammonia and excess aqueous ammonia are added? If it the same, with ammonia acting as a base as opposed to as a nucleophile?