6D.11

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paytonm1H
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Joined: Fri Sep 28, 2018 12:18 am

6D.11

Postby paytonm1H » Fri Dec 07, 2018 11:48 pm

Hi all.

In the 7th edition, question 6D.11 asks whether each salt has a pH greater than, less than, or equal to 7. Part (e) gives AlCl3. Why so small, highly charged cations react with water to form acidic solutions?

The answer key says that the pH is less than 7 and gives the equation:
AlH2O)6 3+(aq) + H2O(l) --> H3O+(aq). + Al(H2O)5OH 2+(aq)

How would you know to change the salt to Al(H2O)6? Also just in general, why do metal cations form acidic solutions?

Thanks!

Chem_Mod
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Re: 6D.11

Postby Chem_Mod » Sun Dec 09, 2018 1:14 pm

You change the salt because water now becomes a very saturating ligand (considering the metal is now surrounded by water) so it replaces chlorine. Metal cations usually act as acids because their positive charge will lead them to form ionic compounds with hydroxide ions in solution, therefore increasing the [H+].


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