I keep getting 3.55 as the pH of 0.0055 M AlCl3. How did you guys get 3.06?
This is what I did.
Al(H2O)6 3+ + H2O(l) <-> H3O+ (aq) + Al(H2O)5OH
I set up my ice table and set my Ka equal to 1.4e-5=x^2/0.0055. When I solve for x, I get 2.8e-4 and the -log of that is 3.55.
Can someone point out where I went wrong? I can't figure it out. Thanks!
6D.15 Help
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Re: 6D.15 Help
Hi there!
I think you just had an extra zero in the concentration of AlCl3!
The concentration should be 0.055 M not 0.0055 M.
Otherwise, everything else looked good.
I think you just had an extra zero in the concentration of AlCl3!
The concentration should be 0.055 M not 0.0055 M.
Otherwise, everything else looked good.
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- Posts: 35
- Joined: Mon Jan 09, 2023 9:08 am
Re: 6D.15 Help
Hey how did you get (H20)6 3+ for AlCl3? How did you write the equation like that? I am a bit confused since AlCl3 does not have an H.
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Re: 6D.15 Help
Naibe Reynoso 2C wrote:Hey how did you get (H20)6 3+ for AlCl3? How did you write the equation like that? I am a bit confused since AlCl3 does not have an H.
AlCl3 interacts with water to dissociate into Al 3+ and three Cl-. Al 3+ then binds with 6 water molecules. I would take a look at coordination compounds to see why but from what I have heard, Al 3+ has all 3-level orbitals emptied, leading to aluminum using 6 of the orbitals to accept lone pairs (hybridizing 3s, 3p, and two 3d orbitals). This leads to the formation of Al(h2O)6 3+
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