Posted by Krystal on Sunday, May 20, 2007 at 6:06pm.
I'm reviewing my notes for a test tomorrow, and I found that I have a question concerning acids that under go multiple protonizations. I know the molarity [or moles, if that's being calculated] of H3O+ of the first protonization can be found using the Ka, etc. I have in my notes that for the next protonization, the molarity of H3O+ on the products side is always the molarity/moles of the H3O+ found in the first prot. + x.
Ex. in case my explanation is confusing:
Prot 1:
H3PO4 + H20 <==> H3O+ + H2PO4-
[H3PO4]= 5.00 - x
[H3O+]= x
Ka= x2/5.00 = .193
Prot 2:
H2PO4- + H2O <==> H3O+ + HPO4-2
[H2PO4-]= .193-x
[H3O+]= .193+x
If I continued this to the third protonization of H3PO4 [HPO4-2 + H20 <==> H+ + PO4-3], would the concentration of H3O+ be [H3O+ in prot 2] + x?
I wonder what you mean by..
Ka= x2/5.00 = .193
You need to solve for x from Ka, so I assume that last equal sign means..
Ka= x2/5.00 => x= .193
Then..
Go to step two..
H2PO4- + H2O <==> H3O+ + HPO4-2
[H2PO4-]= .193-x
[H3O+]= .193+x and so on.
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