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Offline Rutherford

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Solutions
« on: March 04, 2012, 08:33:07 AM »
Why KAl(SO4)2, NaHSO4, NaH2PO4 make acidic solutions? I know protolysis, but I cant' understand these three examples.

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Offline Rutherford

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Re: Solutions
« Reply #2 on: March 04, 2012, 09:39:51 AM »
I know that theory, but I don't understand why those compounds make an acidic solution.
KAl(SO4)2-->?
NaHSO4 made from a strong base (NaOH) and a strong acid (HSO4-) so it has to make a neutral solution.
NaH2PO4 same as above only that the H2PO4- acid is less strong than HSO4- so it should make a basic solution, but they are all acidic.

Offline Borek

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Re: Solutions
« Reply #3 on: March 04, 2012, 10:05:46 AM »
NaHSO4 made from a strong base (NaOH) and a strong acid (HSO4-) so it has to make a neutral solution.

HSO4- is not the acid used to prepare NaHSO4.
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Offline Rutherford

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Re: Solutions
« Reply #4 on: March 04, 2012, 10:06:50 AM »
H2SO4?

Offline AWK

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Re: Solutions
« Reply #5 on: March 04, 2012, 10:45:39 AM »
All: Al(H2O)63+, HSO4- and H2PO4- are Bronsted acids.
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Offline Rutherford

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Re: Solutions
« Reply #6 on: March 04, 2012, 11:00:34 AM »
How does that explain my question? Sorry for bothering, but I really don't understand. After a lot of thinking I came to this: using NaHSO4 for example, if I write the hydrolysis reaction:
Na++H++SO42-+H2O-->Na++SO42-+H3O+
The H3O+ makes the solution acidic.
NaH2PO4 i can't write the H2PO4- anion as 2H++PO43- because it is a weak acid:
Na++H2PO4-+H2O-->Na++HPO4-+H3O+ but why isn't here H3PO4 and OH- made when H2PO4- can be a proton donor and acceptor either and the Ka1 of H3PO4 isn't too big like it is for H2SO4? And what would the hydrolysis reaction for KAl(SO4)2 look like?

Offline Borek

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Re: Solutions
« Reply #7 on: March 04, 2012, 11:36:12 AM »
Na++H++SO42-+H2O-->Na++SO42-+H3O+

This is not hydrolysis. HSO4- is an acid, that's enough. Na+ presence doesn't matter.

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i can't write the H2PO4- anion as 2H++PO43- because it is a weak acid:

It is a weak acid, but still strong enough to acidify the solution. Again, Na+ presence doesn't matter.

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And what would the hydrolysis reaction for KAl(SO4)2 look like?

Al case is a different one.
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Offline AWK

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Offline Rutherford

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Re: Solutions
« Reply #9 on: March 04, 2012, 02:09:39 PM »
You are right Borek, I should remove the Na+ ions because they are on the both side of the equations. How do I know if an acid is strong or weak, what pKa is the limit value?
And pretty much I have to read now :D because of AWK. Thanks to you both!

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