PH calculation once again

  • Thread starter Guillermo
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  • #1
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This pH calculation question is only for those interested, but I wan't to do it, but I have no idea where and how to start...

What is pH of solution of 0.1M diprotic acid if loarithms of overall protonation constants are 4.3 and 5.5.

What is going on? No idea how to start...

Borek, are you tehere?

G
 

Answers and Replies

  • #2
Borek
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What's the problem? That's relatively easy, you have a solution of weak diprotic acid, you may ignore the second step of dissociation, that's not more difficult then the previous question.
 
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  • #3
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Wow, I was afraid you will not check the forum today :)

What are protonation constants?

G
 
  • #4
Borek
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One of the ways to define acid dissociation equlibrium. In case of diprotic acid it will look like this:

reactions:

[tex]H^+ + A^{2-} \leftrightarrow HA^-[/tex]

[tex]2H^+ + A^{2-} \leftrightarrow H_2A[/tex]

and overall constants (there are also stepwise constants):

[tex]B_1 = \frac{[HA^-]}{[H^+][A^{2-}]}[/tex]

[tex]B_2 = \frac{[H_2A]}{[H^+]^2[A^{2-}]}[/tex]

Compare these to Ka1 and Ka2 definitions - you should be able to find out how to calculate them from known B1 and B2 (hint: check Ka1*Ka2).

Protonation constants are rarely used. The idea behind them is that if you use protonation constants calculations are done in exactly the same way as the calculations for complex creation (protons take place of ligands). Math is exactly the same.
 
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  • #5
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What I get is Ka2=1/B1 and Ka1=1/B2Ka2.

Ka2=5.0119*10^-5 and Ka1=6.3096*10^-2

and pH=1.1

G
 
  • #6
Borek
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Ka values seem OK (where did you get so much significant digits from?)

pH is wrong. I think you did the same mistake as previously. I wonder if you will check the forums now :wink:
 
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  • #7
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:blushing:

1.27

G
 

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