How to Calculate the pH of a 0.1M Diprotic Acid Solution?

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SUMMARY

The pH of a 0.1M diprotic acid solution can be calculated using the overall protonation constants, which are logarithmically defined as 4.3 and 5.5. The relevant acid dissociation reactions are H+ + A2- <=> HA- and 2H+ + A2- <=> H2A. The calculated values for the stepwise dissociation constants are Ka1 = 6.3096 x 10-2 and Ka2 = 5.0119 x 10-5, leading to a final pH of 1.27.

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  • Understanding of diprotic acids and their dissociation reactions
  • Familiarity with acid dissociation constants (Ka)
  • Knowledge of logarithmic calculations
  • Basic principles of chemical equilibrium
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  • Learn how to derive Ka values from protonation constants
  • Explore the calculations involved in weak acid dissociation
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Guillermo
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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
 
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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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Wow, I was afraid you will not check the forum today :)

What are protonation constants?

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

reactions:

H^+ + A^{2-} \leftrightarrow HA^-

2H^+ + A^{2-} \leftrightarrow H_2A

and overall constants (there are also stepwise constants):

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

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

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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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
 
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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:blushing:

1.27

G
 

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