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Topic: PH amphiprotic salt approximation  (Read 6666 times)

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Offline Rutherford

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PH amphiprotic salt approximation
« on: December 11, 2012, 01:18:48 PM »
On this link: http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt,  below the equation 12.12 is a sentence "Comparing 12.8 and 12.12 it is obvious that both assumptions that allowed to derive equation 12.8 are equivalent to the assumption that quotient [H2A]/[A2-] is close to 1." How can they be close to 1, if I had for example 0.001M of a weak amphiprotic substance Ka1=10-3 and Ka2=10-7, [H2A]  should be much lower than [A2-]? I need an explanation because that way is shorter and easier to do fast on paper that the one above it.

Offline Borek

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Re: PH amphiprotic salt approximation
« Reply #1 on: December 11, 2012, 02:07:41 PM »
No idea what kind of explanation you are looking for. First derivation uses well defined assumptions - if they are wrong, ratio is probably far from 1.

Your example (0.001M HA-, pKa1,2 = 3 & 7) yields a solution with pH 5.15 (instead of expected 5) and ratio

[tex]\frac{[H_2A]}{[A^{2-}]} = \frac{6.9\times10^{-6}}{1.4\times10^{-5}}=0.49[/tex]

Apparently you can't use approximate formula for this case. Hardly surprising, tables below shows that below 0.01M one has to expect problems.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline Rutherford

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Re: PH amphiprotic salt approximation
« Reply #2 on: December 11, 2012, 02:39:58 PM »
I was aiming at the difference between Kb1(10-11) and Ka2(10-7). I have the reactions:
HA- ::equil:: H++A2- K=10-7
HA-+H2O ::equil:: H2A+OH- K=10-11
I don't understand how possibly, whatever concentration I take, [H2A]≈[A2-] because H2A is made tougher than A2-. This explanation I need.

Offline Borek

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Re: PH amphiprotic salt approximation
« Reply #3 on: December 11, 2012, 04:06:49 PM »
I don't understand how possibly, whatever concentration I take, [H2A]≈[A2-]

Nothing to explain here, as it is not true. It is the other way around - as long as these concentrations are close to each other, you can use formula [itex]pH=\frac 1 2 (pK_{a1}+pK_{a2})[/itex]. When they are not close to each other, the formula fails. As explained on the ChemBuddy page and as I mentioned in the previous post, these concentrations are comparable only as long as starting concentration of HA- is large enough (and both dissociation constants have reasonable values, neither too high nor too low).
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Offline Rutherford

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Re: PH amphiprotic salt approximation
« Reply #4 on: December 12, 2012, 06:50:44 AM »
(and both dissociation constants have reasonable values, neither too high nor too low).
I just wanted to know this. Therefore, I should find a reserve way to calculate the pH. Thanks.

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