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Why can't pH of diprotic acids be calculated like that of amphiprotic?

by krackers
Tags: acids, amphiprotic, calculated, diprotic
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krackers
#1
Jul26-14, 11:17 AM
P: 57
I was reading this website (http://www.chembuddy.com/?left=pH-ca...phiprotic-salt) on calculating pH of amphiprotic salts.

My question is why can't you do something similar for diprotic acids? Take for example the diprotic acid H2S.

Because you have the following reactions:

H+ + HS- ⇔ H2S

HS- ⇔ H+ + S2-

can't you treat HS- like an amphiprotic substance and calculate the pH the same way you calculated pH for other amphiprotic substances?

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I also have another question relating to these topics. In this image,

I don't understand how they came to the conclusion that [H3AsO4-] ≈ [HAsO42-]. Further, why is H2AsO42- considered amphiprotic when the value of K-1, that is the equilibrium constant for H2AsO42- + H+ ⇔ H3AsO4, is much higher than that of the reaction where it acts an acid?
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Borek
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Jul28-14, 04:52 AM
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Quote Quote by krackers View Post
I was reading this website (http://www.chembuddy.com/?left=pH-ca...phiprotic-salt) on calculating pH of amphiprotic salts.

My question is why can't you do something similar for diprotic acids? Take for example the diprotic acid H2S.

Because you have the following reactions:

H+ + HS- ⇔ H2S

HS- ⇔ H+ + S2-

can't you treat HS- like an amphiprotic substance and calculate the pH the same way you calculated pH for other amphiprotic substances?
It can be done, but only if you can be sure HA- dissociation won't change first dissociation step. That can work for a sulfuric acid, where we can safely assume first step is always 100% dissociatied.

I don't understand how they came to the conclusion that [H3AsO4-] ≈ [HAsO42-].
Simple stoichiometry.

Further, why is H2AsO42- considered amphiprotic when the value of K-1, that is the equilibrium constant for H2AsO42- + H+ ⇔ H3AsO4, is much higher than that of the reaction where it acts an acid?
Amphiprotic means it can react both ways, it doesn't tell much about relative strengths of both reactions. And the difference of 4 orders of magnitude in such a case is not much.
krackers
#3
Jul28-14, 11:45 PM
P: 57
Quote Quote by Borek View Post
It can be done, but only if you can be sure HA- dissociation won't change first dissociation step. That can work for a sulfuric acid, where we can safely assume first step is always 100% dissociatied.

How would you know whether this holds true for HS-?

The Ka values are as follows:

H+ + HS- ⇔ H2S (Ka = 107)

HS- ⇔ H+ + S2- (Ka = 10-19)

Quote Quote by Borek View Post
Simple stoichiometry.
The amphiprotic reactions are as follows:

H2AsO4 ⇔ HAsO42- + H+ (Ka = 1.1 *10-7)

H2AsO4- + H+ ⇔ H3AsO4 (Ka = 1.6 * 102)

Doesn't the acidic reaction have a much higher Ka and thus won't the concentration of H3AsO4 greater than that of HAsO42-?

Borek
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Jul29-14, 01:51 AM
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Why can't pH of diprotic acids be calculated like that of amphiprotic?

Quote Quote by krackers View Post
How would you know whether this holds true for HS-?
Calculate pH using this approach, plug it into equilibrium equations and see if they still make sense.

H+ + HS- ⇔ H2S (Ka = 107)
More like 10-7.

Have you seen http://www.chembuddy.com/?left=pH-ca...tic-simplified ?

Doesn't the acidic reaction have a much higher Ka and thus won't the concentration of H3AsO4 greater than that of HAsO42-?
They ASSUME system is dominated by the reaction they wrote and they use it to calculate pH. Assumption doesn't have to be correct - but once again, after calculating pH you can do the simple sanity check, seeing if the numbers fit all known equilibria and mass balances. Turns out they do.

It is discussed at the page you linked to - there are tables showing how accurate the results calculated using equivalent assumption are. Quotient [H2A]/[A2-] being close to 1 is equivalent to the idea that [HAsO42-]≈[H3AsO4] (these may look like completely unrelated, but try to understand the acid/base reaction happening and you will see they are the same).


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