Why Does adding water have no effect on pH?

[ipodtouch]

It says here in the final page of the BR Chem Chapter 4 that adding H2O to a mixture of conjugate pair mixture has No effect on the pH.

Not even by a few molecules?


I was always kind of confused by this conceptually because let's say:
We have a 20mL mixture of HCl at 70 HCL /30 Cl-, and then we add like 1 mL of water, which will cause 50/50 dispersion of H2O and HO-.
I am wondering why this doesn't change the pH, because we are adding a significant amount of pH 7.0 to a solution of low pH.... I would think that the solution pH would rise from

2.0 to 2.0000000000000000000000000001


What if we add like 5000mL of water to the 20mL mixture?
I would think that it would be such an overpowering dilution that the pH would be altered.

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[chiddler]

You're right. From pH = pKa - log(A-/HA), as you add water, the ratios of dissolved species do actually change but it is so slight that it makes virtually no impact on the pH. These changes are so minute (as you wrote, a ridiculously small change) that it's like 0.499999/0.5000001 = 1.

For larger amounts of water, this is not true because some of the conjugate base/conjugate acid will react and form the more stable species of the two so the ratio will change (though still, it will be slight unless significantly large amounts of water). This is a perfect example of what not to think for the mcat! (ie i'm wrong)

 

[milski]

You're right. From pH = pKa - log(A-/HA), as you add water, the ratios of dissolved species do actually change but it is so slight that it makes virtually no impact on the pH. These changes are so minute (as you wrote, a ridiculously small change) that it's like 0.499999/0.5000001 = 1.

For larger amounts of water, this is not true because some of the conjugate base/conjugate acid will react and form the more stable species of the two so the ratio will change (though still, it will be slight unless significantly large amounts of water).

But you're adding water to the same mixture. That means that both [A] and [HA] will decrease by the same factor and [A]/[HA] will stay constant. Since pKa is also a constant, pH does not change.

 

[chiddler]

But you're adding water to the same mixture. That means that both [A] and [HA] will decrease by the same factor and [A]/[HA] will stay constant. Since pKa is also a constant, pH does not change.

Hmm good point. Large amounts of water DO change the pH, right? Why would they do that?

otherwise i'd find it hard to believe that a formulated mixture of a gram of salts could make an ocean of water a buffer.

 

[milski]

Hmm good point. Large amounts of water DO change the pH, right? Why would they do that?

otherwise i'd find it hard to believe that a formulated mixture of a gram of salts could make an ocean of water a buffer.

pH is constant only if you have both conjugates in the mixture. For strong acid that means concentration high enough that you have un-disassociated HCl in it.

Another thing to keep in mind: as you keep adding water, eventually [A] and/or [AH] will get lower than 10e-7 and water will become a significant source of [H]. At that point the pH will start changing.

 

[ipodtouch]

pH is constant only if you have both conjugates in the mixture. For strong acid that means concentration high enough that you have un-disassociated HCl in it.

This seems like the main point. The limit the Henderson Hassel-bach equation works up to is until there is no more undissolved HCl?

But this is also confusing because HH equation is:

pH= pKa+ Log (conjugate base/conjugate acid)

Once we approach 0 conjugate acid, but still a few molecules, remaining, we approach a pH of infinity.



If we had a 10mL sample of HCl/Cl-
and added 1,000,000,000,000 gallons of H2O... I would think that the pH would approach ~7 instead....

what am I missing here?

 

[chiddler]

pH is constant only if you have both conjugates in the mixture. For strong acid that means concentration high enough that you have un-disassociated HCl in it.

Another thing to keep in mind: as you keep adding water, eventually [A] and/or [AH] will get lower than 10e-7 and water will become a significant source of [H]. At that point the pH will start changing.

But how does the HH equation describe that? If pKa is a property of the acid/conj base, and the ratio is set, pH should be determined.

 

[milski]

This seems like the main point. The limit the Henderson Hassel-bach equation works up to is until there is no more undissolved HCl?

But this is also confusing because HH equation is:



Once we approach 0 conjugate acid, but still a few molecules, remaining, we approach a pH of infinity.



If we had a 10mL sample of HCl/Cl-
and added 1,000,000,000,000 gallons of H2O... I would think that the pH would approach ~7 instead....

what am I missing here?

- forget about using HCl as an example here. You need to have both the acid and the conjugate present. You cannot achieve that in any practical way with a strong acid.

- yes, it will start approaching 7. As you add more water you decrease both [HA] and [A]. Once one of these gets close to 10e-7, water will be a significant contributor of [H] and you cannot ignore it in your pH calculations. That will start pushing the pH towards 7.

 

[milski]

But how does the HH equation describe that? If pKa is a property of the acid/conj base, and the ratio is set, pH should be determined.

You have two equilibriums going on, one is HA<->H + A, the other is H2O<->H+OH. If you want to be absolutely precise, you have to consider both. For high enough concentrations of HA/A you can ignore the second one. That's not true if the concentrations are comparable to pKa of water.

 

[chiddler]

milski i hope you're not taking your mcat soon. i don't want you leaving these forums <3

 

[milski]

milski i hope you're not taking your mcat soon. i don't want you leaving these forums <3

You should send some gifts to whoever does scheduling at my school - due to class conflicts I am taking it at least three months later than planned.

 

[chiddler]

You should send some gifts to whoever does scheduling at my school - due to class conflicts I am taking it at least three months later than planned.

nice to see all my bribes are working

 

[milski]

I must have too much time on my hands. I was curious about what sort of changes are we talking here so I did play a bit with a few number. The part that surprised me at first is that stronger acids are more easily affected by such dilution. In a way that makes sense, since the weaker the acid is, the more the water is contributing to the pH, so adding more water should not be so important.

A short summary:
For an acid with pKa=2 and [HA]=1 [A]=1 doubling the water changes pH by 0.4%
If we drop the pKa a bit, say pKa=5, [HA]=1 [A]=1 doubling the water changes pH by 0.00017%

Lower concentrations of acid are more sensitive:
pKa=2 [HA]=0.1 [A]=0.1 has 2.7% change in pH with doubled water
pKa=5 [HA]=0.1 [A]=0.1 has 0.0017% change in pH with doubled water

 

[MedPR]

I must have too much time on my hands. I was curious about what sort of changes are we talking here so I did play a bit with a few number. The part that surprised me at first is that stronger acids are more easily affected by such dilution. In a way that makes sense, since the weaker the acid is, the more the water is contributing to the pH, so adding more water should not be so important.

A short summary:
For an acid with pKa=2 and [HA]=1 [A]=1 doubling the water changes pH by 0.4%
If we drop the pKa a bit, say pKa=5, [HA]=1 [A]=1 doubling the water changes pH by 0.00017%

Lower concentrations of acid are more sensitive:
pKa=2 [HA]=0.1 [A]=0.1 has 2.7% change in pH with doubled water
pKa=5 [HA]=0.1 [A]=0.1 has 0.0017% change in pH with doubled water

Nerd.

 

[milski]

Nerd.