Ka2 << Ka1
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Ka2 << Ka1
6E3. Calculate the pH of each of the following solutions of diprotic acids at 25°C, ignoring second deprotonations only when the approximation is justified.
What does the second part of the question ("ignoring deprotonations only when the approximation is justified") mean exactly?
I know the answer key says because Ka2 << Ka1, the 2nd ionization can be ignored. Why is this? When do we ignore the 2nd ionization? And how small does Ka2 have to be compared to Ka1 to be to warrant ignoring?
What does the second part of the question ("ignoring deprotonations only when the approximation is justified") mean exactly?
I know the answer key says because Ka2 << Ka1, the 2nd ionization can be ignored. Why is this? When do we ignore the 2nd ionization? And how small does Ka2 have to be compared to Ka1 to be to warrant ignoring?
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Re: Ka2 << Ka1
It means that the first deprotonated species contributes so little hydrogen ion that it has a negligible effect on pH. In this case it means that after H2CO3 becomes H+ and HCO3-, technically HCO3- can dissociate into H+ and CO3-2, but that second Ka value (Ka2) is so small that the final [H+] when factoring in Ka2 will be essentially that same as the [H+] when only considering Ka1. The book says when Ka1>>Ka2 it can be ignored but I do not know a threshold that determines how big or how small the difference is, it just says much greater.
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Re: Ka2 << Ka1
It means that the first dissociation is the only one that actually contributes something significant/ alters the pH noticeably. As the Ka decreases, that means not as much dissociation to the point where a Ka gets so small that it doesn't matter anymore. After the first dissociation, there aren't much H+ ions or OH- ions that can alter the solution we have.
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Re: Ka2 << Ka1
As Ian & Sara mentioned, the change in [H3O+] from the 2nd deprotonation is negligible. If it helps, it's pretty much the same reason we ignore water's contribution to [H3O+] in general acid/base equilibria. As for any sort of threshold to determine whether Ka2 is small enough, I found a post asking a similar question a while back & Prof. Lavelle replied "You will not be tested on calculating the effects of multiple protonation/deprotonation events. Typically, the second dissociation is so small that it can be ignored to a good approximation." So I'm pretty sure we don't need to know how to tell if a Ka2 is small enough; we can generally assume it is.
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Re: Ka2 << Ka1
The textbook states that as long as Ka2 is less than Ka1/1000 it is safe to approximate that the second deprotonation does not significantly affect pH. I believe this is the threshold that you are looking for.
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Re: Ka2 << Ka1
The previous answers all explained it very well! I just wanted to add that you can ignore Ka2 for most acids except H2SO4 (this shows up in a textbook problem)
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Re: Ka2 << Ka1
Andrew Wang 1C wrote:I just wanted to add that you can ignore Ka2 for most acids except H2SO4
Oh, is that because the Ka2 for H2SO4 is rather large/significant (in addition to its already large Ka1 value)?
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Re: Ka2 << Ka1
annabelchen2a wrote:Oh, is that because the Ka2 for H2SO4 is rather large/significant (in addition to its already large Ka1 value)?
Yeah, the Ka2 of H2SO4 is large enough to affect the pH so you'd include it in the calculations.
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Re: Ka2 << Ka1
You can ignore subsequent deprotonations when the acidity/basicity constant is extremely small, meaning that the protonated form is more favored, and deprotonation is extremely unlikely. I hope this helped!
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Re: Ka2 << Ka1
Usually if Ka2 is at least 1000x less thank Ka1, then it can be ignored and the approximation is valid. It is smaller in most polyprotic acids because the second/third/fourth hydrogen is harder to remove than the first, thus making it a weaker acid than its previous Ka1 counterpart and resulting in a lower Ka2 value. Hope this helped clear things up for ya!
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Re: Ka2 << Ka1
You can ignore Ka2 when it is significantly smaller than Ka1 (hence the double <<) :) The contribution that the second deprotonation makes towards the pH is negligible when added to the first.
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Re: Ka2 << Ka1
Take H3PO4 as an example, Ka1=7.1 × 10-3, Ka2=6.3 × 10-8, Ka3=4.5 × 10-13. The difference between Ka1 and Ka2 is at the scale of 105. Thus, in these cases, the second deprotonation can often be ignored.
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