EK GChem Lect 6, question# 131, 2007 ed.

[ODorDO]

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Sup guys, here is the question:

Carbonic acid has a Ka of 4.3 x 10^-7. What is the pH when 1 mole of NaHCO3 is dissolved in 1 liter of water?

This may seem like an easy question for yall but I just don't understand why we should Kb instead of Ka to calculate the pH here 😛 I mean NaHCO3 is amphoteric so it can be either acidic or basic... right?

Thank you

 

[altogether]

So water is pH=7=pOH
pkb of na.bicarbonate is ~7.4
So if pOH<pKb, the solution is will act as an acid and your molecule (bicarbonate) will act as a base.
Thus, your equilibrium expression should reflect bicarbonate acting as a base for which you need Kb to solve. Hope that hleps!

 

[ODorDO]

So if i didnt know the pka value of sodium bicarbonate will this question still be solvable?

Thank you for the explaination.

 

[Rabolisk]

You don't know the pKa value of sodium bicarbonate in this question. If you meant to say pKa of carbonic acid, the question is not solvable.

However, I think this is a bad question from EK. This question is not a simple acid/base question because bicarbonate is indeed amphoteric. You would need a much more rigorous and systematic calculation of pH, which can then be simplified...

http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt

 

[BerkReviewTeach]

This question is not a simple acid/base question because bicarbonate is indeed amphoteric. You would need a much more rigorous and systematic calculation of pH, which can then be simplified...

Actually, it is amazingly simple and you don't need anything rigorous at all. This is the MCAT, so it's always worth the time to find the quickest way.

Just like pI values are actually equivalence points on a titration curve where the zwitterion is in highest concentration, the pH at equivalence is where the species of interest is in highest ceonctration (in this case HCO3-).

To get the pH of NaHCO3 in water (at any reasonable concentration), you simply need to average pKa1 and pKa2. Based on what the OP gave in terms of info, they cannot solve it, but if pKa2 (= 10.8) was given, then it's a matter of averaging 6.4 and 10.8 to get 8.6.

 

[SKation]

Hi,

I am still confused as to based off what we were given in the question, how do we know that we need Kb to solve the problem?

Thanks!