How many liters of stomach acid (pH = 4.000) will be neutralized by a 1.50 g tablet of "Bicar", a commercial tablet which is 50% flavored and 50% NaHCO3?

Question

Note that:

NaHCO3(s) --> Na+(aq) + HCO3-(aq)

and:

H+(aq) + HCO3-(aq) --> H20(l) + CO2(g)

So far I've determined that for the stomach acid [H+] = 1x10^-4 M, and [OH-] = 2.5x10^-11 M.

For the Bicar tablet I found there to be about 0.89 moles, assuming that 50% of the mass is NaHCO3.

Compared to other related problems I've done, I'm not sure on how to proceed with this question. It's probably relatively easy and I could be making it more complicated than it really is.

I do know, however, that the answer is supposed to be about 89 L of stomach acid neutralized, just for clarification.

Thanks for the help

Answer 1

Yes, I think you're making it tougher than it is.
You have an error in mols of NaHCO3.
grams NaHCO3 = 1.5/2 = 0.75
mols NaHCO3 = 0.75/84 = 0.0089
Then M = mols/L. You know M = 1E-4 and you know mols = 0.0089 so
L = 0.0089/1E-4 = 89 L. If you carry all of the digits that is 89.28 L but you're limited to 2 significant figures by the 1.5 g so that is rounded to 89.L.