TBR OChem S8 Q41

[ajumobim]

---

Members do not see this ad.

The explanation says that only base is added to the water and not acid. Can't HCO3- be considered an acid and protonate 3-methylaniline, causing it to go into the aqueous layer (Tube 4)?

 

[AlteredScale]

The explanation says that only base is added to the water and not acid. Can't HCO3- be considered an acid and protonate 3-methylaniline, causing it to go into the aqueous layer (Tube 4)?

HCO3- is weakly acidic (low hydrogen dissociation constant) and will only protonate a molecule that is strongly basic. 3-methylaniline is weakly basic (due to the amino group) and so protonation by HCO3- wouldn't occur.

Comes back to the idea that behavior of a polyprotic molecule that has already "lost" a hydrogen will be at best weakly basic because it is exponentially harder to "pull away" the second or third hydrogen.

 

[NextStepTutor_3]

Totally agree, @AlteredScale ! Also, just wanted to add that HCO3- is significantly more basic than a molecule like HSO4-, even though both are polyprotic acids that have lost one hydrogen already.

I like to think of it like this. HCO3- is the conjugate base of a weak acid, carbonic acid. If we wrote out its dissociation equilibrium, we'd see H2CO3 ---> H+ + HCO3- ----> H+ + CO3 (2-). For HCO3- to act as a base, it would have to lose a second proton. However, H2CO3 itself is already a very weak acid! If we had HCO3- present in an aqueous solution, it would be much more likely to regain its lost proton (re-forming carbonic acid) than to lose another.

HSO4-, on the other hand, is the conjugate base of a strong acid. While it's still unlikely to lose its one remaining proton, sulfuric acid at least completely dissociates in that aqueous solution. So HSO4-, not being prone to regaining that lost proton, has a notably higher Ka than HCO3-.

 

[AlteredScale]

Totally agree, @AlteredScale ! Also, just wanted to add that HCO3- is significantly more basic than a molecule like HSO4-, even though both are polyprotic acids that have lost one hydrogen already.

I like to think of it like this. HCO3- is the conjugate base of a weak acid, carbonic acid. If we wrote out its dissociation equilibrium, we'd see H2CO3 ---> H+ + HCO3- ----> H+ + CO3 (2-). For HCO3- to act as a base, it would have to lose a second proton. However, H2CO3 itself is already a very weak acid! If we had HCO3- present in an aqueous solution, it would be much more likely to regain its lost proton (re-forming carbonic acid) than to lose another.

HSO4-, on the other hand, is the conjugate base of a strong acid. While it's still unlikely to lose its one remaining proton, sulfuric acid at least completely dissociates in that aqueous solution. So HSO4-, not being prone to regaining that lost proton, has a notably higher Ka than HCO3-.

I definitely learned something new (or perhaps a refresher on memories of come) today, thanks!

 

[ajumobim]

Thank you to the both of you! I have a better understanding of this now. @NextStepTutor_3 Does NextStep offer anything like unlimited tutoring hours just to ask instructors questions?