>>> I thought when CO3^2- + H+ < ---- >
HCO3^2-?
How did you get H+ on the right hand side of the expression when
carbonate is behaving as a base? <<<
Using the equation given by the question, for considering
carbonate ion acting as a base, it's RHS ---> LHS, so (from RHS
to LHS) carbonate ion + proton = monohydrogen carbonate
ion.<<<<<<<<<
This is the part I dont get it in relation to the Kb expression
where you said,
Kb = [H+][conjugate acid] / [conjugate base]
Isn't Kb=[OH-][conjugate acid] / [conjugate base] ??
Ah yes it is, my bad.
Also, I reviewed my steps and realized that I also
made several other errors - one of them being that we
neglected the dilution effect; another being we neglected to
consider the phenoxide as a competing base.
However, I also spotted an error in the question, which makes it
impossible to do the question correctly in any
case (regardless of other errors committed). Common sense
dictates that the Ka value for the monohydrogen carbonate ion
should be smaller (ie. weaker acid) than the Ka value for
dihydrogen carbonate aka carbonic acid (obviously a stronger
acid).
At 25 deg C, the correct Ka values should be 4.2 x 10^-7 for
carbonic acid, and 4.8 x 10^-11 for the monohydrogen carbonate
acid. The correct Kb values should hence be 2.4 x 10^-8 for
the monohydrogen carbonate base, and 2.1 x 10^-4 for the carbonate
ion.
At 25 deg C, the Ka value for phenol should be 1.3 x 10^-10, and
the Kb value for phenoxide ion is 7.7 x 10^-5.
Notice that the strongest acid is carbonic acid, then phenol,
then monohydrogen carbonate ion.
Notice that the strongest base is carbonate ion, then
phenoxide ion, then monohydrogen carbonate ion.
Based on the dilution effect,
the new molarity of carbonate ions would be 0.5 mol/dm3.
the new molarity of phenol would be 0.3565 mol/dm3.
Imagine hypothetically that, disregarding Ka value (for the sake
of explaining an important concept here), ALL of the
acidic protons from phenol (the only significant source of
acidic protons) were dissociated into solution; we would have
0.3565 mol/dm3 of protons that would be snatched up by two
competing bases - 0.5 mol of carbonate ions (from strong
electrolyte aqueous sodium carbonate) and 0.3565 mol of phenoxide
ions.
Since carbonate ions are stronger bases than phenoxide
ions, all of the acidic protons would be snatched by the
carbonate ions to form monohydrogen carbonate ions (which being a
weaker acid than phenol, would not transfer protons over to
the phenoxide ions to form phenol). In fact, there would still be
0.5 - 0.3565 = 0.1435 mol of carbonate ions present, that are
stronger bases than the 0.3565 mol of phenoxide ions present.
That being the case (the fact that there is still a significant
percentage of carbonate ions (the strongest base of the 3 species)
that are still unprotonated), you would certainly not expect any of
the monohydrogen carbonate ions to have the opportunity
to be protonated to form dihydrogen carbonate aka carbonic
acid, which is our only hope of obtaining / releasing from
solution carbon dioxide guess.