Why does a saturated aqueous solution of 0.713moldm-3 phenol not cause evolution of carbon dioxide when added to 1.0moldm-3 sodium carbonate solution?
Carry out short calculations to aid your explanations.
Ka: H20 +CO2 <---> H+ HCO3- 4.5x10^-7
Ka: HCO3- < -----> H+ CO32- 2.0x10^-4
pKa of phenol = 10.0
How to start?
Calculate ionic product?
(errors in both solution working as well as in question; both sides' bad... see post below)
Hmm..but I dont understand the expression for Kb of the conjugate bases.
If say for monohydrogencarbonate and carbonate equilibirum,
Kb expression = 1 / Ka expression
= [ acid monohydrogen carbonate] /[H+] [conjugate base carbonate ion]
Is that how you get your expression? but then it is not the same as the one you got.
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?
So we are comparing if the [H+] from phenol and the [H+] required for protonation of the carbonate and hydrogencarbonate ions?
WA...how on earth would we know we are looking for this during exam? haha
(errors in both solution working as well as in question; both sides' bad... see post below)
>>> 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] ??
Originally posted by contemporarydancer:>>> 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.
Uhh... "gas". Not "guess".