Let $\ce{R-OH}$ be a group with a $pK_a$ of 16. Are the following two statements correct?

  • In principle, the group can protonate another acid with $pK_a > 16$, and so (put differently), any "acid" with $pK_a > 16$ can be used to "deprotonate" the group

  • If I had a table with bases were only $pK_b$ values were given, in order to select these bases from the table which are able to deprotonate the group, I would need to choose the bases with $pK_b < -2$ because the $pK_a$ values of their conjugate acids (i.e. $pK_{a\text{H}}$) will be 16 or higher.

  • $\begingroup$ I don't think this is sufficient for a full answer, so I'll leave a comment. Re: your first statement, in order for the acid-base reaction to be product-favored, it is necessary that the pKa of the conj. acid be greater than the pKa of the acid on the reactant side. As for your second point, the relationship pKa + pKb = 14 is only valid for aqueous solutions under standard conditions. I don't think it's valid for all conj. acid/base pairs in other solvents, but I could be mistaken. $\endgroup$
    – Greg E.
    Dec 3, 2013 at 19:27
  • $\begingroup$ I would hardly call a substance with pKa > 16 an acid, but the first statement seems right in principle. Remember that acid-base reactions are equilibria, so if your pKa is say, 16.2, only a slight deprotonation will occur. $\endgroup$
    – chipbuster
    Dec 4, 2013 at 7:32
  • $\begingroup$ @GregE. That's what I'm saying, isn't it? pKa(R-OH)=16 and so it can protonate another acid AH with pKa(AH) > 16. Your comment about my second statement is right, pKa + pKb = 14 is only valid in H2O (because of the ionization product limit). $\endgroup$
    – TMOTTM
    Dec 4, 2013 at 9:23
  • $\begingroup$ @chipbuster Sure, it's not an acid in the classical meaning, but technically speaking, it acts as an acid w.r.t. the other substance. $\endgroup$
    – TMOTTM
    Dec 4, 2013 at 9:24
  • $\begingroup$ @TMOTTM, yes, I wasn't disagreeing with your first statement, sorry if I was unclear $\endgroup$
    – Greg E.
    Dec 4, 2013 at 11:45

1 Answer 1


Your first statement is correct. The basicity increases with $pK_a$, so a base with $pK_a > 16$ will be able to deprotonate your alcohol.

Your second statement is also in principle correct; however, the value of the sum $pK_a+pK_b$ is solvent-dependent, and 14 is the value for water at standard conditions. $pK_a$ and $pK_b$ values of acids and bases depend on the solvent as well. For example, sodium amide is a strong base in liquid ammonia, but not so in water due to the leveling effect of the solvent. Dissolving sodium amide in water will give a solution of sodium hydroxide, which is not basic enough ($pK_a=13$) to deprotonate the alcohol.

Furthermore, the amount of deprotonated alcohol will depend on the difference between $pK_a$ values of alcohol and base, since the acid-base reaction is an equilibrium. For example, a base with $pK_a=33$ will deprotonate your alcohol to a greater extent than one with a $pK_a$ of 16.5 .


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