Where x represents the conjugate base, which of the acids listed is expected to have the smallest h-x bond dissociation energy? a. acid a, pka = 4.10 b. acid b, pka = 4.75 c. acid c, pkb = 9.63 d. acid d, pkb = 10.90?

Remember that pKa comes from the equilibrium equation HX(aq) <==> H+(aq) + X-(aq). Using the law of mass action, we get the equation:
Ka = Products/Reactants = [H+][X-]/[HX]
The larger the Ka value, the more acidic HX is, because more of it has formed HX. However, in terms of pKa, we take the -log(Ka).
So the LOWER pKa will be the more acidic molecule. So we know the lowest pKa is the most acidic proton.
Next, we have to know the relationship between pKb and pKa. Since Kw for water is 10^-14, and Kw = Ka * Kb, then if we take the -log of both sides, we get 14 = pKa + pKb (technically at STP, but deviations are not that important for right now). Now we solve for each molecule's pKa, and the lowest pKa is the most acidic, and therefore has the weakest H-X bond dissociation energy.
Acid A's pKa is 4.10
Acid B's pKa is 4.75
Acid C's pKb is 9.63, so its pKa is 14 - 9.63 = 4.37
Acid D's pKb is 10.90, so its pKa is 14 - 10.90 = 3.10
Since Acid D's pKa is the lowest, it is most willing to lose its proton from the H-X bond, and therefore Acid D is expected to have the smallest H-X bond dissociation energy.

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