Consider a very weak base in solution. The pKb of this base would likely be:
A. equal to the pOH of the solution.
B. higher than the pOH of the solution.
C. lower than the pOH of the solution.
D. near 7 at 25ºC.
Explanation: The correct answer is B. Normally, to calculate pH or pOH values, we require some numerical information about the concentration of the solution. In this question, this information is not given, so we will have to go off of the concepts alone. The MCAT often requires similar conceptual thinking, and this question is still very doable – so do not panic when you notice that numbers are not provided! Let us write a generic reaction to represent our base in water:
B + H2O <—-> HB+ + OH–
Kb = [OH–][HB+]/[B]
Using an alternate form of the Henderson-Hasselbalch equation, expressed in terms of pOH and pKb:
pOH = pKb + log [HB+]/[B]
where [HB+] and [B] represent the concentrations of the conjugate acid and original base molecule at equilibrium, respectively. A very weak base will have a very small Kb. This means that equilibrium favors the reactants (including [B]) over the products (including [HB+]). Thus, we can infer that the value of [HB+]/[B] << 1, making the value of log [HB+]/[[B] << 0. This would result in a pOH value that is less than the pKb. Thus, pKb > pOH.