M Prep | MCAT question 8/11/17

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SweetBurger

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I realize this might be a little strange to post in this forum, but it seemed like the best fit.
Heres the q: A base is dissolved in solution such that the concentration of the base (X) is 100 mM and the concentration of the conjugate acid (HX+) is 1 M at equilibrium. If the pKb is 2.89, which of the following is closest to the pH of the solution?
A. 2
B. 3
C. 11
D. 12

I initially ignored A and B as the pKb is approx 3 so the pH should be much higher. Now I know you can use pOH=pKb+log(Acid/Base) ---> pOH=2.89+log(1/.1) = 2.89+1 = 3.89 =pOH. So the pH should be roughly 10 (which isn't even an answer choice). However, the explanation they gave uses what looks like a wrong combination of the Henderson Hasselbach formula and leads them to an answer of 12.
Screenshot heres a screenshot of the question + explanation

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You are not going wrong. You are exactly right. The pH using the Kb equation or Henderson-Hasselbach (pOH version) should be 10.11.
 
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