Ch.16 - Aqueous Equilibria: Acids & Bases
Chapter 16, Problem 122
Nicotine 1C10H14N22 can accept two protons because it has two basic N atoms 1Kb1 = 1.0 * 10-6; Kb2 = 1.3 * 10-112. Calculate the values of Ka for the conjugate acids C10H14N2H+ and C10H14N2H22 + .
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