Chem 334 - Fall 1999
Organic Chemistry I
Dr. Carl C. Wamser

Chapter 3 Homework Answers

Brown & Foote, pages 115 - 118 :
Problems 3.5 - 29 (all)

1. Given that Kw = [H+] [OH-] = 10^-14 , explain why the pKa of H2O is cited as 15.7 and the pKa of H3O+ as -1.7 .

In order to compare Ka in the same way as for other acids, the equilibrium constant should include concentration of water, [H2O]. So Ka = Kw / [H2O]. Since [H2O] in pure water is 55.5 M, Ka = 1.8 x 10E-16, or pKa = 15.7 .

The Ka for H3O+ would be defined as Ka = [H+] [H2O] / [H3O+] . But [H+] is the same as [H3O+] , so Ka = [H2O] = 55.5 , or pKa = -1.7 .

2. The pKa of CH3OH2+ (protonated methanol) is -2.2. Predict the preferred direction of an equilibrium acid-base reaction between H3O+ and CH3OH.

CH3OH + H3O+ <===> CH3OH2+ + H2O

The stronger acid is CH3OH2+ (more negative pKa), so the equilibrium will lie farther to the left.