Professor Carl C. Wamser
Homework 4 Answers
Carey, pages 182 - 186 :
Problems 4.19 - 4.46
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.
3. Write a complete mechanism for the SN1 reaction of t-butyl alcohol with HBr.
4. Write a complete mechanism for the SN2 reaction of HI with methanol.