I can't figure this one out
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Answers (2)
pOH = 14 - pH = 14 - 5.40 = 8.6
pOH = -log[OH-] = 8.6
[OH-] = 10^-8.6 = 2.51x10^-9
Inital: 0.10M of (CH3)3NHCL
Change: +2.51x10^-9
Kb = [OH-][(CH3)3N+] / [(CH3)3NHCL]
= (2.51x10^-9)^2 / 0.10
Kb = 6.30x10-5
(CH3)3NHCl is a strong salt >> (CH3)3NH+ + Cl-
pH = 5.40 >> [H+] = 3.98 x 10^-6
(CH3)3NH+ << >> (CH3)3N + H+
at equilibrium
0.10- 3.98 x 10^-6..... 3.98 x10^-6....3.98x10^-6
K = 1.58 x 10^-10 = Kw / Kb
Kb = 6.33 x 10^-5