pH = 11.98
11.98 = -log(H^+)
(H^+) = 1.05E-12
...(CH3)2NH + HOH ==>(CH3)2NH2^+ + OH^-
I....0.164.............0...........0
C.....-x...............x............x
E....0.164-x...........x............x
Since (H^+)(OH^-) = Kw = 1E-14, then
(OH^-)= 1E-14/1.05E-12 = 9.6E-3
Therefore, x = about 9.6E-3 = (OH^-) = [(CH3)2NH].
Substitute those values into Kb expression and solve for Kb, then
pKb = -log(Kb)
Kb = [(CH3)2NH2^+][OH^-]/[(CH3)2NH]
The pH of a 0.164 M aqueous solution of dimethylamine is 11.98. Write the ionization equation,
calculate the values of Kb and pKb and comment on strength of the base....
1 answer
1 answer