Asked by geeta
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....
calculate the values of Kb and pKb and comment on strength of the base....
Answers
Answered by
DrBob222
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]
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]