Asked by Jessica
Calculate the pH of a 0.2M solution of Triethylamine, (C2H5)3N, with a Kb of 4.0x10^-4
Answers
Answered by
DrBob222
..(C2H5)3N + HOH ==> (C2H5)3NH^+ + OH^-
I...0.2M................0...........0
C....-x.................x...........x
E...0.2-x...............x...........x
Substitute the E line into the Kb expression and solve for x = (OH^-), then convert that to H^+ by
(H^+)(OH^-) = Kw = 1E-14
Then pH = -log(H^+)
I...0.2M................0...........0
C....-x.................x...........x
E...0.2-x...............x...........x
Substitute the E line into the Kb expression and solve for x = (OH^-), then convert that to H^+ by
(H^+)(OH^-) = Kw = 1E-14
Then pH = -log(H^+)
Answered by
Anonymous
[OH^-]=8.94x10^-3
pOH=2.05
pH=11.95
[H^+]=1.12x10^-12
pOH=2.05
pH=11.95
[H^+]=1.12x10^-12
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