A.) In the laboratory, a general chemistry student measured the pH of a 0.497 M

Question

A.) In the laboratory, a general chemistry student measured the pH of a 0.497 M aqueous solution of trimethylamine, (CH3)3N to be 11.730.

Use the information she obtained to determine the Kb for this base.

Kb(experiment) =

B.) In the laboratory, a general chemistry student measured the pH of a 0.497 M aqueous solution of caffeine, C8H10N4O2 to be 12.137.

Use the information she obtained to determine the Kb for this base.

Kb(experiment) =

C.) In the laboratory, a general chemistry student measured the pH of a 0.497 M aqueous solution of ammonia to be11.492.

Use the information she obtained to determine the Kb for this base.

Kb(experiment) =

Explanation / Answer

Trimethylamine, caffeine and ammonia are simple bases.

Acid-base chemical reaction for the three bases:
B + H2O <--> HB+ + OH-
B, base
H2O, water
HB+, conjugated acid
OH-, hydroxyl
Equation to calculate Kb:
Kb = [HB+][OH-]/[B]
If the solutions are prepare only by adding the base B to water,
[HB+] = [OH-]
So,
Kb = [OH-]2/[B]

pH + pOH = 14
pOH = 14 - pH
-pOH = pH - 14
pOH =-log([OH-])
-pOH = log([OH-])
[OH-] = 10-pOH
[OH-] = 10pH - 14

Kb = (10pH - 14)2/[B]
Use this equation to solve the 3 questions A, B and C, where [B] is the concentration of the base.

______________________________

A)
An aqueous solution of trimethylamine 0.497 M that has a pH 11.730.
Kb = (10pH - 14)2/[B]
pH = 11.730
[B] = 0.497 M

Kb = (1011.730 - 14)2/0.497 M
Kb = 5.803x10-5

______________________________

B)
An aqueous solution of caffeine 0.497 M that has a pH 12.137.
Kb = (10pH - 14)2/[B]
pH = 12.137
[B] = 0.497 M

Kb = (1012.137 - 14)2/0.497 M
Kb = 3.781x10-4

______________________________

C)
An aqueous solution of ammonia 0.497 M that has a pH 11.492.
Kb = (10pH - 14)2/[B]
pH = 11.492
[B] = 0.497 M

Kb = (1011.492 - 14)2/0.497 M
Kb = 1.939x10-5

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