The odor of a dead fish is attributed to trimethylamine, (CH3)3N, one of a serie

Question

The odor of a dead fish is attributed to trimethylamine, (CH3)3N, one of a series of compounds of nitrogen, carbon, and hydrogen with a general formula of (CH3)nNH3-n where n = 0-3 (the n = 0 compound is ammonia).

a) The Kb of trimethylamine [(CH3)3N)] is 6.5E-5 at 25 C. Calculate the pH of 3.00nE-4 M solution of dimethylamine.

b) The Kb of mythylamine [(CH3)NH2] is 4.4E-4 at 25 C. Calculate the pH of 2.88E-3 M solution of mythylamine.

c) the Kb of dimethylamine [(CH3)2NH] is 5.9E-4 at 25 C. What concentration of dimethylamine is needed for the solution to have the same pH as the solution in part B?

Explanation / Answer

We must calculate the Ka for the above reaction. We will use the Kb of dimethylamine to do this:

1.00 x 10¯14 = (x) (6.5 x 10¯5)

x = 1.00 x 10¯14 / 6.5 x 10¯5 = 1.54 x 10¯10

Please note that I left some guard digits on the Ka value. I will round off the final answer to the appropriate number of significant figures.

We now need to calculate the [H+] using the Ka expression:

1.54 x 10¯10 = [(x) (x)] / (3.0 x 10-4 - x)

neglect the minus x

x = 2.15 x 10¯7 M

Since this is the [H+], our last step will be to calculate the pH:

pH = - log 2.15 x 10¯7 = 6.67

b.)

1.00 x 10¯14 = (x) (4.4 x 10¯4)

x = 1.00 x 10¯14 / 4.4 x 10¯4 = 2.27 x 10¯11

Please note that I left some guard digits on the Ka value. I will round off the final answer to the appropriate number of significant figures.

We now need to calculate the [H+] using the Ka expression:

2.27 x 10¯11= [(x) (x)] / (2.88 x 10-3 - x)

neglect the minus x

x = 2.5568 x 10¯7 M

Since this is the [H+], our last step will be to calculate the pH:

pH = - log 2.5568 x 10¯7 = 6.62

c.

1.00 x 10¯14 = (x) (5.9 x 10¯4)

x = 1.00 x 10¯14 / 5.9 x 10¯4 = 1.69 x 10¯11

Please note that I left some guard digits on the Ka value. I will round off the final answer to the appropriate number of significant figures.

We now need to calculate the [H+] using the Ka expression:

1.69 x 10¯11= 2.5568 x 10¯7 ] /Unknown conc

Unknown conc = 1.512 x 10¯4 M

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