Question

In: Chemistry

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

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?

Solutions

Expert Solution

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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