[antoinetta]

How do you calculate the pH of a 3.60×10-1 M aqueous solution of triethylamine (C2H5)_3N, given that Kb = 4.0×10-4?

I don't seem to be getting the right answer.  Here's what I did:

I set up an ICE table and calculate the [OH-] to be 1.18*10^-2.

Then I found the pOH using -log(OH-) = 1.928

So, the pH is 14-1.928 = 12.072

But, the answer in the text book said that this is wrong. ... can somebody please help?

[chiralic]

Hi Antoinetta:

Read this...http://butane.chem.uiuc.edu/cyerkes/chem104A_S07/Lecture_Notes_104/lect26c.html

Look at on section Reactions

[AWK]

seems to be OK with these data.
Note, in example given by chiralic a quite different Kb is used for triethylamine.

[Borek]

Theory:

http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base

Practical example:

http://www.chembuddy.com/?left=pH-calculation-questions&right=pH-weak-base-hydrolysis-q1

Cheat sheet in case nothing else works:

http://www.chembuddy.com/?left=BATE&right=pH-cheat-sheet

And finally - as AWK wrote - 12.07 is a correct result.

[antoinetta]

Thanks everyone for their input.

[english]

I got 12.08.  You have rounding error. 

Check to see if that is the reason.


The assumptions you made are jusitifed in ignoring the small change in x. 

That is,

[TEtA]       > 400
    K_b