Posted by Laura on Thursday, June 14, 2012 at 7:00pm.
This question I really don't get it. I tried tried tried to figure out how to start solve the problem but still couldn't figure out. So can you help me with one question PLEASE? If so, can you please explain to me how you get the answer? Also can you show me step by step of solving the problem? I need to know how to do this problem, I will have test on Monday so I need to understand it. Thank!
Here the question:
what is the pH of a solution that is 0.30 M in aniniline, C6H5NH2 and 0.15M in anilinium chloride, C6H5NH3Cl? Show the equation for ionziation

chemistry  DrBob222, Thursday, June 14, 2012 at 8:12pm
Have you had a class on the HendersonHasselbalch equation?
pH = pKa + log[(base)/(acid)]
The base is aniline (0.30M) and the acid is the chloride(0.15M). The pKa is the pKa of aniline. In my book the Kb for aniline is 3.94E10 which makes pKb = 9.40 and that subtracted from 14 makes pKa = 4.60.
The ionization equation for aniline is
C6H5NH2 + HOH ==> C6H5NH3^+ + OH^ 
chemistry  Laura, Thursday, June 14, 2012 at 10:28pm
No, my professor didn't teach student on HendersonHasselbalch equation. My professor only show us the HendersonHasselblach forumla but not show how to use it. Also she didn't give us what is pKa for this problem. Anyway, thank you so much for your help.

chemistry  Laura, Thursday, June 14, 2012 at 10:33pm
How did you get the 3.94E10?

chemistry  DrBob222, Thursday, June 14, 2012 at 11:28pm
As I stated in my response I looked up Kb for aniline in one of my texts and obtained 3.94E10. You should look it up in your text; your text may have a slightly different value. However, log Kb = pKb; therefore, pKb for aniline from my value is 9.40 and since pKa + pKb = 14, then pKa = 149.40 = 4.60.
Then substitute into the HH equation this way,
pH = 4.60 + log(0.30)/(0.15) and solve for pH. I get 4.90.
There is another way to do it and not use the HH equation.
....C6H5NH2 + HOH ==> C6H5NH3^+ + OH^
I.....0.30.............0...........0
C......x...............x..........x
E.....0.30x.............x.........x
Then Kb = (C6H5NH3^+)(OH^)/(6H5NH2)
....C6H5NH3Cl^ ==> C6H5NH3^+ + Cl^
I....0.15..........0..........0
C...0.15..........0.15.......0.15
E......0..........0.15.........0.15
Now substitute into the Kb equation this way:
(C6H5NH3^+) = x + 0.15 (x from the ionization of C6H5NH2 and 0.15 from the anilinium chloride.
(OH^) = x
(C6H5NH2) = (0.30x) and it look this way.
3.94E10 = (x+0.15)(x)/(0.30x)
Assume x+0.15 = 0.15 and assume 0.30x = 0.30. You can do that since x is small. That gives you
3.94E10 = (0.15)(x)/(0.30) and solve for x which is OH^.
I get 7.88E10 for OH and pOH = 9.10 and pH = 149.10 = 4.90.
You can see why we use the HendersonHasselbalch equation; it is so much easier to use for buffer problems. 
chemistry  Laura, Saturday, June 16, 2012 at 10:17pm
Yea, I realized that HH is more easy and faster. I'm going use HH better instead other way. Anyway thank you so much for explain.