Posted by Laura on Thursday, June 14, 2012 at 7:00pm.
Have you had a class on the Henderson-Hasselbalch 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 3.94E-10 which makes pKb = 9.40 and that subtracted from 14 makes pKa = 4.60.
The ionization equation for aniline is
C6H5NH2 + HOH ==> C6H5NH3^+ + OH^-
No, my professor didn't teach student on Henderson-Hasselbalch equation. My professor only show us the Henderson-Hasselblach 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.
How did you get the 3.94E-10?
As I stated in my response I looked up Kb for aniline in one of my texts and obtained 3.94E-10. 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 = 14-9.40 = 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.....0.30-x.............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) = (0.30-x) and it look this way.
3.94E-10 = (x+0.15)(x)/(0.30-x)
Assume x+0.15 = 0.15 and assume 0.30-x = 0.30. You can do that since x is small. That gives you
3.94E-10 = (0.15)(x)/(0.30) and solve for x which is OH^-.
I get 7.88E-10 for OH and pOH = 9.10 and pH = 14-9.10 = 4.90.
You can see why we use the Henderson-Hasselbalch equation; it is so much easier to use for buffer problems.
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.