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May 2, 2016
Posted by **Laura** on Thursday, June 14, 2012 at 7:00pm.

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 8:12pmHave 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^- - chemistry -
**Laura**, Thursday, June 14, 2012 at 10:28pmNo, 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.

- chemistry -
**Laura**, Thursday, June 14, 2012 at 10:33pmHow did you get the 3.94E-10?

- chemistry -
**DrBob222**, Thursday, June 14, 2012 at 11:28pmAs 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. - chemistry -
**Laura**, Saturday, June 16, 2012 at 10:17pmYea, 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.