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^-
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
5 answers
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.
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.
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