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Topic: Help writing buffer reactions?  (Read 12730 times)

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cabaal

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Help writing buffer reactions?
« on: March 26, 2010, 03:22:45 PM »
Calculate the pH of a 1.00 L of the buffer 0.80 M CH3NH2/1.00 M CH3NH3Cl before and after the addition of (a) 0.070 mol NaOH and (b) 0.11 mol HCl. KB = 4.4x10-11.

CH3NH2 + H2 CH3NH3+ + OH-
0.80 mol                1.00 mol     0
-x                        +x            +x
0.80-x                  1.00+x       x

(x+x2)/(0.80-x) = 4.4x10-11
x = 3.52x10-4 = [OH]-
pH = 14+log(3.52x10-4) = 10.55

CH3NH3Cl + NaOH NaCl + CH3NH2 + H2O
Adding NaOH would consume CH3NH3Cl, and therefore shift the reaction to the left.

CH3NH2 + HCl  CH3NH3+ + Cl-
Adding HCl would consume CH3NH2 to produce CH3NH3+, and therefore shift the reaction to the left.

My questions
Are my reactions all correctly written and are my reaction shifts correct?
How do I use the Henderson Hasselbalch equation? I don't know the volumes of the NaOH or HCl added, so how can I calculate the concentration of the conjugate base or the acid?

Thank you.

Borek

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Re: Help writing buffer reactions?
« Reply #1 on: March 26, 2010, 03:48:00 PM »
Why do you use ICE table to calculate pH of the buffer? Use HH equation.

You may assume volume of the added acid/base was so small it has not changed volume of the solution in a significant way.

Besides, strange as it sounds, concentrations of the acid and conjugate base in the solution don't matter. What is important is ratio of number of moles, as in ratio of concentrations volumes cancel out.
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cabaal

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Re: Help writing buffer reactions?
« Reply #2 on: March 26, 2010, 06:38:31 PM »
How would the HH equation work in place of the ICE table? I get a completely different value than pH = 10.55.

edit: Oh, I understand now. I forgot I was looking at a basic solution, so I was thinking it gave me the pH instead of the pOH.

-log(4.4x10-4) + log(1 M CH3NH3-/0.8 M CH3NH2) = pOH of 3.45 = pH of 10.55

cabaal

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Re: Help writing buffer reactions?
« Reply #3 on: March 26, 2010, 07:19:54 PM »
CH3NH3Cl + NaOH NaCl + CH3NH2 + H2O
Adding NaOH would consume CH3NH3Cl, and therefore shift the reaction to the left.
[CH3NH2] = 0.80 mol + 0.070 mol = 0.87 mol
[CH3NH3Cl] = 1.00 mol - 0.070 mol = 0.93 mol
pH = 14-(-log(4.4x10-4) + log(0.87 mol / 0.93 mol)) = 10.67

CH3NH2 + HCl   CH3NH3+ + Cl-
Adding HCl would consume CH3NH2 to produce CH3NH3+, and therefore shift the reaction to the left.
[CH3NH2] = 0.80 mol - 0.11 mol = 0.69 mol
[CH3NH3+] = 1.00 mol + 0.11 mol = 1.11 mol
pH = -log(2.7x10-11) + log(0.69 mol / 1.11 mol) = 10.36

Does this look correct?

cabaal

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Re: Help writing buffer reactions?
« Reply #4 on: March 26, 2010, 08:01:14 PM »
Wait, A. be pH = 14-(-log(4.4x10-4) + log(0.93 mol / 0.87 mol)) = 10.61?

Borek

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Re: Help writing buffer reactions?
« Reply #5 on: March 26, 2010, 08:01:58 PM »
KB = 4.4x10-11.

Kb or Ka?

As far as I can tell none of your results is correct. Please recheck if you are using correct version of the HH equation and correct value of pKa (or pKb). You are on the right track, just some math went wrong.

1 am here, more tomorrow.
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cabaal

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Re: Help writing buffer reactions?
« Reply #6 on: March 26, 2010, 08:13:12 PM »
The value in my first post is incorrect, but I can't edit it. KB = 4.4x10-4. KA = 2.7x10-11.

Part A: -log(4.4x10-4)+log(0.93 mol / 0.87 mol) = 3.39 pOH = 10.61 pH
Part B: -log(4.4x10-4)+log(1.11 mol / 0.69 mol) = 3.56 pOH = 10.44 pH

I think this is correct since:
HH for bases pOH = -log(KB)+log(B+/BOH)
HH for acids: pH = -log(KA)+log(A-/HA)

I was getting confused with the charges and the KA vs KB.

Borek

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Re: Help writing buffer reactions?
« Reply #7 on: March 27, 2010, 05:21:34 AM »
Seems like you are finally there.

Most of the time I could not reproduce your calculations because of that wrong dissociation constant, that slowed things down.
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cabaal

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Re: Help writing buffer reactions?
« Reply #8 on: March 27, 2010, 09:12:52 AM »