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Offline susdujcrd

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buffer solution question
« on: April 20, 2012, 05:49:37 AM »
A question from my first analytical chemistry course, and I'm having a hard time fully comprehending the subject:

How many moles of Ammonium Sulfate should be added to a 320mL 0.105M solution of NH3 to get a solution with a pH of 9.35?    Kb(NH3)=1.8e-5 M

Attempt:

I used the buffer equation:

POH=PKb+log([NH4+]/[NH3])

4.65=4.74+log([NH4+]/0.105]

[NH4+]=0.084M

the way I see it, this is the desired concentration of NH4+ in the solution.
The problem is I already have NH4+ forming in the solution even before adding the salt, because of the reaction NH3+H20 <-> NH4+ + OH-
so I just cant say "I have Y, I'll add X, and I'll have X+Y" because its an equilibrium.

Help would be appreciated

Offline AWK

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Re: buffer solution question
« Reply #1 on: April 20, 2012, 06:09:38 AM »
Forget about this reaction:
Quote
NH3+H20 <-> NH4+ + OH-
And remember about stoichiometry of ammonium sulfate.
AWK

Offline susdujcrd

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Re: buffer solution question
« Reply #2 on: April 20, 2012, 06:20:37 AM »
I'm aware that (NH4)2SO4 --> 2NH4+SO4
but I dont think the answer is just "0.084M*320mL gives 0.02 moles of NH4, that should come from 0.01 moles of ammonium sulfate"

Offline Borek

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Re: buffer solution question
« Reply #3 on: April 20, 2012, 07:22:19 AM »
The problem is I already have NH4+ forming in the solution even before adding the salt, because of the reaction NH3+H20 <-> NH4+ + OH-
so I just cant say "I have Y, I'll add X, and I'll have X+Y" because its an equilibrium.

Yes, but no.

You are right there is an equilibrium, but the shift in concentrations due to equilibrium is negligible. In most cases when preparing buffers you can safely assume concentrations of acid and conjugate base to be given by the stoichiometry of the components. This approach fails for too strong and too weak acids/bases, but as a rule of thumb it will work for everything with pKa/pKb between 3.5-11.5 (assuming reasonable concentration, not 10-5M).
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline susdujcrd

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Re: buffer solution question
« Reply #4 on: April 20, 2012, 07:44:09 AM »

If you calculate the concentration of NH4+ in a 320mL 0.105M solution of NH3, you get approx 1.37e-3 M
According to the buffer equation, a solution with a pH of 9.35 will have 8.4e-2 M of NH4+

is one scale of magnitude is enough to rely on stoichiometry alone?

Offline Borek

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Re: buffer solution question
« Reply #5 on: April 20, 2012, 08:26:06 AM »
Apparently - yes. I am not sure it is a good approach. When you add NH4+ you shift the dissociation, so you can't easily compare initial and final concentrations. Rule of thumb I listed earlier is tried and checked on many examples, so I trust it.

By my calculations your buffer should be prepared by adding 1.77 g of ammonium sulfate. This gives 0.08369M solution in NH4+. Exact calculation shows that the equilibrium concentration of NH4+ in such a solution is 0.08371M. Difference is pretty small.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline susdujcrd

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Re: buffer solution question
« Reply #6 on: April 20, 2012, 08:38:58 AM »
I see. Thank you guys very much

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