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Topic: Buffer solution calculation help please  (Read 1244 times)

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Offline zoh899

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Buffer solution calculation help please
« on: October 12, 2020, 09:04:15 AM »
A buffer solution is to be made using 1.00 mol dm–3 ethanoic acid, CH3CO2H, and
1.00 mol dm–3 sodium ethanoate, CH3CO2Na.
Calculate to the nearest 1 cm3 the volumes of each solution that would be required to
make 100 cm3 of a buffer solution with pH 5.50.
Clearly show all steps in your working.

Ka (CH3CO2H) = 1.79 × 10–5 mol dm–3

Ok so I used the equation pH=pka+log([base]/[acid])
I got [base]/[acid]=5.66, but I have no idea how to proceed further, since I don't know the number of moles of the ethanoic acid or of sodium ethanoate, however in the answer key they have simply put it like this
B+A=100, B=100-A (B=base, A=acid)
(100-A)/A=5.66 and they got the volume of acid as 15cm^3 and of base 85cm^3.
Can someone please explain?


Offline Babcock_Hall

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Re: Buffer solution calculation help please
« Reply #1 on: October 12, 2020, 10:02:42 AM »
You know the volume of the buffer and other critical pieces of information.

Offline zoh899

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Re: Buffer solution calculation help please
« Reply #2 on: October 12, 2020, 11:26:55 AM »
You know the volume of the buffer and other critical pieces of information.
but there are no number of moles given and conc formula is n/v, the original volume is not given so how do I calculate the moles? 

Online Borek

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Re: Buffer solution calculation help please
« Reply #3 on: October 12, 2020, 01:25:30 PM »
A an B are volumes, aren't they?
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Offline zoh899

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Re: Buffer solution calculation help please
« Reply #4 on: October 12, 2020, 02:15:29 PM »
A an B are volumes, aren't they?
arent they concentrations of the respective base and acid???

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Re: Buffer solution calculation help please
« Reply #5 on: October 13, 2020, 03:24:49 AM »
They can be, but in this case the concentrations in mixture are directly proportional to the volumes used, so these are equivalent ways of thinking. Try to derive that remembering that if A and B are volumes A+B=100.
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