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Topic: Creation of a buffer and calculations of pH  (Read 4896 times)

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Offline pHooo

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Creation of a buffer and calculations of pH
« on: February 24, 2013, 06:11:53 PM »
1. Calculate the theoretical pH of a solution that is .200 M in both acetic acid and sodium acetate?

2. Diprotic acid: Oxalic Acid H2C2O4 has a pKa1 = 1.27 and a pKa2 = 4.28.

This data can be used to prepare a buffer at pH 4.7

Calculate the mass of sodium hydrogen oxalate (NaHC2O4 (weak acid in buffer)) and sodium oxalate (Na2C2O4 (conjugate base)) necessary to prepare 100mL the buffer at the pH listed (4.7) if the final concentration of the weak acid HC2O4- is ~0.10M


Offline Arkcon

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Re: Creation of a buffer and calculations of pH
« Reply #1 on: February 24, 2013, 06:51:05 PM »
Can you start with the formula that relates pKa and pH?
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline pHooo

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Re: Creation of a buffer and calculations of pH
« Reply #2 on: February 24, 2013, 08:10:42 PM »
Yeah the formula is pH = pKa + log ([A-]/[HA])

I guess I'm just unsure of how to approach the problem when there are two pKa values

Offline Borek

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Re: Creation of a buffer and calculations of pH
« Reply #3 on: February 25, 2013, 04:23:43 AM »
If the difference between pKas is higher that 3 you can ignore the other pKa value and treat the solution as if it contained only a monoprotic acid.
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Offline pHooo

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Re: Creation of a buffer and calculations of pH
« Reply #4 on: February 25, 2013, 08:28:39 AM »
So I have
4.7 = 4.28 + Log (x/.10)

After doing this i got x = .263

now that I have the conc of both HC2O4- and Na2C2O4 I can find the moles then the mass

my next question would be since I found the conc of HC2O4- would I have to do any extra math to account for the conc of Na in NaHC2O4 or would I still just use .1 M

Offline Borek

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Re: Creation of a buffer and calculations of pH
« Reply #5 on: February 25, 2013, 04:53:18 PM »
my next question would be since I found the conc of HC2O4- would I have to do any extra math to account for the conc of Na in NaHC2O4 or would I still just use .1 M

No idea what you are asking about.
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