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Topic: Addition of acid to a phosphate buffer  (Read 6969 times)

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Offline thederivah

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Addition of acid to a phosphate buffer
« on: September 17, 2007, 04:35:26 PM »
a flask contains 100 mL of 0.1 M phosphate buffer that has a pH of 7.4. What volume of 0.1 M Hcl should be added to drop the pH to 6.7.

At pH 7.4 the pKa for the buffer solution is 6.86


i have no idea what to do

Offline Borek

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Re: Addition of acid to a phosphate buffer
« Reply #1 on: September 17, 2007, 04:50:41 PM »
Use Henderson-Hasselbalch equation. Assume adding HCl you are stoichiometrically protonating HPO42-.

TBH I am not sure what

At pH 7.4 the pKa for the buffer solution is 6.86

means. pKa2 of phosphoric acid is 7.2. Most likely 6.86 is a thermodynamically corrected value, but I don't like the wording.
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Offline thederivah

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Re: Addition of acid to a phosphate buffer
« Reply #2 on: September 17, 2007, 05:09:08 PM »
im sorry, it means the pKa2 for the phosphate buffer is 6.86 (the one that has been given to us to use)  I used the Henderson Hasselbalch equation to get the ratios for the pH when its at 7.4 and 6.7 but I dont know what to do from there.  I kind of worked it out, not sure if its correct and got 37 mL of HCl to be added.

Borek if its not too much could you explain how i should go about doing this?

Offline Borek

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Re: Addition of acid to a phosphate buffer
« Reply #3 on: September 17, 2007, 06:03:05 PM »
If you know both ratios you are close - just calculate how much HPO42- must be protonated to H2PO4-. Then it is pretty simple stoichiometry.
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Offline thederivah

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Re: Addition of acid to a phosphate buffer
« Reply #4 on: September 17, 2007, 08:50:43 PM »
I got both ratios and got the concentrations of both H2PO4 (from the 2 different pH's) from there I subtracted the two and multiplied by .1L to get moles.  Then I used that number of moles of H2PO4 which should be equal to the moles of HCl and divided it by .1M HCl to get Liters of HCl which was .037L sound right?

Offline Borek

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Re: Addition of acid to a phosphate buffer
« Reply #5 on: September 18, 2007, 02:36:37 AM »
Method sounds about right, but I haven't checked numbers.
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