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All Message Boards : Phosphate Buffer
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 Message 1 of 3 in Discussion 
From: MSN NicknameGuy_SoCa  (Original Message)Sent: 10/21/2008 11:35 PM
How many moles of HCl are required to change the pH of 500 ml of 0.100 M Phosphate buffer, pH 12.6, to a final pH of 2.00? The pKa's of phosphate are 2.12, 7.21, and 12.32?

I'm trying to solve this problem in 3 different steps
1) use the H-H to go from PO4 (-3) to the -2 species
2) number of moles of -2 species = number of moles of acid needed to go to -1 species
3) Use the H-H to go from -1 to the acidic form..

Thanx!


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 Message 2 of 3 in Discussion 
From: MSN Nickname·Steve·Sent: 10/22/2008 3:04 AM
Your approach looks OK.  In the first solution at pH 12.6, there is essentially only HPO42�?/SUP> and PO43�?/SUP> present, so I used HH to calculate the ratio PO43�?/SUP> / HPO42�?/SUP>.  Then, since HPO42�?/SUP> + PO43�?/SUP> = 0.100 M, I could calculate each concentration.

Next, I did the same thing with the second solution at pH 2.00.  This time there is essentially only H3PO4 and H2PO4�?/SUP> present.  I used HH to calculate the ratio H2PO4�?/SUP> / H3PO4, and since H2PO4�?/SUP> + H3PO4 = 0.100 M, I could calculate each concentration.

Multiplying each of the four concentrations by the total volume, 0.500 L, gives the moles of each species, which in each solution must add up to 0.0500 mol.

Then I calculated how many moles of H+ to add to convert all of the phosphate in the first solution to H3PO4, which was 2 X moles of HPO42�?/SUP>  +  3 X moles of PO43�?/SUP>.  Then I calculated how many moles of H+ to subtract from the second solution, now containing 0.0500 mol of H3PO4 only, to give the moles of H2PO4�?/SUP> calculated above.  The difference in these amounts of H+ gives the moles of H+ needed to convert the phosphate species in the first solution to those in the second solution.

I got 0.104359 mol of H+ overall to add to the first solution to form the second solution.  Let me know if you get the same answer.  My method was a little different from yours, as I did not use HH at the end, but I think the net result will still be the same.
 

Steve

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 Message 3 of 3 in Discussion 
From: MSN Nickname·Steve·Sent: 10/22/2008 2:01 PM
>>  I got 0.104359 mol of H+  <<
 
I forgot to add the "free" H+ ion that must be present to have a pH of 2.00.  The solution must be 0.010 M in free (dissociated) H+ ion to give a pH of 2.00.  The moles of free H+ ion is 0.500 L X 0.010 mol/L  =  0.0050 mol.  Adding this to 0.104359 mol gives 0.109359 mol, or about 0.11 mol H+ to add.  Maybe!  I need to re-examine this approach in case I am oversimplifying.