From: rmcory@uwovax.uwo.ca
Subject: Re: Acid-base equilibrium (triprotic)
Organization: University of Western Ont, London
Date: Thu, 17 Dec 1992 01:30:26 GMT

In article <1992Dec16.195658.27637@pellns.alleg.edu>, dblackbu@alleg.edu (Dave Blackburn) writes:
> In article <c60a-gx.724486364@danube.Berkeley.EDU> c60a-gx@danube.Berkeley.EDU  
> (Jeffrey Twu) writes:
>>rmcory@uwovax.uwo.ca writes:
>>>In article <c60a-gx.724438370@danube.Berkeley.EDU>,  
> c60a-gx@danube.Berkeley.EDU (Jeffrey Twu) writes:
>>>> Alright, I've been pondering over this question for the last few days,
>>>> and have deduced it impossible to solve (with respect to the constraints,
>>>> not the difficulty).  I would like your comments on it.
>>>> 
>>>> You are given four stock solutions, each one molar (1 mol/L) in their
>>>> concentrations:  H3PO4, NaH2PO4, Na2HPO4, Na3PO4 (charges omitted).  From
>>>> only these four solutions (no H2O lying around), you will make a buffer
>>
>>>Sorry, I'm afraid you have to have some H2O! Someone is pulling your leg.
>>
>>>> solution with the following restraints:
>>>> 
>>>> 1) pH = 6.8 (reasonably buffered around there)
>>>> 2) total volume of new solution = 1 liter
>>>> 3) Free phosphate concentration = [PO4] = 0.10 M
>>
>>>What is undoubtedly required here is ***TOTAL*** "phosphate" conc = 0.10 M, 
>>>which in this case would be [H2PO4] + [HPO4] since at pH 6.8 the concs of
>>>H3PO4 and PO4 would be negligible.
>>
>>>Otherwise, the problem is impossible, since [PO4] cannot be 0.10 M at pH  
> 6.8.
>>
>>>> 
>>>> How much of each solution will you use?
>>>> 
>>>> You are given the following data:
>>>> _Acid_	       _pKa_	_Ka_
>>>> H3PO4		2.148	7.11 * 10^-3
>>>> H2PO4		7.199	6.32 * 10^-8
>>>> HPO4		       12.15	7.1  * 10^-13
>>>> 
>>>> Now, if this problem is actually impossible as written, would it be
>>>> solvable if you removed one or more of the constraints?  (This question
>>>> isn't part of the actual problem.)
>>
>>Ok, let's say you do allow total phosphate concentration = [H3PO4] +
>>[H2PO4] + [HPO4] = 0.10 M ([PO4] negligible).  But, will we still need
>>water to dilute?  (Remember, the total volume must be one liter.)
>>
>>Jeff
>>+-Just remember:------------------------------------------------------------+
>>|	Take it easy.							    |
>>|		...and don't forget that human touch.	   	            |
>>+<c60a-gx@danube.berkeley.edu>_________________________________<Jeffrey Twu>+
> 
> Well, if each of your stock solutions is 1.0 M in X3PO4 (X= H or Na) then  
> without dilution the total phosphate will be 1.0 M no matter which of the  
> solutions you mix together. (Phosphate with the appropriate number of protons,  
> or as the biochemists out there say, P-sub-i)
> --
> 
> * Dave Blackburn  dblackbu@alleg.edu  flames to self@home.dev.null *          
> * Chemistry, Macs, Hypercard, cats,  <>< religion, & the Dead      *
> * (Anyone else who held these opinions would be as crazy as I am.) *
> *   "Sometimes the songs that we hear are just songs of our own"   *
-- 

Like I said: "You have to have some H2O!"

--------------------------------------------------------------
Prof. Robert M. Cory, Chemistry Dept, U. Western Ontario, London N6A 5B7 CANADA
Office (519)679-2111x6339; Home (519)472-8866; FAX (519)661-3022; cory@uwo.ca