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 Message 2 of 2 in Discussion 
From: MSN Nickname·Steve·  in response to Message 1Sent: 2/26/2005 10:37 PM
In the equilibrium ratio [OAc-] / [HOAc], the units of the numerator and denominator are moles / L.  Since everything is in the same container, the volume L is the same in the numerator and denominator, so the value of the ratio in molarity units is the same value as moles of OAc- / moles of HOAc.
 
To make a solution containing a specific number of moles of OAc- and HOAc from inital solutions of OAc- and HOAc at molarities M1 and M2, remember
 
moles of OAc-      =    V1  X  M1  and
moles of HOAc     =    V2  X  M2
 
where V1 is the volume of OAc- solution and V2 is the volume of HOAc solution.
 
Therefore,
 
 moles OAc-      =     V1 M1
 moles HOAc            V2 M2
 
But since M1  =  M2  =  0.500 M, M1 and M2 cancel to give
 
 moles OAc-     =    V1
 moles HOAc          V2
 
If VT  =  the total volume V1  +  V2,
 
 moles OAc- / V    =       [OAc-]      =    V1
 moles HOAc / VT            [HOAc]           V2
 
because the VT terms can cancel.
 
 
Steve