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posted by  blue8 on 4/26/2008 7:00:18 PM  |  status: Live  

titrations... Please help.... DUE TOMORROW....

Course Textbook Chapter Problem
General Chemistry Chemistry: The Central Science (10th) by Brown, Lemay, Bursten 17 N/A
Question Details:
 Suppose you have added 100.0 ml of a solution containing 0.5000 moles of acetic acid per liter to 400.0 ml 0.5000 of M NaOH. What is the final pH? The Ka of acetic acid is 1.770 x 10-5??
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posted by reghs on 4/26/2008 9:08:53 PM  |  status: Live
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Response Details:
henderson-hasselbach equation      
 
pH= pka + log () 1
 
moles of nb
=
                                 
 
moles of na
                                       
from equation 1
 
pH= -log ( 1.770x10-5) + log ()
     = 5.3541
 
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