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Scholar
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Date Posted: 7/23/2008 3:27:01 PM  Status: Closed
Titration Question
Course Textbook Chapter Problem
N/A N/A N/A N/A
Question Details:
A 40.0 mL sample of 0.100M HNO2 is titrated with 0.200 M KOH.  Determine the pH at the equivalence point for the titration.
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Oracle
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Date Posted: 7/23/2008 4:19:34 PM  Status: Live
Asker's Rating: Lifesaver   
Response:
I'll direct you along the path.............
 
Both are monoprotic and monobasic so the reaction is one to one.
 
You can calculate the number of moles of nitrous acid (MV = 0.00400 mol).  From this you will say the same number of moles of base will be needed to reach the equivalence point, meaning you will need a certain volume of base (mol/M=Vbase = 0.00400 mol/0.200M = 0.0200 L of base added).
 
OK, consider the reaction:
HNO2 + KOH  ---> KNO2 + HOH
A solution of KNO2 is produced!  This is a solution of a weak base, NO2-, i.e.,  the conjugate base of HNO2 .
 
How many moles of KNO2 ? 0.00400 mol
What is the final volume? 40 + 20 = 60 mL = 0.060 L
What is the concentration of weak base, NO2- ? 0.0040 mol/0.060L = 0.0667 M
 
OK, now the problem is simplified to finding the pH of a solution of 0.0667 M NO2-.
 
You can look up in your table the Ka of HNO2, so Kb for NO2-, will be Kw/Ka.
 
Finally the same old same old tricks..........
 
Kb = x2/(0.0667-x)......when you get x, you have [OH-] .....which leads easily to pOH....and finally to pH.

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Date Posted: 7/23/2008 10:37:29 PM  Status: Live
Asker's Rating: N/A-Posted by Person Asking Question   
Response:
Ka is 4.6*10-4, so Kb is 2.17 * 10-11
I am getting a pH of 8.08.....my textbook says 7.99....not sure how



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