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Miggi Groupbie
Miggi
Miggi

Joined: 02 June 2006
Posts: 142

Posted: 21 March 2011 at 9:26pm | IP Logged
Thank you so much for your help.Smile
for the first question ,I got [A-] = 0.0296M is that right?
 
and in the last question it says to find out
Ka for those two points too. So,shouldn't the Ka be different?
I used the formula [H3O+]=Ka[HA]/[A-] and got Ka of  9.212*10^-5 for first one and 7.804*10^-5 for
2nd point. So, does that mean it's wrong or is it right?
 
 


Edited by Miggi - 21 March 2011 at 9:49pm

akhl IF-Rockerz
akhl
akhl

Joined: 30 March 2007
Posts: 5248

Posted: 22 March 2011 at 1:53am | IP Logged
You have to calculate Ka values at both points. But both should be roughly equal, otherwise something wrong in collecting the data or reading the graph.
 
Ignore what I wrote earlier. In this I am giving you complete answer using ICE table.
 
MW of ascorbic acid = 176.13
Moles of HA added = 0.8994/176.13 = 0.005106
 
At the first point:
Before titration:-
Total volume = 50 ml + 4.00 ml = 54 ml
[NaOH] = moles of NaOH/total volume = 0.40 M * 4.00 ml/54 ml
[NaOH] = 0.02963 M
Moles of HA = initial moles of HA - moles of NaOH = 0.005106 - 0.40 * (4.00 * 10^-3)
Moles of HA = 0.003506
[HA] = 0.003506 moles/(54 ml) = 0.003506/(54*10^-3)
[HA] = 0.0649 M
 
HA  --------> H+  +  A-
0.0649          0         0      (I)
-x               +x        +x    (C)
0.0649-x       x         x       (E)
 
x = [H+] = 10^-pH = 10^-3.7 = 2.00 * 10^-4 = 0.0002
[A-] = x + [NaOH] = 0.0002 + 0.02963 = 0.02983 M
[HA] = 0.0649 - x = 0.0649 - 0.0002 = 0.0647 M
Ka = [H+] [A-]/[HA] = 0.0002 * 0.02983/0.0647 = 9.2 * 10^-5
 
At the second point:
Before titration:-
Total volume = 50 ml + 9.00 ml = 59 ml
[NaOH] = moles of NaOH/total volume = 0.40 M * 9.00 ml/59 ml
[NaOH] = 0.06102 M
Moles of HA = initial moles of HA - moles of NaOH = 0.005106 - 0.40 * (9.00 * 10^-3)
Moles of HA = 0.001506
[HA] = 0.001506 moles/(59 ml) = 0.001506/(59*10^-3)
[HA] = 0.0255 M
 
HA  ---------> H+  +  A-
0.0255          0         0      (I)
-x               +x        +x    (C)
0.0255-x         x         x       (E)
 
x = 10^-pH = 10^-4.5 =  3.1623 * 10^-5 M
 
[A-] = x + [NaOH] = 3.1623 * 10^-5 + 0.06102 = 0.0611 M
[HA] = 0.0255 - x = 0.0255 - 3.1623 * 10^-5 = 0.02547 M
Ka = [H+] [A-]/[HA] = (3.1623 * 10^-5) (0.0611) / 0.02547 =
7.59 * 10^-5 


Edited by akhl - 22 March 2011 at 8:19am
akhl IF-Rockerz
akhl
akhl

Joined: 30 March 2007
Posts: 5248

Posted: 22 March 2011 at 8:21am | IP Logged
See the above solution. There is some difference between the two Ka values. This means there is some error in taking the data.
But it is ok. Some small error will always be there. Just confirm once more that the data are correct.
 


Edited by akhl - 22 March 2011 at 1:19pm
Miggi Groupbie
Miggi
Miggi

Joined: 02 June 2006
Posts: 142

Posted: 22 March 2011 at 10:51pm | IP Logged
hey thank you so much and i have got the same answers as yours by the different method  but i have written down this method too.
Thanks again :) 
Miggi Groupbie
Miggi
Miggi

Joined: 02 June 2006
Posts: 142

Posted: 23 March 2011 at 9:07pm | IP Logged
 
 
Thanks in advance:)
akhl IF-Rockerz
akhl
akhl

Joined: 30 March 2007
Posts: 5248

Posted: 23 March 2011 at 11:57pm | IP Logged
Show the same H atom bonded to both N atoms. It will be covalently bonded to one N atom (solid line) and hydrogen bonded to another N atom (dashed line). Draw as shown below:-
 
akhl IF-Rockerz
akhl
akhl

Joined: 30 March 2007
Posts: 5248

Posted: 24 March 2011 at 6:26am | IP Logged
 
 
akhl IF-Rockerz
akhl
akhl

Joined: 30 March 2007
Posts: 5248

Posted: 24 March 2011 at 8:48am | IP Logged
Answer to the last problem:
In the last problem you need three compounds. I am giving their structures below.
 
 
 

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