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shree93 IF-Dazzler
shree93
shree93

Joined: 07 October 2006
Posts: 3264

Posted: 16 March 2011 at 1:45pm | IP Logged
A voltage of 16.0 V is applied across the two ends of a Tungsten wire with a temperature coefficient of resistivity of 4.5010-3 1/K. Initially the wire is kept at room temperature. If the wire is cooled down to freezing temperature, then the power dissipated by the wire will: 
 decrease 
 stay the same 
 increase 

i think it will increase, but I am not 100% sure.
akhl IF-Rockerz
akhl
akhl

Joined: 30 March 2007
Posts: 5264

Posted: 16 March 2011 at 3:20pm | IP Logged
Originally posted by shree93

A voltage of 16.0 V is applied across the two ends of a Tungsten wire with a temperature coefficient of resistivity of 4.5010-3 1/K. Initially the wire is kept at room temperature. If the wire is cooled down to freezing temperature, then the power dissipated by the wire will: 
 decrease 
 stay the same 
 increase 

i think it will increase, but I am not 100% sure.
 
The temperature is lowered. Therefore the resistance of the wire decreases.
P = V^2/R
V remains constant and R decreases. Therefore, P will increase.
 
You are right. It will increase.
Miggi Groupbie
Miggi
Miggi

Joined: 02 June 2006
Posts: 142

Posted: 21 March 2011 at 11:25am | IP Logged
Hi,
This is a chemistry question. I did a lab in which a strong base was added in a weak acid to make a buffer solution.
 
from my graph@ Half Titration point pH=4.1 and volume of NaOH=6.7ml.
Mass of ascorbic acid was 0.8994g and NaOH=0.40M
The question was to choose to points equally spaced on either side of the half titration point in the buffer zone and calculate both [HA] and
[A-] for both points?Assume graduated cylinder dilivered 50ml of water in total and assume volume of acid is neglible. 
 
One point I chose was when pH=3.7 and  volume of NaOH=4.00ml.
 and the other was where pH =4.5 and volume=9.00ml.
 
 
To solve: I found out the moles of ascorbic acid to be 5.06*10^-3
and added 0.0004Lof base and 0.05L of water to  0.09L
from this I found out the concentration of [HA] to be 0.056M but I don't know how to find the equilibrim concentration using ICE table.
 
another approach was to use formula [HA]=(moles of acid originally-moles of base added)/total volume  and for [A-]= moles OH-/total vol
but it gives me a negative number here is what I did 5.06*10^-3-0.016/0.09 and it gives -0.122.
I don't know what to do please help.
  
akhl IF-Rockerz
akhl
akhl

Joined: 30 March 2007
Posts: 5264

Posted: 21 March 2011 at 3:08pm | IP Logged

Initial moles of the acid = 5.06 * 10^-3 = 0.00506

At the first point, moles of NaOH = 0.40 * 4.00 * 10^-3 =  0.0016
Total volume = 50 ml + 4.00 ml = 54 ml = 0.054 L
[HA] = (original moles of acid  - moles of base)/total volume
[HA] = (0.00506  -  0.0016) / 0.054
[HA] = 0.0641 M
 
At the second point, moles of NaOH = 0.40 * 9.00 * 10^-3 = 0.0036
Total volume = 50 ml + 9.00 ml = 59 ml = 0.059 L
[HA] = (0.00506  -  0.0036)/0.059
[HA] = 0.0247 M
 
You wrote 0.016. It will actually be 0.0016.
 
Finding [A-] values:
Note that pKa = pH at half equivalence point.
So pKa = 4.1
Now you know pKa, pH and [HA]. So can you find [A-] using Henderson equation?
 


Edited by akhl - 21 March 2011 at 4:33pm
Miggi Groupbie
Miggi
Miggi

Joined: 02 June 2006
Posts: 142

Posted: 21 March 2011 at 6:07pm | IP Logged
Hey, thanks for your help. I just have one more question. When finding the [H+] for the above two points, do we still assume pH=pKa and Ka = [H3O+] or would it be different as the points are before and after half equivalence point?

Edited by Miggi - 21 March 2011 at 6:07pm
akhl IF-Rockerz
akhl
akhl

Joined: 30 March 2007
Posts: 5264

Posted: 21 March 2011 at 6:17pm | IP Logged

pH and [H3O+] values change but pKa and Ka values do not change.

pKa = pH at half titration point
 
So at all points,
pKa = 4.1
and Ka = 10^-4.1
Ka = 7.9433 * 10^-5
 
You already know pH values. So find [H3O+] using the formula
[H3O+] = 10^-pH
 
At one point, pH = 3.7
Therefore [H3O+] = 10^-3.7 = 2.00 * 10^-4
 
At the second point, pH = 4.5
Therefore [H3O+] = 10^-4.5 = 3.16 * 10^-5
 


Edited by akhl - 21 March 2011 at 6:35pm
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: 5264

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

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