Acheiver test 2 ghem chem problem ??
Started by munecaDDS
Here they solve it with a little bit different numbers but they explained very good
http://forums.studentdoctor.net/showthread.php?t=406065
http://forums.studentdoctor.net/showthread.php?t=406065
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From what i can make of this. i think, i am most probably wrong, since both solutions are the same concentration upon adding 20ml of HCl the solution should become neutral. therefore, to get a pH of 2 u would need a hydronium [H3O+]concentration of 10^(-2)which is 1/100. im kinda lost from there but this is how i think its done. if u add 10 ml of HCl to the already present 40ml of the natural solution u will get a 50mil solution.the concentration of the 10ml solution is .05. once placed in the new solution u have added 0.05MX0.01L= .0005 moles in the total solution that is 50 ml. 0.0005/0.05=0.01 the concentration we are looking for. i hope that made sense lol im in class and i tried to throw this together as quickly as possible
#10 joonkimdds
Senior Member
Status: Pre-Dental
Join Date: Jun 2005
Location: Fairfax, VA
Posts: 1,428
How many milliliters of 0.05 M HCl are required to turn 20 ml of 0.05 M KOH into a solution of pH 2.00?
E. 30
The answer is E.
This is what I found from search for titration problem.
I just thought that adding this problem here could help you studying 😀
HCl and KOH have the same molarity so adding 20ml of HCL will make the solution into neutral.
Now the fun begins.
M1V1 = M2V2
(0.05M)(x) = (0.01M)(40ml + x)
0.05x = 0.4 + 0.01x
0.04x=0.4
x= 10 mL
therefore,
50ml - 20ml = 30
__________________
DATDATDATDATDATDATDATDATDATDATDATDATDATDAT
this is what i found from what joonmdds had written.. but yea this is a tough problem.. is there any in the destroyer like this.. because i have not come across it yet.
Senior Member
Status: Pre-Dental
Join Date: Jun 2005
Location: Fairfax, VA
Posts: 1,428
How many milliliters of 0.05 M HCl are required to turn 20 ml of 0.05 M KOH into a solution of pH 2.00?
E. 30
The answer is E.
This is what I found from search for titration problem.
I just thought that adding this problem here could help you studying 😀
HCl and KOH have the same molarity so adding 20ml of HCL will make the solution into neutral.
Now the fun begins.
M1V1 = M2V2
(0.05M)(x) = (0.01M)(40ml + x)
0.05x = 0.4 + 0.01x
0.04x=0.4
x= 10 mL
therefore,
50ml - 20ml = 30
__________________
DATDATDATDATDATDATDATDATDATDATDATDATDATDAT
this is what i found from what joonmdds had written.. but yea this is a tough problem.. is there any in the destroyer like this.. because i have not come across it yet.
#10 joonkimdds
Senior Member
Status: Pre-Dental
Join Date: Jun 2005
Location: Fairfax, VA
Posts: 1,428
How many milliliters of 0.05 M HCl are required to turn 20 ml of 0.05 M KOH into a solution of pH 2.00?
E. 30
The answer is E.
This is what I found from search for titration problem.
I just thought that adding this problem here could help you studying 😀
HCl and KOH have the same molarity so adding 20ml of HCL will make the solution into neutral.
Now the fun begins.
M1V1 = M2V2
(0.05M)(x) = (0.01M)(40ml + x)
0.05x = 0.4 + 0.01x
0.04x=0.4
x= 10 mL
therefore,
50ml - 20ml = 30
__________________
DATDATDATDATDATDATDATDATDATDATDATDATDATDAT
this is what i found from what joonmdds had written.. but yea this is a tough problem.. is there any in the destroyer like this.. because i have not come across it yet.
Can some body explain this a little more. How does the pH factor into this? I know pH=2 so, -log [H+]=2 so, 1x10^-2 = [H+]= .01M
How do you apply this to the problem?
