Acheiver test 2 ghem chem problem ??

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munecaDDS

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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?
A. 10 B. 15 C. 20 D. 25 E. 30
 
what is the answer...?? I am trying to solve it but don't know how to use the pH in this problem.. just using the molality formula is not workign out.. anyone else know?
 
Yea thats my problem also with the molarity formula you get 20 but the answer is 30 for this problem and i didn't understand the explanation achiever had
 
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
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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.
 
No i haven't seen anything in destroyer like this before. And thanks for your help
 
#10 joonkimdds
Senior Member



Status: Pre-Dental
Join Date: Jun 2005
Location: Fairfax, VA
Posts: 1,428
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icon1.gif

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?
 
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