- Joined
- Feb 12, 2008
- Messages
- 87
- Reaction score
- 0
- Points
- 0
- Pre-Dental
Can someone explain...
- Thread starter Predentknight
- Start date
i think its E
first you know using 20 mL of HCl will neutralize it since both are 1 M
then i think
(1 M)(x mL) = (.01 M)((20+20)+x)
the .01 M came from the fact that you can pH 2...so [H+] =1 X10-2
solve for x and I got .4ish
and add that to 20 so 20.4?!?!
can someone confirm
first you know using 20 mL of HCl will neutralize it since both are 1 M
then i think
(1 M)(x mL) = (.01 M)((20+20)+x)
the .01 M came from the fact that you can pH 2...so [H+] =1 X10-2
solve for x and I got .4ish
and add that to 20 so 20.4?!?!
can someone confirm
answer is A
go to the link provided
it is your exact question and it has a nice descriptive answer. I had problems with the same question
http://forums.studentdoctor.net/showthread.php?t=530427
go to the link provided
it is your exact question and it has a nice descriptive answer. I had problems with the same question
http://forums.studentdoctor.net/showthread.php?t=530427
- Joined
- Oct 16, 2007
- Messages
- 79
- Reaction score
- 0
- Points
- 0
- Pre-Dental
same general idea in the problem you linked to, but with half the volume of acid, so i think the 2nd poster is right and the answer is e. is it?
okay i just searched it since it bothered me and it was pretty simple. i feel stupid for not coming up with it my self.
answer is B btw.
since 20 ml of HCl will completely titrate with 20 ml of NaOH, what we are looking for here is what amount of that extra [H+] coming from HCl would bring your pH down to 2 from 7.
M1V1 = M2V2 (since normality of HCl and NaOH are equal)
M1 = 1
V1 = 20 ml
M2 = 0.01 (which is pH = 2; [H+] = 10^-2)
V2 = what we are looking for (the extra amount of Volume that changed to new pH level)
(1)(20) = (0.01)(x)
x = 20 / (0.01) = 0.2 ml
so now we add 0.2 ml to our original amount of 20 ml of HCl so our answer would be 20.2
answer is B btw.
since 20 ml of HCl will completely titrate with 20 ml of NaOH, what we are looking for here is what amount of that extra [H+] coming from HCl would bring your pH down to 2 from 7.
M1V1 = M2V2 (since normality of HCl and NaOH are equal)
M1 = 1
V1 = 20 ml
M2 = 0.01 (which is pH = 2; [H+] = 10^-2)
V2 = what we are looking for (the extra amount of Volume that changed to new pH level)
(1)(20) = (0.01)(x)
x = 20 / (0.01) = 0.2 ml
so now we add 0.2 ml to our original amount of 20 ml of HCl so our answer would be 20.2
sorry didnt see the volumes were different
- Joined
- Feb 12, 2008
- Messages
- 87
- Reaction score
- 0
- Points
- 0
- Pre-Dental
Ak47 is correct the answer is E. Thanks for the explanation.
Here is where you went wrong. When you add the two together, you have a total volume of 40 mL solution at a pH of 7. So we are looking for how much HCl we need to add to a neutral solution to get a total H+ ion concentration of 0.01.
Therefore, M1 = 0.01 V1 = 40 mL M2=1 V2 = ?
V2 = .04 + 20 mL so the answer is 20.4 mL
Similar threads
- Question
