Can someone explain...

[Predentknight]

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What volume of HCl was added if 20ml of 1M NaOH is titrated with 1M HCL to produce a pH of 2?

a.10.2ml
b.20.2ml
c.30.4ml
d.35.5ml
e. none of these

 

[ak47]

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

 

[250rsavage]

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

 

[happyasaclam88]

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?

 

[lphiewok]

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

 

[250rsavage]

sorry didnt see the volumes were different

 

[Predentknight]

Ak47 is correct the answer is E. Thanks for the explanation.

 

[DentalDeac]

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 you set this up wrong, this answer would give you 2000 for x

so now we add 0.2 ml to our original amount of 20 ml of HCl so our answer would be 20.2

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