dontbam

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Can anyone help me with this question?

What volume of HCl was added if 20mL of 1M NaOh is titrated with 1M of HCl to produce a pH=2 solution?

according to topscore the answer is 20.4 mL
 

thehipster

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The answer doesn't happen to be 200 mL, does it?
 
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asckwan

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I got it to be .2L. Anyone else?

(forgot my dot there :oops: )
 

lovechild404

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i got the answer to be .2L = 20 mL. i'm not sure why the answer would be 20.4mL...where did the .4 come from?
 
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dontbam

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hey can you guys show me how you calculated the 20ml? in top score they set it up where it's like 0.01 * (40 + x) = x

solving, that's where the .4 comes from, how did you guys solve it?
 

Chga

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you have to convert to normality
1 mole NaOH/1L = x mole NaOH/.02L
x= .02 moles
(.02 moles)(1 equiv of -OH)= .02 N
since OH and H in HCl are 1 to 1 to equalout
there is .02N of H+ ions meaning .02M of HCl
since its 1M HCl there are .02L of it.
.02L = 20ml
have no idea how some people got .2L or where the 20.4 comes in.
it doesnt make sense
 

djeffreyt

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dontbam said:
Can anyone help me with this question?

What volume of HCl was added if 20mL of 1M NaOh is titrated with 1M of HCl to produce a pH=2 solution?

according to topscore the answer is 20.4 mL

first...you have 20mL of a 1M Strong Base that will take only 1 proton per molecule (NaOH). To neutralize that, add an equal amount of a strong acid (HCl) at 1 M. So to neutralize the 20mL of 1 M NaOH, we add 20 mL of 1 M HCl.

That was step 1, and you do that in your head just by looking at the problem. Then the question is how much more do we have to add of 1M HCl to our now 40mL of neutral solution to get the pH = 2

You need to know this formula below and how to work with it
pH = 2
pH = -log [H+]
2 = -log [H+]
10^-2 = [H+]
or
[H+] = 0.01 M

if we call X the amount of HCl we need to add to our 40ml solution to reach a pH = 2 then you can put it together like this. It's kind of a dilution question. You start with 1 M HCl. You are diluting it by adding some of this 1M HCl to a 40mL neutral solution. When will the solution's pH drop of 0.01 M HCL 9which is pH =2)

(1 M HCl) * (X mL) = (0.01 M HCl) * (40 + X ml)

I'm not gonna do the math here...but the answer comes out right.
 

asckwan

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Oh dang! I see now, 20mL is the amount to neutralize the solution, but then you have to take it one step further to get the pH = 2.

Achiever also has a question like this:

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?

Practice practice practice!!
 

fancymylotus

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asckwan said:
Oh dang! I see now, 20mL is the amount to neutralize the solution, but then you have to take it one step further to get the pH = 2.

Achiever also has a question like this:

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?

Practice practice practice!!
Can someone do this problem out? I'm having a ****** moment.
 

cocokeala

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finishing up Dr. Jeffs math hope its right cuz i came out with the right answer :)

V= .4+.01V
.99V= .4
V= .4

.4 is the extra amount of HCL you need to add to the neutralized soln which was 40 (remember 20 HCL + 20NaOH)

well we only want how much HCL was added
so we know that 20 HCL was added just to neutralize the NaOH, and we added .4 HCL to get it to pH=2.

so 20+.4= 20.4
 

lovechild404

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BRILLIANT!...we're all on our way to dental school now! :laugh:
 

tom_servo_dds

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amsie said:
Can someone do this problem out? I'm having a ****** moment.
I came up with 30 mL for HCl needed:

(0.05M HCl)(X mL) = (0.01)(40mL + X mL)
X = 10 mL + 20mL to neutralize = 30 mL total
 

Envision

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tom_servo_dds said:
I came up with 30 mL for HCl needed:

(0.05M HCl)(X mL) = (0.01)(40mL + X mL)
X = 10 mL + 20mL to neutralize = 30 mL total
Hey tom,

doing that problem i got 28. is your 30 round off? or is my math incorrect? thanx.
 

dat_student

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dontbam said:
Can anyone help me with this question?

What volume of HCl was added if 20mL of 1M NaOh is titrated with 1M of HCl to produce a pH=2 solution?

according to topscore the answer is 20.4 mL
ok, I solve it again:

V is the total volume of HCl

1 * V - 1 * 20
__________ = 10^-2
V + 20

V = 20.2 / 0.99

V = 22.4

or you can do the following:

20 ml of HCl is needed to neutralize 20 ml of 1 M NaOH. So, total volume of the final solution is V + 20 + 20 (or V + 40). V is the extra volume of HCl that is needed to produce a pH=2 solution.

1 * V
_______ = 10^-2
V + 40

V = 0.4

20 (to neutralize) + 0.4 (extra) = 20.4

*Again, I assumed that at pH = 7 we have little H+.
 

tom_servo_dds

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Envision said:
Hey tom,

doing that problem i got 28. is your 30 round off? or is my math incorrect? thanx.
You're right, 28 - thanks for catching that!
 

Envision

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tom_servo_dds said:
You're right, 28 - thanks for catching that!
no wait, your right! i did it again and i got thirty! haha :D
 

tom_servo_dds

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Envision said:
no wait, your right! i did it again and i got thirty! haha :D
Looks like we both made the same mistake :laugh: - it is 30, which is good since the options were:

A. 10
B. 15
C. 20
D. 25
E. 30
 
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