proline

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I might be asking a stupid question, but I'm a little rusty on acid base stuff.

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

I know how to neutralize the solution, but I don't really know how to solve the problem :-(
 

cryptozoologist

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ph2 means excess acid.

20 ml 1M NaOH neutralized with 20 ml 1M HCL.
Ph=2, so you have excess HCL. for Ph=2, [H+] conc = .01 or 1e-2 (from your log equation of ph).

so you do a second comparison:
HCL you can add * molarity of HCL = target molarity * (total solution volume)
X * 1M = .01M * (40+X)
X = approximately .4

so amt of HCL added to achieve PH =2 is 20 ml + .4 ml

= 20.4 ml
 
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proline

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thanks for your response, though I am still a little confused

if you add 1M acid and 1M based, it becomes a neutral solution. if its neutral doesn't it have a pH of 7? And if so, why wouldn't we use M = .0000001 instead M = 1

I guess my problem is I really don't understand what the molarity of a solution is when you add 1M acid and 1M base. I think I'm doing stupid stuff and mixing up numerous concepts. Why do we use 1M instead?
 
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cryptozoologist

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you use 1M because that is what they tell you to use in the problem.
 

OSUDDS

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cryptozoologist said:
you use 1M because that is what they tell you to use in the problem.
Did this question come straight from Topscore?
 

dontbam

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yeah a neutral solution is at pH 7 but the question is asking for the pH of the solution at 2.0 which is 1 x 10^-2.

i asked this same question in an earlier post!
 

DonExodus

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Again, not to be stupid, but 1M HCL has a pH of 1, correct? pH is the [H+] = M, not the mol H+.
Whats confusing about this to me is that a 2M solution would have a pH of 2, which makes no sense, so I know my logic is flawed.

I guess what Im getting at is how do I determine the pH of a solution from the molarity?
M= mol/liter, I realize this. Im in one of those "I had it, but studied too much and confused myself" states.
Unless, Im incorrect in assuming a 1M solution of HCl also has 1mol H+ / liter....

Thanks, sorry for the stupid question, but Ive studied myself ******ed it would appear :p
 
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