Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: RogerJH on March 17, 2011, 01:57:17 PM
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I'm trying to solve this problem for a class activity.
If I measure out 1 L of pure water and I have 5% solution of acetic acid (household vinegar), what is the amount of vinegar I need to change the pH of water from 7 to 6?
I'd appreciate any help figuring this out. ???
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pH 6 means 1 *10-6 mol/l of H+
5% acetic acid cotains 0,84 mol/l (table book)
what means 1* 10-6/ 0.84 = 1.2 * 10-6 l = 1,2 µl
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Thanks Nobby. I really appreciate the help. Just to clarify...
So 1.2 microliters of 5% acetic acid solution will change the pH of 1 L of pure water from pH 7 to pH 6?
That seems like a very small amount of vinegar.
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Maybe I have to change my opinion, because acetic acid is a weak acid.
pH = ½( p K a - log [ HA ] 0 )
6 = 0.5 (4,75 -log [ HA ] 0)
But anyway its only 1 drop
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Sure that makes sense. Now I want to calculate the amount of acid rain needed to change the pH of a body of water such as a lake.
Assume the water body is 1.0 cubic km = 10^12 liters.
Assume its pH is 7.0.
Assume acid rain has a pH 5.0. (I think this acts as a weak acid...)
What is the volume of acid rainwater needed to change the pH of the water body from 7.0 to 6.5?
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First 1 km3 is 1 *1015 l
because 1 m3 = 1000 l
we use the same equation:
6.5 = 0.5 * (4,75 -log [ HA ] 0)
cHA = 5,62 *10-9 mol/l
5,62 *10 6 mol
The rain has pH 5 what means
5 = 0.5 * (4,75 -log [ HA ] 0)
cHA = 5,62 *10-6 mol/l
V = 5,62 *10 6 mol/5,62 *10-6 mol/l
= 1 *1012 l.
So about 0,1 % of the volume rain has to be added.
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Cool. You really know your stuff!
I am confused though since Wolfram Alpha (which I find helpful for these kinds of things calculates) 1 km3 as 1012 l. http://www.wolframalpha.com/input/?i=1+cubic+kilometer
For instance 1 m3 = 1000 l, and there are 1000 * 1000 * 1000 = 109 meters in a cubic kilometer = 109 * 1000 = 1012.
Unless I am not getting it.
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pH = ½( p K a - log [ HA ] 0 )
Do you know when this equation an be used? There is a condition that must be meet.
Note: you are giving final answers. That's against forum rules (http://www.chemicalforums.com/index.php?topic=33740.0).
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Just to throw real world out there, its really hard to get pure water and water that is pH 7. Even lab grade dionized or distilled water is usually pH 6 or lower (from CO2 in air), fyi.
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Good point. I am trying to demonstrate the effect of acid rain, viz. a small amount of acid can change water chemistry dramatically.
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I would then suggest, for a demonstration to show acid rain, would be to get some pH 7 water (make it pH 7 with small amounts of base).
Have the students (or whoever) check the pH of the water (which should be at 7 which you set). Then just have them blow through a straw into the water, and the CO2 from their breath will make it acidic, and then check the pH after doing it for 30 seconds.
This is also why students should be told to NOT BREATH out directly onto acid/base quantitative titrations like they always seems to do!
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Do you know when this equation an be used? There is a condition that must be meet.
Note: you are giving final answers. That's against forum rules.
@ Mr. Borek
whats wrong in my calculation? You are free to correct it. Please explain. And where is it written in your forum rules that I am not permitted to give final answers? Is this only allowed for some admin people here? Or is it meant with the Socratic method. Never heard this.
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Cool. You really know your stuff!
I am confused though since Wolfram Alpha (which I find helpful for these kinds of things calculates) 1 km3 as 1012 l. http://www.wolframalpha.com/input/?i=1+cubic+kilometer
For instance 1 m3 = 1000 l, and there are 1000 * 1000 * 1000 = 109 meters in a cubic kilometer = 109 * 1000 = 1012.
Unless I am not getting it.
You are right
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whats wrong in my calculation?
I already pointed you in the right direction. Equation you used is a simplified one, that has limited applicability, after calculating result you should check if the condition that allows use of the equation is meet.
See http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base for more details.
And where is it written in your forum rules that I am not permitted to give final answers? Is this only allowed for some admin people here? Or is it meant with the Socratic method. Never heard this.
Yes, it is part of Socratic method, but I see it was not spelled out directly in the forum rules. Time to edit them then.
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Yes, it is part of Socratic method, but I see it was not spelled out directly in the forum rules. Time to edit them then.
I dont agree with this. Why do we have this strange rule here. Everybody should have his own way to answer questions or to solve problems. Why only one method should be valid.
Ok the student should be think by himself, but sometimes its easier to give a finished solution. If this solution is wrong other people can correct it and show how it will go better.
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I dont agree with this.
You are free to find other forum, where it is possible to give final answers. In our experience it doesn't help - students copy the solution, but they don't understand it.
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This I understand.
We close it now, ok.
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@Borek
Actually I am designing a class activity for a colleague. This is not for a test or exam. I very much appreciate the help and expertise in this forum. I really just needed the answers to help me design the activity.