vanBassum
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NaCl Titration
Hello guys,
I realize this is probably a high school question, but please bear in mind I never had this in school. Everything I know about chemistry is
self-taught. Now that is out of the way, the question: 
My goal is to roughly determine the amount of chloride in a solution. A precision around 10g/L would be sufficient. The method I would like to use is
the following: http://www.chlorates.exrockets.com/cl_titr.html
The recipe tells me to get 1ml of sample solution. (Solution A)
Then dilute this by 349ml of water. (Solution B)
Then take 10ml of sol. B and add 10ml of water. (Solution C)
Now C is used for the testing.
This will give me a solution of 1/700 from the original sample, correct?
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The problem with this, one titration costs 350ml of water. The second problem, when I want to start with a different estimate, I need to make the
first solution again. So I came up with the following:
Take 1ml of the sample.
Add to 19ml of water, which gives 1/20. Let's call this solution D.
Now prepare 4 beakers with 34ml, 19ml, 9ml and 4ml.
Add 1ml of solution D to all 4 beakers.
This will give me all the different estimate solutions:
1/700 for 350g/L
1/400 for 200g/L
1/200 for 100g/L
1/100 for 50g/L
This way I save on Demi water and can do all the different estimates with solution D as starting point. I probably only have to do 2 titrations if I
start with the 350g/L one and then zero it down with one of the others if required.
What do you guys think, is my reasoning correct?
I have ordered everything I need for this test, once I have the things I need, I'll test this with a known solution of NaCl.
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Rainwater
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http://www.sciencemadness.org/smwiki/index.php/Silver_nitrat...
"You can't do that" - challenge accepted
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Texium
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If you’re using some high-grade expensive
water for this, don’t. Regular distilled water ($2/gallon last I checked) should be absolutely fine for this titration, especially considering you
only want a resolution of 10 g/L.
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vanBassum
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Hello,
I can just buy the water ofc. but thats not the point. 
The goal for me is to learn. I wanted to try this to see whether my reasoning is correct. Since I don't have a lot of experience, It could be that
there is a specific reason why they do it their way. Another possibility is that they didn't give it mutch thought and this was the easiest method for
them. I can think of a few reasons they work this way. e.g. Accuracy, reproducability, ect. The thing is, I can't verify this, its al assumptions from
my end. One test that I can think of is to do a titration of a known solution. The problem is, how do I know that the measurements that I get aren't
correct by coincidence. Or maybe the measurements are a slightly off, but I discard this as my lack of experience when its actually a fault in my
calculations before?
I still plan to just try this on a known solution, but any help in advance is welcome.
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vanBassum
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So, I just got the items I need and did some experiments.
I have figured out that my theory above has a fault. The original recipe takes 1cc and adds this to 349cc of water. Then they take 10cc and add this
to 10cc of water. This is the mixture that is used for the titration. In my original theory, I counted the 2nd dilution, but that is of course wrong.
It is diluted, but all 10cc's are in the mixture. Therefore, the titration is executed on 10cc of 1/350 of the sample, not 10cc of 1/700 as I thought.
This makes things a lot simpler, I can just do 1/10 the scale to save on water but pay with accuracy.
So take 1cc of sample, add this to 34cc of water for 1/35. Then take 1cc instead of 10cc and add this to enough water to be able to see the reaction.
This gives 1cc of 1/35 the original sample. This is the same amount as 10cc of 1/350 the sample. If that is not enough accuracy, it will at least give
a direction for the next titration.
---- EDIT ----
Awesome, I just did a titration of a saturated solution of NaCl. (I know, not the most accurate) But it took me 1cc of AgNO3 with 1cc of 1/35 sample.
That corresponds to 350g/L. That is very close to the 360g/L @ 25C theoretical maximum. So this seems to be a good indication this is working. I might
do a more accurate titration on a known sample tomorrow. Just to get the hang of it.
[Edited on 29-6-2022 by vanBassum]
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