zenosx
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Titration. I suck at the math. Help
Ok. I'm trying to titrate my HNO3 using NaOH.
I made up a 2M / 100ml sol. by dissolving 7.9g (best my scale can do) in 100 mL of H2O for a 2 molar sol.
From estimates, I figured in had around 15M ( is fuming red) so 2 mL diluted to 20 mL would give me a 1.5 molar sol.
It took 22ml or 239 drops to reach endpoint with phenolphthalein. Being this is red fuming (made from the KNO3+H2SO4 distillation process under
reduced pressure) I will say again I figured greater than 90% or 15M min.
So from my horrific math that I won't post due to I'm on my iPhone. I calculated a 15.8 molar concentration. Does this work out right for those better
at this than me? I couldn't remember the exact endpoint PH of phenolphthalein on pink endpoint so did my best with PH paper around 7.3 I think. (I so
need a ph meter high range paper sucks).
All feedback appreciated. Btw I used the Macid Vacid = Mbase Vbase formula.
Thanks to anyone that can advise. I really need to get these titrations down but without a chem instructor it's sometimes hard to know for sure you
are in the right ballpark.
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IanCaio
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Its irrelevant, but you would actually need 8g of NaOH to get a 2M solution on 100mL.
But guess the difference is gonna be insignificant on the final molarity calculation.
Guess its also not really necessary to take in consideration the pH of solution since we are talking about a not really precise calculation and the pH
changes really fast close the equivalence point.
So basically you're just gonna calculate how much NaOH is on 22mL and since the reaction reacts 1 Mol with 1 Mol, the HNO3 concentration is
the same as the NaOH.
22mL of 2M solution means 0.044 Mols of NaOH. Assuming the NaOH that didnt reacted is irrelevant, we'll assume we have 0.044 Mols of HNO3
in the 20mL. So 2mL of the titrated solution actually haves 0.044 Mols, meaning the concentration of your nitric acid is 22M.
I might be wrong though, I'll review calculations now..
Obs: pH always with 'p' on low case 
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tetrahedron
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your calculation checks out. probably the NaOH was stale having absorbed moisture and CO2 from the atmosphere and/or the water used was slightly
acidic, both of which could have led to an overreport.
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simba
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Quote: Originally posted by IanCaio  | Its irrelevant, but you would actually need 8g of NaOH to get a 2M solution on 100mL.
But guess the difference is gonna be insignificant on the final molarity calculation.
Guess its also not really necessary to take in consideration the pH of solution since we are talking about a not really precise calculation and the pH
changes really fast close the equivalence point.
So basically you're just gonna calculate how much NaOH is on 22mL and since the reaction reacts 1 Mol with 1 Mol, the HNO3 concentration is
the same as the NaOH.
22mL of 2M solution means 0.044 Mols of NaOH. Assuming the NaOH that didnt reacted is irrelevant, we'll assume we have 0.044 Mols of HNO3
in the 20mL. So 2mL of the titrated solution actually haves 0.044 Mols, meaning the concentration of your nitric acid is 22M.
I might be wrong though, I'll review calculations now..
Obs: pH always with 'p' on low case |
You're very wrong there. 22M nitric acid? Not in this world at least.
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IanCaio
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| Quote: |
your calculation checks out. probably the NaOH was stale having absorbed moisture and CO2 from the atmosphere and/or the water used was slightly
acidic, both of which could have led to an overreport.
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You're probably right, it did sound quite wierd to have such a high molarity. 70% nitric acid is about 15M, I was assuming it might be almost fuming
nitric acid..
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You're very wrong there. 22M nitric acid? Not in this world at least.
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Yeah, I thought about that too. I assumed it must be fuming because as far as I know 15M HNO3 solution is about 70%. But even though, I
just took the numbers and did the math (which I hope is right  )
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weiming1998
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| Quote: Originally posted by simba | | Quote: Originally posted by IanCaio | Its irrelevant, but you would actually need 8g of NaOH to get a 2M solution on 100mL.
But guess the difference is gonna be insignificant on the final molarity calculation.
Guess its also not really necessary to take in consideration the pH of solution since we are talking about a not really precise calculation and the pH
changes really fast close the equivalence point.
So basically you're just gonna calculate how much NaOH is on 22mL and since the reaction reacts 1 Mol with 1 Mol, the HNO3 concentration is
the same as the NaOH.
22mL of 2M solution means 0.044 Mols of NaOH. Assuming the NaOH that didnt reacted is irrelevant, we'll assume we have 0.044 Mols of HNO3
in the 20mL. So 2mL of the titrated solution actually haves 0.044 Mols, meaning the concentration of your nitric acid is 22M.
I might be wrong though, I'll review calculations now..
Obs: pH always with 'p' on low case |
You're very wrong there. 22M nitric acid? Not in this world at least.
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It's possible to get 22M nitric acid on Earth. In fact, my calculation gave a figure of 22M as well.
Molarity is the number of moles of a particular substance per litre of solution. You forgot to note that nitric acid is denser than water. 22M gives
1386 grams/L of acid. Wikipedia gives a density of 1.5129 grams/ml for pure nitric acid, so it is possible that nitric acid can be 22M.
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zenosx
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Thanks everyone, I got 15.8 molarity and I also got the .44 result in my math as well.
This is most certainly RFNA made from 0% H20 with higher purity KNO3 + 98% H2SO4, so it's most certainly above 70% HNO3. I was going to put it under
vacuum to try and remove the NO2 from the acid, but have read that this might cause it to degrade faster.
Upon atmospheric exposure, this stuff dumps some NO2 and white fumes in a major way, I live in the SE so the humidity here is quite high even with a
24/7 de-humidifier running in the lab. I deal with this stuff rarely, and if I do, I usually quickly dilute it to ~70%.
I was trying to get an absolute titration of it, and being that it is probably greater than 96% HNO3, (way above the 68% azeotrope with H2O that HNO3
creates).
Going from wiki. A commercial grade of fuming nitric acid, referred to in the trade as "strong nitric acid" contains 90% HNO3 and has a density of
1.50 g/mL. This grade is much used in the explosives industry. It is not as volatile nor as corrosive as the anhydrous acid and has the approximate
concentration of 21.4 molar.
So 22M HNO3 is very possible, only NOT from distilling 70% or other lower grades of HNO3, due to the azeotrope. Just like ethanol, other ways of going
higher than the azeotrope are necc, (like adding benzene to ethanol to get higher than 95% E.G. "Dry" ethanol)
I think tomorrow I will weight an exact 1mL of my acid and see what I get. Unfortunately my best scale only does 0.1 accuracy.
I would like to see what math you did for the 22 molar, since that seems to be the correct value.
On the NaOH, 8g is a rounding, the actual precise calc is 7.9997 or something, 8g is a good rounding), and yes I probably had some atmospheric water
absorption, and also was using a dropping funnel and graduated cylinder to measure dispensed volumes so there is some inaccuracy. I'd say at least +/-
5%.
However, the acid (2 ml) was diluted to 20 ml with H2O,,, that messed me up.
Show the math please I'd be appreciative for future titrations.
I needed 15.8 M HNO3 for one of my reactions tonight, and even running slightly lower than Stoichiometric I had excess acid rather than unreacted
product, so even with some dilution, it's higher than I expected.
Thanks again everyone for your work! Once I go back down to the lab I will post the math I have done, however the woman is now calling me to bed,,,
sigh.....
[Edited on 15-10-2012 by zenosx]
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simba
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| Quote: Originally posted by weiming1998 | | Quote: Originally posted by simba | | Quote: Originally posted by IanCaio | Its irrelevant, but you would actually need 8g of NaOH to get a 2M solution on 100mL.
But guess the difference is gonna be insignificant on the final molarity calculation.
Guess its also not really necessary to take in consideration the pH of solution since we are talking about a not really precise calculation and the pH
changes really fast close the equivalence point.
So basically you're just gonna calculate how much NaOH is on 22mL and since the reaction reacts 1 Mol with 1 Mol, the HNO3 concentration is
the same as the NaOH.
22mL of 2M solution means 0.044 Mols of NaOH. Assuming the NaOH that didnt reacted is irrelevant, we'll assume we have 0.044 Mols of HNO3
in the 20mL. So 2mL of the titrated solution actually haves 0.044 Mols, meaning the concentration of your nitric acid is 22M.
I might be wrong though, I'll review calculations now..
Obs: pH always with 'p' on low case |
You're very wrong there. 22M nitric acid? Not in this world at least.
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It's possible to get 22M nitric acid on Earth. In fact, my calculation gave a figure of 22M as well.
Molarity is the number of moles of a particular substance per litre of solution. You forgot to note that nitric acid is denser than water. 22M gives
1386 grams/L of acid. Wikipedia gives a density of 1.5129 grams/ml for pure nitric acid, so it is possible that nitric acid can be 22M.
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Indeed, I forgot taking the higher density into account, my bad! Thanks for correcting.
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IanCaio
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| Quote: |
I would like to see what math you did for the 22 molar, since that seems to be the correct value. On the NaOH, 8g is a rounding, the actual precise
calc is 7.9997 or something, 8g is a good rounding), and yes I probably had some atmospheric water absorption, and also was using a dropping funnel
and graduated cylinder to measure dispensed volumes so there is some inaccuracy. I'd say at least +/- 5%. However, the acid (2 ml) was diluted to 20
ml with H2O,,, that messed me up. Show the math please I'd be appreciative for future titrations.
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Indeed the base weight error is really small, it wont make much difference..
The 22mL of base that reacted have 0.044 mols of NaOH (you said that you got the same value). So we know that approximately 0.044 mols of
HNO3 reacted.
I made my calculation using the 2mL because its the only source of acid in the 20mL solution. So if 0.044 mols of HNO3 reacted, we know it
came from this 2mL, thus we can assume its concentration by simple dividing 0.044mol/0.002L.
Hope I made it understandable, english is not my main language 
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zenosx
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Thanks, that clears it up. So I'm looking at ~22 M concentration, which is way higher than the 15.8, which would be for 70% concentration. The numbers
check out. Thanks everyone
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