Apparently I don't understand percent concentrations... a little help?
May 17, 2016 4:13 PM Subscribe
In my workplace there is a reference to a "12% NH4OH" reagent that is prepared by diluting 20 mL of NH4OH to 50mL with H2O. It's obvious to me that this isn't a v/v concentration but I cant see how the formula makes it either a 12% w/w or a 12% w/v concentration. Assuming that we're starting with reagent grade (29% assay) NH4OH, which is it... and how does it work out to 12%?
This isn't "homework" - not sure how to articulate that in a better fashion.
This isn't "homework" - not sure how to articulate that in a better fashion.
Best answer: For a dilution, use:
(starting concentration)(starting volume) = (final concentration)(final volume)
So for this:
(29%)(20 mL) = (final concentration)(50 mL)
(final concentration) = (29%)(20 mL) / (50 mL) = 11.6%
posted by Durin's Bane at 4:23 PM on May 17, 2016 [1 favorite]
(starting concentration)(starting volume) = (final concentration)(final volume)
So for this:
(29%)(20 mL) = (final concentration)(50 mL)
(final concentration) = (29%)(20 mL) / (50 mL) = 11.6%
posted by Durin's Bane at 4:23 PM on May 17, 2016 [1 favorite]
Response by poster: Thanks! w/w or w/v? And why? Sorry-- just trying to make sure I understand.
posted by whatisish at 4:30 PM on May 17, 2016
posted by whatisish at 4:30 PM on May 17, 2016
v/v
You're measuring the quantity of both the water and the base by volume.
posted by mr_roboto at 4:32 PM on May 17, 2016
You're measuring the quantity of both the water and the base by volume.
posted by mr_roboto at 4:32 PM on May 17, 2016
Best answer: No. Sorry, I'm wrong. It depends on how the NH4OH solution was originally prepared. It'll be the same as that.
posted by mr_roboto at 4:41 PM on May 17, 2016 [1 favorite]
posted by mr_roboto at 4:41 PM on May 17, 2016 [1 favorite]
Best answer: 29% NH4OH is 29g of NH4OH in 100g. SG is 0.9 (according to the internet), so it's 29g NH4OH in 111mL. Subsample 20mL then dilute to 50mL and you get 5.23g NH4OH in 50mL, so 10.5gm NH4OH in 100mL.
Problem - you don't know the SG of the final solution. It's probalby somewhere between 0.9 and 1, so the final conc is going to be between 10.5% and 11.6%. So 12ish % w/w, more likely 11%. (Note that whoever decided on this dilution scheme, almost certainly didn't do this maths).
posted by kjs4 at 5:03 PM on May 17, 2016
Problem - you don't know the SG of the final solution. It's probalby somewhere between 0.9 and 1, so the final conc is going to be between 10.5% and 11.6%. So 12ish % w/w, more likely 11%. (Note that whoever decided on this dilution scheme, almost certainly didn't do this maths).
posted by kjs4 at 5:03 PM on May 17, 2016
Best answer: To further clarify, the original solution is (check this on the bottle) 29% w/w. Generally commercial reagents are w/w.
w/v is mass NH4OH/100mL solution, which in your final solution is 10.5% w/v. You can't calculate the w/w % in the final solution without knowing the SG.
This sort of calculation is a bit tricky, so most people don't bother for in house reagents. They just make it up using the same method every time. If you're the poor sod responsible for labeling the new NH4OH bottle, you could change it to 10.5% w/v to be more accurate. But it probably doesn't matter.
posted by kjs4 at 6:31 PM on May 17, 2016
w/v is mass NH4OH/100mL solution, which in your final solution is 10.5% w/v. You can't calculate the w/w % in the final solution without knowing the SG.
This sort of calculation is a bit tricky, so most people don't bother for in house reagents. They just make it up using the same method every time. If you're the poor sod responsible for labeling the new NH4OH bottle, you could change it to 10.5% w/v to be more accurate. But it probably doesn't matter.
posted by kjs4 at 6:31 PM on May 17, 2016
Response by poster: >> But it probably doesn't matter.
It doesn't. It's all academic at this point.
Yes. The bottle of the original solution is marked 29% w/w with a SG of .90.
So... to recap... I understand why 10.5% w/v is appropriate. I'm not quite as clear, though, on the w/w concentration. As a practical matter, is it just understood that when a w/w reagent grade solution is diluted as the first posters articulate, the resulting 11.6% concentration is nominally w/w as opposed to an exact figure? I mean, if a competent chemist were doing the dilution as described, he'd label that dilution's concentration as 11.6% w/w - even though the actual w/w concentration is a bit different?
In hindsight, it seems like a more appropriate way to tackle it would be to just call it a 40% v/v (concentrated) NH4OH solution since nobody really cares if the original is 28, 29, or 30% w/w.
Thanks again folks.
posted by whatisish at 9:35 AM on May 18, 2016
It doesn't. It's all academic at this point.
Yes. The bottle of the original solution is marked 29% w/w with a SG of .90.
So... to recap... I understand why 10.5% w/v is appropriate. I'm not quite as clear, though, on the w/w concentration. As a practical matter, is it just understood that when a w/w reagent grade solution is diluted as the first posters articulate, the resulting 11.6% concentration is nominally w/w as opposed to an exact figure? I mean, if a competent chemist were doing the dilution as described, he'd label that dilution's concentration as 11.6% w/w - even though the actual w/w concentration is a bit different?
In hindsight, it seems like a more appropriate way to tackle it would be to just call it a 40% v/v (concentrated) NH4OH solution since nobody really cares if the original is 28, 29, or 30% w/w.
Thanks again folks.
posted by whatisish at 9:35 AM on May 18, 2016
Don't label it 40% NH4OH! That's not even close to right, and you start to head into issues with safety (incorrect labelling etc.). Label it as "~12 % NH4OH, 20mL of 29% NH4OH in 50mL".
To answer you question, a competent chemist would label the diluted NH4OH in mol/L (I cannot be arsed calculating what this would be). That's the standard concentration unit for most laboratory reagents made up by volume. Why they sell conc reagents by % w/w I do not know, and it drives me nuts as I have to do this ridiculous maths everytime I want to dilute something to a specific molarity. If the chemist be bothered doing the maths, because it doesn't matter, then they'll describe the dilution method on the label as well as an approximate concentration (also, in both cases, they would label the bottle with the appropriate safety statements and hazard pictograms).
Solids can be made into % solutions, as they can be easily made up by weight (weigh x grams of NaOH, make up to y grams with H2O). I have seen some reagents that are labelled as %, but made up by volume, matrix solutions for ICP's for example. So long as they are made up the same way every time, it's fine. Though HNO3 is quite nifty, as the density (1.41) and the concentration 69% multiply together equal 1, so you need 50mL HNO3 to make up 1L of 5% HNO3, or close enough anyway. % w/v I've only ever seen in a couple of commercial products, H2O2 and pool chlorine. I imagine it's because they are (or were) made up to a specified volume. It's really a bit dodge, as %'s should be of the same unit, like parts per x.
If you wanted to actually make up 12% NH4OH w/w, the obvious way would be to dilute by weight (weigh 29% NH4OH, make up to 50gm with MQ), but this would stink out the lab with ammonia fumes, as balances and fumehoods don't mix. So you would have to calculate the volume of 29% needed (Mass needed for 50g of 12% solution = 0.12%*50g/0.29% = 20.7g. Volume needed for 50g 12% solution = 20.7g/0.9g/mL = 23mL), measure out the 23mL of NH3 in the fumehood, then measure (separately*) 27mL H2O (i.e. 27g) and add that and mix. Then you would have 12% NH4OH. But you'll have a bit more than 50mL.
(*The reason all this pedantry is sometimes necessary is that the final density of the solution is not predictable from the densities of the initial solutions. Sigh.)
There are also a lot of random unit conventions, some of which are also regional (for example, in Australia conc ammonia is sold as 25% (w/w) NH3), or specific to a type of lab. Some are wrong and/or misleading, but are generally coming from a situation where close enough is good enough and correct is annoying. E.g. some chemists quite happily use ppm when they actually mean mg/L, which can be misleading to the uninitiated (it's easier to say though). And if a physicist or biologist wrote the method, all bets are off as to what the units mean.
posted by kjs4 at 5:45 AM on May 19, 2016
To answer you question, a competent chemist would label the diluted NH4OH in mol/L (I cannot be arsed calculating what this would be). That's the standard concentration unit for most laboratory reagents made up by volume. Why they sell conc reagents by % w/w I do not know, and it drives me nuts as I have to do this ridiculous maths everytime I want to dilute something to a specific molarity. If the chemist be bothered doing the maths, because it doesn't matter, then they'll describe the dilution method on the label as well as an approximate concentration (also, in both cases, they would label the bottle with the appropriate safety statements and hazard pictograms).
Solids can be made into % solutions, as they can be easily made up by weight (weigh x grams of NaOH, make up to y grams with H2O). I have seen some reagents that are labelled as %, but made up by volume, matrix solutions for ICP's for example. So long as they are made up the same way every time, it's fine. Though HNO3 is quite nifty, as the density (1.41) and the concentration 69% multiply together equal 1, so you need 50mL HNO3 to make up 1L of 5% HNO3, or close enough anyway. % w/v I've only ever seen in a couple of commercial products, H2O2 and pool chlorine. I imagine it's because they are (or were) made up to a specified volume. It's really a bit dodge, as %'s should be of the same unit, like parts per x.
If you wanted to actually make up 12% NH4OH w/w, the obvious way would be to dilute by weight (weigh 29% NH4OH, make up to 50gm with MQ), but this would stink out the lab with ammonia fumes, as balances and fumehoods don't mix. So you would have to calculate the volume of 29% needed (Mass needed for 50g of 12% solution = 0.12%*50g/0.29% = 20.7g. Volume needed for 50g 12% solution = 20.7g/0.9g/mL = 23mL), measure out the 23mL of NH3 in the fumehood, then measure (separately*) 27mL H2O (i.e. 27g) and add that and mix. Then you would have 12% NH4OH. But you'll have a bit more than 50mL.
(*The reason all this pedantry is sometimes necessary is that the final density of the solution is not predictable from the densities of the initial solutions. Sigh.)
There are also a lot of random unit conventions, some of which are also regional (for example, in Australia conc ammonia is sold as 25% (w/w) NH3), or specific to a type of lab. Some are wrong and/or misleading, but are generally coming from a situation where close enough is good enough and correct is annoying. E.g. some chemists quite happily use ppm when they actually mean mg/L, which can be misleading to the uninitiated (it's easier to say though). And if a physicist or biologist wrote the method, all bets are off as to what the units mean.
posted by kjs4 at 5:45 AM on May 19, 2016
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posted by supercres at 4:22 PM on May 17, 2016