Poppy
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Volume and concentration calculations
Hi.
I want to play a game. The destino of which will decide you'll be a chemist or not.
I have a sample of HCl lablled as 33%, I want to make it a 20% solution. How much water should I add to a 100mL sample of the acid considering its
density as 1.2g/L?
[Edited on 8-4-2013 by ScienceSquirrel]
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elementcollector1
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This is a basic calculation. C1V1=C2V2
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Vargouille
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True, El, but it doesn't quite work with percentages, because of the differing densities. One way to get around that is to convert from
percentages to molarity: 33% HCl becomes 10.86 M; 20% HCl becomes 6.03 M. Then you can use M1V1=M2V2, so (10.86 M)(100 mL)/(6.03 M) = 180.10 mL.
That's the final volume of the diluted solution, 180.10 mL - 100 mL = 80.1 mL. If I was working with sig figs, I would say "80. mL".
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elementcollector1
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Wouldn't you say 80.1mL for 3 SF's? I don't see anything that limits it to 2.
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silvergrahm
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You thought the game was over. But it has just begun...
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Vargouille
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All of the given values (33%, 20%, 1.2 g/mL) only have two sig figs. I just keep units to two degrees after the decimal for calculations, then round
it for an answer.
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elementcollector1
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My mistake. See, this is why I will never get a 100% on a chem test - it's those darn SF's!
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Poppy
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I use the a rule which applies on picking a sample
33/100 = X/ M where X stands for the mass of HCl and M the mass of the sample
100mL water weighs 100g
So lets put 33/100 = X / 100 wherein X comes to be 33g..
Now 33 / 100 + M = 20/100
Easy as that xD?
[Edited on 4-7-2013 by Poppy]
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chemrox
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Poppy, ahem, you're not supposed to bring your home work here..try mixing a little ephedra with your poppy paste ;^)
"When you let the dumbasses vote you end up with populism followed by autocracy and getting back is a bitch." Plato (sort of)
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DutchChemistryBox
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The density of the acid solutions:
ρ(20°C 33%)=1.1642g/ml
ρ(20°C 20%)=1.0980g/ml
m(HCl 33% total)= ρ(20°C 33%) * V(HCl 33%) = 1.1642g/ml * 1000ml = 1164.2g
m(HCl 20% total)= ρ(20°C 20%) * V(HCl 33%) = 1.0980g/ml * 1000ml = 1098.0g
So 33% of 1164.2gram = 384gram
And 20% of 1098.0gram = 219.6gram
n(HCl ) =m(HCl)/M(HCl)
n(HCl 33%) =384g / 36.46g/mol = 10.532mol
n(HCl 20%) =219.6g / 36.46g/mol = 6.023mol
C(HCl) = n(HCl) / V(solution)
C(HCl 33%) = 10.532mol / 1000ml = 10.532M
C(HCl 20%) = 6.023mol / 1000ml = 6.023M
C(HCl 33%) * V (HCl 33%) = C(HCl 20%) * V (HCl 20%)
10.532M * 100ml = 6.023M * V (HCl 20%)
V (HCl 20%) = (10.532M * 100ml) / 6.023M = 174.9ml
V (to add) = V (HCl 20%) - V (HCl 33%) = 174.9ml – 100ml = 74.9ml = 75ml
So I would say 75ml
Edit because of a stupid fault
[Edited on 7-4-2013 by DutchChemistryBox]
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Vargouille
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You must use density because 33% means 33 g HCl per 100 g solution, and the density of the solution is different from the density of pure water. As
much as I dislike the Courtier's Reply, it may be necessary in this case. Try starting the calculations over with that in mind.
EDIT: I'm not sure what point you had in attaching a picture of two MSDSes, but I think the point is obvious from them nevertheless: the concentration
of HCl is in proportion to that of the entire solution, as with that of water.
[Edited on 7-4-2013 by Vargouille]
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Poppy
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Quote: Originally posted by chemrox | Poppy, ahem, you're not supposed to bring your home work here..try mixing a little ephedra with your poppy paste ;^) |
Hahaha
Okay.
But we are not getting a consesus here
X / (X+M) = 33/100
100mL x 1.2g/L = 120g
X + M = 120g --> X = 120 - M
X / (120 - M + M) = 33/100
X/120 = 33/100
X = 39.6
Otherwise: being the mass of the sample 120g:
being M the additional mass of water to achieve a supposed concentration:
X/ (120 + M) = 20/100
X/120 = 33/100 --> X = 39.6
39.6/ (120 + M) = 20/100
3960 = 2400 + 20M
20M = 1560
M = 78
Things are diverging here. Hows that one may come with a simple explanation?
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DutchChemistryBox
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Quote: Originally posted by Vargouille | You must use density because 33% means 33 g HCl per 100 g solution, and the density of the solution is different from the density of pure water. As
much as I dislike the Courtier's Reply, it may be necessary in this case. Try starting the calculations over with that in mind.
EDIT: I'm not sure what point you had in attaching a picture of two MSDSes, but I think the point is obvious from them nevertheless: the concentration
of HCl is in proportion to that of the entire solution, as with that of water.
[Edited on 7-4-2013 by Vargouille] |
You're right! Thankyou, I edited it.
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Vargouille
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I think the differences between all of the answers are due to the varying methods used and the original numbers used. Dutch, the disagreement between
our answers is because I use 1.2 g/mL as the density of 33% HCl, and 1.1 g/mL as the density of 20% HCl, while you use more accurate figures.
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Poppy
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Prodigious, neared comrades, prodigious.
This typical impplication on the additional masses to be put in place transcended the factual uncertainty behind chemical prognostics, as clearance to
be made exept in field and terms.
Which, from the adapted bias, matches accurately both standards and actual experience regardless the shadow of doubt?
Many as yet radicals noted their ponderations and hindered stupefaction and so are many yet to be unseen.
As for the concentration issue, it has not been uncommon to point w/V ratios as well as V/V which is very inconclusive
as per Wiki's (and everybody flerts about it) disposes solubility in terms of 100g water, not on 100mL solution.
Ideas?
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Magpie
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This seems like such a simple problem that I want to play also:
100mL @1.2g/mL = 120g
(120g)(0.33) = (0.20)(W + 120g)
W +120 = (120)(0.33)/0.20 = 198
W = 198 - 120 = 78g of water
The single most important condition for a successful synthesis is good mixing - Nicodem
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Vargouille
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Poppy, what are you even...?
There's a difference between sounding erudite and sounding like you opened up a thesaurus.
In any case, V/V is not used for solutions of gases. You may be able to guess why. Moreover, wt/vol is used less often as well, and the giving of a
density is an indication that the concentration is indeed wt/wt. The comment on solubility is a trivial one; pointing out that solubility is measured
in a different way from concentration is nigh meaningless when the topic at hand is the concentration of an unsaturated solution. The answers have
been given, and the discrepancies explained. If this is indeed a game, it is surely a poor one.
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Endimion17
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So, a guy has some homework to do and he makes a "I want to play a game" thread. Cool.
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Poppy
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So what have we learn't today?
Watch out, when possibly teaching at college, that C.V rule originally comprises more of a term than simply a numerical substitution.
Magpie wins.
Thank you guys, the game is over.
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elementcollector1
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But I have only begun to play!
(And I haven't gotten all the achievements yet...)
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