prole
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Making solutions from solutions
Hello all, I'm asking for help clearing up a small math problem regarding making dilute solutions from stock solutions.
I have a 34% H2SO4 solution that I'd like to dilute to 5% H2SO4. I have the formula for doing this as V1C1=V2C2. Here's how I wrote it up:
(100mL)(.34)=x(.05). For x I get 680mL. My question is, 'is 680mL my final solution, meaning do I add 580mL H2O to the initial solution to get a
total of 680mL for the second solution (which feels right to me), or do I add 680mL H2O to get a total of 780mL'?
I'm 99% sure that I add the 580 to the 100, but I want to be 100%. Thank you.
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woelen
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Unfortunately things are more difficult. Adding 100 ml to 580 ml of liquid usually does not result in formation of 680 ml of liquid. The resulting
liquid may have a smaller volume (or a larger volume). The addition of volumes only is valid at low concentrations.
A simpler way is the following. Don't work with volumes, but with weights. Take a known weight of your 34% solution. E.g. take 100 gram. This solution
will contain 34 grams of H2SO4. Now if you want a 5% solution, then the total weight must be 34/0.05 = 680 grams. So, you have to add 580 gram of
water, which happens to be 580 ml of water.
So, making 5% H2SO4 can be done by taking 100 gram of solution of 34% (which is quite a lot less than 100 ml of liquid, I think
around 75 ml or so) and adding 580 gram (=580 ml) of water.
[Edited on 4-1-07 by woelen]
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prole
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Thanks, woelen. That simplifies it and makes it more clear to me.
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Vitriolo
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It's more easy: put the 100mL in a graduated cylinder, and rise to 680mL. For this will be necesary more then 580mL, but the final solution will have
the correct concentration
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Levi
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| Quote: | Originally posted by woelen
Unfortunately things are more difficult. Adding 100 ml to 580 ml of liquid usually does not result in formation of 680 ml of liquid.
[Edited on 4-1-07 by woelen] |
Is this phenomena a result of different partial pressures or is it due to the dehydrating nature of
H<sub>2</sub>SO<sub>4</sub>?
Chemcrime does not entail death. Chemcrime is death.
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garage chemist
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It's a result of different molecule size, among other things.
The phenomenon is also observed with Ethanol and water.
Our teacher once demonstrated this to us- 50ml water and 50ml Ethanol are put into a measuring cylinder without mixing, the volume is 100ml. Then it
is stirred to ensure mixing, and the volume shrinks to IIRC 96ml. Quite impressive.
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cadimodo
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After stirring, 4 ml ethanol had been vaporised. 
[Edited on 1-2-2007 by cadimodo]
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unionised
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I'm sure that folk here will form their own opinions.
Was that last, inaccurate, post more useful of less useful than this one?
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BromicAcid
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One of the questions on an exam in P-Chem was written as an essay question where it said that we were a Bar Tender and had to make a solution of
ethanol and water at exactly 100 ml and wanted to know how much of each to get the proper concentration and the exact amount. We joked about it for
most of the year that if chemistry didn't work out for us at least we could mix drinks.
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pantone159
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I second the suggestion of measuring things by weight, not volume.
Volumes don't always combine linearly, they vary with temperature, and you have to use a special piece of glassware (cylinder, pipette, etc) to
measure by volume, which you then need to clean.
Mass changes not a bit under these conditions, and an electronic scale is a lot faster and easier to use than measuring volume.
Historically, chemistry has measured concentration mostly as moles/liter, but it is much easier and I think almost always more accurate to measure by
weight, so I use moles/kg of solution.
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numbcomfort
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| Quote: | Originally posted by woelen
Unfortunately things are more difficult. Adding 100 ml to 580 ml of liquid usually does not result in formation of 680 ml of liquid. The resulting
liquid may have a smaller volume (or a larger volume). The addition of volumes only is valid at low concentrations.
A simpler way is the following. Don't work with volumes, but with weights. Take a known weight of your 34% solution. E.g. take 100 gram. This solution
will contain 34 grams of H2SO4. Now if you want a 5% solution, then the total weight must be 34/0.05 = 680 grams. So, you have to add 580 gram of
water, which happens to be 580 ml of water.
So, making 5% H2SO4 can be done by taking 100 gram of solution of 34% (which is quite a lot less than 100 ml of liquid, I think
around 75 ml or so) and adding 580 gram (=580 ml) of water.
[Edited on 4-1-07 by woelen] |
I'm sorry I think I'm math stupid today. Why did you devide 34 by .05? I see that you want to convert the solution from 34G H2SO4 per 100G to .05G per
100G, so is your thinking like this: (34GH2SO4/100GSolution)(100GSolution/.05G)?
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