degmo
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Doubling Molarity
Hi,
If I have a reaction mixture with 10ml 1.0M HCL and 10ml 4.0M acetone, How could I double the concentration of acetone keeping the total volume at
20ml and same concentration of HCL?
Thanks.
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BromicAcid
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Usually
molarity initial X volume inital = molarity final X volume final
So if you want to double the concentration of acetone and keep the volume the same and the concentration of acid the same you're going to want to
alter the molarity of the acetone solution you're initially adding. Doubling it, so you need to use the calculation to alter the acetone
solution that you're going to feed into it. To make an 8 Molar solution you would take 8 mol of acetone (368 ml) and add enough water to make 1 L
(632 ml). Then you can take those numbers and divide them down by the same numbers to lower the quantity you're making, as long as you keep the
ratio the same.
P.S. The HCl will go up in concentration now that it is only being dilluted with 8M acetone/water and not 4M but then again you might not be using
acetone and water and these calculations would be wrong. Also, this should be in the beginings section.
[Edited on 2/24/2004 by BromicAcid]
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Mumbles
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You must be shitting me, or in my chem class. We just got that question on one of our labs yesterday.
You should put down use 10mL of 8 M solution of acetone. I highly doubt that any beginning chem teaching is going to be that strick. I brought this
up to my teacher and he told me to quit looking into it so much.
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Biochem
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C1V1=C2V2 simple as that plug and chug
c1 = concentration of substance 1
v1 = volume of substance 1
c2 = concentration fo substance 2 or concentratin for substance you are trying to get to
v2 = volume of subsstance 2 or volume you are trying to get to
Bromic acid gives detailed explaination but I think this formula is more widely recognized even thought it is essentially one and the same...heh just
working on my "harmless" status hope it helps
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