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Author: Subject: 18M H2SO4 to 5%
T=HC2
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[*] posted on 31-1-2006 at 13:03
18M H2SO4 to 5%


How do I figure out exactly how much 18M H2SO4 needs to be added to the water in order to make 150 ml of a 5% solution? I have been searching for a calculator to no avail.
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12AX7
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[*] posted on 31-1-2006 at 13:37


Well if you assume density of the resulting solution is close to 1 and you don't mind 5% error, then just add 150 * 0.05 = 7.5 grams of H2SO4 (I forget what 18M is, but that's pretty well 98%, isn't it?).

That's close enough for government work, but your teacher probably wants it exact. Should've payed attention to your textbook when you were going over solutions.

Tim




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T=HC2
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[*] posted on 31-1-2006 at 14:15


When it came it just said "Lab Grade 18M", but I did a search online while looking for info on solutions and it said that 18M (36N) H2SO4 was 96%. So, I just weigh 7.5gm and then add it to the water? What % would that leave me with? I see in the Merck Index that the density of H2SO4 is ~1.84, but I'm still somewhat confused about density, solution by weight, and solution by volume. So, what about adding 10 ml of H2SO4 to 100 ml of water..would that leave me with a 9.6% solution by weight? Then I could add 100 ml more water and have a 4.8% solution? Should I be adding ~18.4 ml to the 100 ml instead of 10 ml given the density of H2SO4? Thanks for the help.
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[*] posted on 31-1-2006 at 14:18
5%


That's about 4 ml 18M H2SO4 to 146 ml H20 assuming the density of the H2SO4 is 1.841.
May want to make it 4.1 to 145.9 if your teacher is a real stickler for precision.
Just be sure to add the acid SLOWLY to the water with constant stirring. Not the other
way around unless you like getting splattered with hot acid. I've watched concentrated H2SO4
pull the water molecules out of table sugar leaving a hot, spongy, steaming carbon mass
behind.

[Edited on 31-1-2006 by MadHatter]




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T=HC2
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[*] posted on 31-1-2006 at 15:32


Thanks! I was searching online and based on what I read I was about to add 30 ml of H2SO4 to 957 ml of H2O, but your answer is much better for me. I was also reading a website that said 1M is 5.3%..this doesn't seem right. I was pouring H2SO4 over some sugar cubes and watching them turn like that..sure is fun. :D
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[*] posted on 31-1-2006 at 19:07
Percent


Actually, by weight 1 M would be closer to 9.4%. 1 mole of H2SO4 is 98.08 grams. This is 53.3 ml,
approximately of course, given a density of 1.841. Mix with 946.7 ml of H2O to yield 1 litre. Total
weight of the solution is 1044.78. Simple division 98.08/1044.78 = 9.38%.




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12AX7
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[*] posted on 1-2-2006 at 07:24


Quote:
Originally posted by T=HC2
but I'm still somewhat confused about density, solution by weight, and solution by volume.


Tell me about it.

Can ANYONE even give me a *reasonable* answer as to why solubility products are in what units they are and not just fricking weight per solute?

Honestly... to measure a solubility constant... you make a concentrated solution, then measure how much dissolved... Fair enough. Then you convert that to molar concentration, THEN to the Ksp value. So that when you have to use the value... you convert BACK to molar, and FINALLY to plain old solubility... W T F?

Tim




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[*] posted on 1-2-2006 at 13:59


"Can ANYONE even give me a *reasonable* answer as to why solubility products are in what units they are and not just fricking weight per solute?"

Yes, but I don't think this is the right thread for it.
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