itchyfruit
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standard solutions/vol/mol
I'm confused, yes again!!
Is their a thread about making solutions/dilute acids/bases?
If not can someone explain this, if wanted to make say a 10% solution of cuso4 would i have to heat the cuso4 to remove the h2o weigh 10g of the
anhydrous crystals and then add that to 90ml/g of h20?
also how does this work when diluting say 37% hcl or 98% h2so4 is this worked down from a 100% theoretical concentrate eg if i had say a 50% h2so4
would i have to put 20ml of that to 90ml of h2o to obtain a 10% solution?
% volume and mol whats the difference?
sorry to ask so many ? in one hit but the complicated bit for me is the different forms of measurement !!
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sparkgap
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"can someone explain this, if wanted to make say a 10% solution of cuso4 would i have to heat the cuso4 to remove the h2o weigh 10g of the anhydrous
crystals and then add that to 90ml/g of h20?"
You don't have to heat just to simplify the arithmetic.  Figure out how much of
the pentahydrate is equivalent to ten grams of the anhydrous salt, and add water as needed (subtract the amount of water from the pentahydrate from
the total amount of water you were supposed to add to the anhydrous salt).
Also, there's the formula C(a)V(a)==C(b)V(b) where C(a) and V(a) are the concentration and volume of solution a and similarly for solution b. It's
very useful. 
sparky (~_~)
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Saerynide
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% vol usually means (vol of solute)/(vol of solution)*100%. So a 10% vol ethanol means in 100 ml, 10 ml is EtOH. That, however, does not necessarily
mean there is 90 ml of H2O, though, as volumes are not always additive due to molecular interactions (like EtOH and H2O decrease in volume when
mixed). So to make a 10% ethanol solution, you would pour out 10 ml of EtOH, and add water until the total volume is 100 ml.
Molarity is (mols solute)/(L solvent). This is a solute-free basis, meaning it is a measure that is indepent of all the other stuff that might be in
the solution (other than the solvent). Because the denominator is L of solvent and not L solution, we dont have to worry about non-additive
properties like how much space the solute(s) takes up.
So a 1 M NaCl solution has 1 mol of NaCl in 1 L of H2O.
For a 1 M solution of Na+ (aq), we can have:
1 mol NaCl in 1 L H2O, or
0.5 mol NaOH and 0.5 NaCl in 1 L H20, or
1 mol NaOH and 2 mol KCl in 1 L H2O, etc you get the point.
As long as there is 1 mol of Na+ ions in 1 L of H20, that is a 1 M Na+ (aq) solution, regardless of all the other crap in there, like the KCl or the
OH- ions.
Isn't molarity great? 
[Edit]: And then there's molality... look that one up. It's even more useful.
[Edited on 6/26/2009 by Saerynide]
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entropy51
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Here are some links you can check out Itchy. And remember Google is your friend!
http://faculty.ksu.edu.sa/alhomida/Documents/BCH%20231%20Lec...
http://www.ruf.rice.edu/~bioslabs/methods/solutions/formulas...
http://www.sfu.ca/biology/courses/bisc367/handouts/lab01.pdf
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itchyfruit
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I realise you're getting a bit fed up with all my questions so if you could tell me if i have this correct or not.
Having read this thread and the links provided (Thanks again for those) i think i'm about ready to spend a couple of evenings making up my
solutions,my plan is to make 50 10% solutions (i only have 50 bottles) and here's how i intend to do it, but i need to know if i have right or i'd
have cocked up a lot of solutions by Monday
10% copper sulphate :- cu 64 s 32 4x o 64 = 160 5(h2o) 2x h 2 o 16 = 18 x 5 = 90 so according to my calculations copper sulphate pentahydrate is
about 35% h2o so to make a 10% solution i would need to weigh out about 13g and then add h2o to 100ml or 52g then up to 400ml to make 400ml of
solution.
10% hydrochloric acid :- if i use 36% conc hcl i would need to measure 29ml of conc hcl then add h20 to 100ml or 116ml conc then h20 to 400ml to make
400ml of solution.
10% nitric acid :- if i use 70% conc hno3 i would measure 15ml of conc hno3 and then add h20 to 100ml and so on!!
For sulphuric acid at 98% it should be approx 10ml then to 100ml.
For anhydrous solids like potassium chloride/ferrocyanide/permanganate etc it should be 10g solid then add h2o to 100ml.
I've rounded up/down so please don't point out the lack of precision i just need to know if i have got the hang of working it out.
BTW I haven't been working on this for a whole month (just in case you think i'm really slow)
I'm going to post a list of all the solutions i intend to make and would be grateful for any don't bother with that one or you should make this one
stronger/weaker and don't forget this one type suggestions.
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Magpie
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Some general guidelines might help here:
1. Solutions can be made up by weight % or volume %. Know what you want and don't mix them up during the calculations.
2. Solutions can be made up to a desired molarity. Molarity is the number of moles/liter. Don't get this mixed up with the wt% and vol%
calculations either.
Example of wt%: Desired: 100g of 16wt% aqueous NaCl soln. Mix 16g of NaCl with 84g of water.
Example of volume %: Desired: 120 ml of 13 vol% ethanol in water. Mix 15.6 ml of ethanol with water bringing the level up to the 120 ml mark.
Example of molarity: Desired: 0.5 liter of 1.2 molar aqueous solution of NaCl. Wt NaCl = (0.5)(1.2)(23 +35.5) = 35.1g; Mix 35.1g of NaCl with
water bringing the level up to the 0.5 liter mark.
Note: I try my best to avoid dealing with volume %.
The single most important condition for a successful synthesis is good mixing - Nicodem
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itchyfruit
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I think i get what your saying,by weighing the solute and then topping up to 100/400ml i'm mixing weight% and volume %.
So i weigh to solute and add the remainder in solvent in ml rather than adding to 100/400ml this is to eliminate the difference in volume
but did i calculate the pentahydrate and conc/dilute acid proportions correctly ?
I think i'm using mols to calculate the amount of h2o in cuso4 5(h20) or have i got this wrong to.
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ammonium isocyanate
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Honestly for many things I think a molar solution is more useful than a %weight/volume becasue it simplifies calculations and you can't get confused
about the meanings (unless your M's look like m's). For example, today I am mixing up two standardized solutions. The first will be a .5M copper
sulfate solution which I will use for destroying H2S, and the second will be a .1M KOH solution for titrations.
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itchyfruit
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I don't mind which one i use!! I just want to be sure that i have worked it out correctly and know how to convert one to another.
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DJF90
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Ammonium _: Three points. I agree that molar solutions are much more useful. You can always write the % on the label too if you want. I dont see how
you intend on using copper sulfate to "destroy" H2S; at best I see it just forming a (probably insoluble) sulfide. Perhaps theres a redox reaction
there that I am unaware of. I'd also like to add that you should probably use a saturated solution to deal with something as nasty as H2S - you dont
want to be "caught short" if you know what I mean. And KOH isn't a great standard, being that it forms carbonate with carbon dioxide in the air. A
better solution would be one based on a salt with no/known water of crystallisation and no reactivity which will change the concentration in the
standard solution. Perhaps NaHCO3 would be a better candidate.
itchyfruit: I second ammonium's suggestion of using molar concentrations. Working by weight or volume is very variable: the concentration of the
solution in chemical terms will depend on the density/molar mass of the reagent. Therefore a 10wt% NaOH solution will be stronger than 10wt% solution
of KOH, and alot stronger than a 10wt% solution of H2SO4. (RMM's are 40, 56, and 98 respectively). But a 1M solution of each would have the same
concentration (1 mole of reagent per liter).
[Edited on 24-7-2009 by DJF90]
itchyfruit: I advocate the use of 1M or 2M solutions. There is the formula "Moles = conc. x vol.", where the concentration is in mol dm^-3 and the
volume is in dm^3 (i.e. liters). If you want to make a liter of 1M copper sulfate solution, you can dissolve 1 mole of copper sulfate (anhydrous, RMM
= 64+32+(4x16) = 160g/mole, OR pentahydrate, RMM = 64+32+(4x16)+(5x18)[this is 5 moles of water] = 250g/mole) in 1L.
To recap: If you want to make 1L of 1M copper sulfate solution, you can either dissolve 160g of anhydrous copper sulfate, or 250g of copper sulfate
pentahydrate, in 1 liter of water. Using equation above, you can figure out that if you half the volume of water, the concentration of the solution
would double. So 250g of CuSO4.5H2O in 500mls would give a 2M copper sulfate solution. You can also use the equation to work out how many moles of
substance X you need to make a solution of Y volume and Z concentration. (e.g. if you want 250ml of 4M solution of substance X, you need 4 mol dm^-3 x
0.25dm^3 = 1 mol of substance X.
[Edited on 25-7-2009 by DJF90]
[Edited on 25-7-2009 by DJF90]
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itchyfruit
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So in simple terms (0.5)(1.2)(23+35.5) is 0.5x1.2x58.5 = 35.1
My intention is to work through the experiments in Woolens site and they seem to be in weight % so if i use the molar system to make them how do i
convert them to weight % and visa versa ?
I know i'm missing something obvious here !!
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ammonium isocyanate
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DJF90,
Copper(II) sulfate reacts with hydrogen sulfide to precipitate highly insoluble copper(II) sulfide, which is what I meant by "destroy" (I should have
said detoxify or something like that), as copper(II) sulfide is nontoxic. I plan on using a huge volume of .5M solution to treat the hydrogen sulfide
in order to make sure the gas has time to be absorbed and reacted, given the low solubility of H2S in water (but maybe I'm totally off on this). I
might take you up on your suggestion of using sodium bicarb- as I guess it would be mostly converted to trisodium phosphorus.
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Magpie
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| Quote: |
My intention is to work through the experiments in Woolens site and they seem to be in weight %
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This is exactly the point in knowing how to make solutions by any of the common methods I showed. Although molarity is indeed usually more useful for
chemistry, directions will sometimes specify solutions in terms of wt%. Therefore, you need to learn how to use molarity and wt%. Vol%, not so much.
The single most important condition for a successful synthesis is good mixing - Nicodem
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itchyfruit
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I think i have it
1mol/l of cuso4 would have 160g of solute in 1000ml of solution and be a 16% solution.
1mol/l of feso4 would have 152g of solute in 1000ml of solution and be a 15.2% solution.
1mol/l of nacl would have 58.5g of solute in 1000ml of solution and be a 5.85% solution.
1mol/l of lico3 would have 67g of solute in 1000ml of solution and be a 6.7% solution.
And so on.
This has made acids a bit confusing ie h2so4 should be 100 but i've never come across anything over 98,hno3 should be 63 but mine is 70 and fuming is
99, hcl seems to work out ok.
But i think i'll ponder that one tomorrow!!
Thanks for the help chaps
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Magpie
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| Quote: |
1mol/l of cuso4 would have 160g of solute in 1000ml of solution and be a 16% solution.
1mol/l of feso4 would have 152g of solute in 1000ml of solution and be a 15.2% solution.
1mol/l of nacl would have 58.5g of solute in 1000ml of solution and be a 5.85% solution.
1mol/l of lico3 would have 67g of solute in 1000ml of solution and be a 6.7% solution
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Your molarity calculations are correct but not the % calculations. The weight of a liter of each of those solutions will be greater than 1000g.
To convert from molarity to wt% you must know the density of the solution. This is found in handbooks like CRC.
The single most important condition for a successful synthesis is good mixing - Nicodem
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itchyfruit
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I see what you mean, i'm assuming that 160g is 160ml but thats not the case so i know the solvent(water) is 1g to 1ml and i can work out the volume of
the solute via the density of the individual elements(i have all these written on one of my periodic tables) this should give me the density? that
don't seem right i'm of to look up CRC
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itchyfruit
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I have a cunning plan(i don't have a CRC hand book and can't find a copy on line) i'm going to make my 400ml solution(the cuso4 is currently
dissolving) then measure out 100ml weigh it and as i know their is 16g of solute once i have the total weight i can work out the weight percentage.
I love it when a plan comes together 
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ammonium isocyanate
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Another important thing to remember is that volumes don't add linearly. For example, If you place crystals of copper sulfate pentahydrate in a beaker
and add water until the volume reaches x, the volume will actually be greater than x once all the crystals have dissolved.
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itchyfruit
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I didn't know that, but added water to 350ml and then to 400ml just to be sure
What do you think of my system ?
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Magpie
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Your system is good. In fact it is better than using a handbook as handbooks don't have tables for all salts.
There's an even easier way:
When you dilute your weighed salt to the mark for a certain molarity, supply the water from a known amount in a graduated cylinder. The water left in
the cylinder tells you what you added to the salt. Since 1 ml = 1g, no further weighing is required.
The single most important condition for a successful synthesis is good mixing - Nicodem
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itchyfruit
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Cunning!!
I tried to make a 1mol/l of potassium dichromate,i don't think it's going to work
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itchyfruit
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Same problem with potassium permanganate, so i've changed it to a .2 mol/l and i'm still having trouble getting all the solute to dissolve.
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DJF90
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As I seem to remember, the standard solution I used to use for permanganate was 0.05M. And it was still intensely coloured. Dichromate solution IIRC
is made up to 2M in 2M sulfuric acid. (i.e. 2 moles of sulfuric acid and 2 moles of K2Cr2O7 in 1L of water)
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itchyfruit
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Is their a list of useful solutions with recommended concentrations?
I've tried a couple of searches but can't seem to find one,if their isn't perhaps i could start one, and then people that actually know what their
talking about could correct/add to it.
I suppose 2 lists one for inorganic and one for organic !!
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