Sciencemadness Discussion Board - standard solutions/vol/mol

standard solutions/vol/mol

itchyfruit - 26-6-2009 at 08:19

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 !!

sparkgap - 26-6-2009 at 08:24

"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 (~_~)

Saerynide - 26-6-2009 at 10:29

% 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]

entropy51 - 26-6-2009 at 11:46

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

itchyfruit - 24-7-2009 at 12:18

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.

Magpie - 24-7-2009 at 12:59

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 %.

itchyfruit - 24-7-2009 at 14:28

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.

ammonium isocyanate - 24-7-2009 at 15:22

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.

itchyfruit - 24-7-2009 at 15:49

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.

DJF90 - 24-7-2009 at 15:49

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]

itchyfruit - 24-7-2009 at 16:33

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 !!

ammonium isocyanate - 24-7-2009 at 17:22

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.

Magpie - 24-7-2009 at 18:21

Quote:

My intention is to work through the experiments in Woolens site and they seem to be in weight %

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.

itchyfruit - 24-7-2009 at 18:41

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

Magpie - 24-7-2009 at 22:36

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

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.

itchyfruit - 25-7-2009 at 04:10

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

itchyfruit - 25-7-2009 at 07:41

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

ammonium isocyanate - 25-7-2009 at 12:55

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.

itchyfruit - 25-7-2009 at 15:51

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 ?

Magpie - 25-7-2009 at 17:27

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.

itchyfruit - 26-7-2009 at 03:00

Cunning!!

I tried to make a 1mol/l of potassium dichromate,i don't think it's going to work

itchyfruit - 26-7-2009 at 11:01

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.

DJF90 - 26-7-2009 at 14:59

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)

itchyfruit - 27-7-2009 at 03:47

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 !!