Helix Posted June 20, 2006 Posted June 20, 2006 Hey everyone. I'm starting an internship in a few weeks and I'm a little confused by one aspect of what I'll be doing. I don't know how to prepare (stock) solutions, such as 28% NaCl. (Actually I'm not even sure that's the correct way to note that.) Can anyone tell me how to prepare solutions or direct me to a website? Thanks. Helix
Skye Posted June 20, 2006 Posted June 20, 2006 Yes, when you write 28% NaCl you would normally need to specify whether it is w/w, w/v or v/v, where w=weight and v=volume. The first is solute (NaCl), the second is the total solution. So the first (w/w) would be 28 grams of NaCl in a total of 100 grams of solution. This isn't used very much. What is used alot is w/v, so you would put 28 grams of NaCl in 100 mL of solution. If you don't see a specification you generally assume it's w/v. When you make these up weigh out the solute first and add it to the vessel (eg. volumetric flask), then fill the vessel to 100 mL with water. Of course it doesn't have to 100 mL, I just used that for ease.
Helix Posted June 20, 2006 Author Posted June 20, 2006 Thanks so much, Skye. I understand now, which is a huge help.
scicop Posted June 20, 2006 Posted June 20, 2006 Um, just to clarify. Usually you wouldn't really see solutions calling for 28% sodium choride, rather they'll give the millimolar concentrations. That is function of its molecular weight and volume of liquid. Review your chemistry 101 book to figure out molar concentrations. If you see a 28% value and you know you're gonna weigh a power then assume w/v (1% is 1g/100mL). If its a liquid or viscous detergent assume (v/v). When doing dilutions from a stock, and you need to know how much to take, I usally divide the stock concentration by the desired concentration. That gives you a dilution factor. Take the final volume you want and devide that by your dilution factor and viola! you get the amount you need to take of the stock. So, in the simplest case, if you have a 100mM stock solution (lets say NaCl) and you want to make 10mM in 150mls. You devide 100mM by 10nM and you get the dilution factor of 10! Now take the 150mls divide by that dilution factor (10) and you get 15! That tells you that you need to take 15 mls of that 100mM solution and add it to 85ml of whatever you're diluting into. You have to remember to subtract your the amount you will add from the final volume otherwise you concentration will be wrong (think about it). It also works for antibody (or protein) preparation. If you have 100mg/mL stock solution and you want to make a 50ug/mL in 5mL then, devide 100mg by 0.050mg (REMEMBER..CONVERT to divide same magnitude units, not a biggy if you're using a sci calc) get your dilution factor, divide 5mL by that, and voila!! thats the amount you need to take. You can figure it out. So..lets say you get 100ul (notice the unit) then, you would add that to 4.9mL This is how US scientist do it. Review your chemistry to figure out Molarity. which is by the way mols/L. So for salt, if the MW is 28 (check me on this) then one mol is 28g. Put that into 1 Liter of H20 and you have a 1M solution. so what if you want a 500mM? that's 14g in 1L. or what if I only want 500mL? then thats 7g in 500mL. Ok..good luck and welcome to the world of solution making!! ALOT of this will become intuitive to you and after a while, you want even need to think about it, you'll just do it off the top of your head like that!
ecoli Posted June 20, 2006 Posted June 20, 2006 I agree scicop... I often freeze up when trying to do dilution problems on a test. I have a much easier time in the lab. Perhaps because I can visualize things.
Skye Posted June 20, 2006 Posted June 20, 2006 You'll see 28% NaCl sometimes. Not so much in a molecular biology lab, but alot in analytical chemistry. (It's '28% NaCl' this, '28% NaCl' that, all day long). In any case I presumed he's making a stock from salt crystals and water. If you already had a stock your method would be what you'd want to use though. Na= 23, Cl= 35, so NaCl= 58.
scicop Posted June 20, 2006 Posted June 20, 2006 yeah too lazy to look up the MW, its been a while since I've made NaCl solution! Ok, cool deal on the 28% on analytical. I've worked in industrial analytical chem labs before, though we used the molar method An Ecoli, me too, I really had a hard time when it came to testing, for some reason i'm more hands on and can grasp something better following practical exposure.
CharonY Posted June 21, 2006 Posted June 21, 2006 Well, and if you work with solvents you will find % (v/v) a lot. Even in molecular biology labs However, a good question to keep students occupied is to tell them to make 5 M HCl starting from a 32% HCl solution...
zyncod Posted June 22, 2006 Posted June 22, 2006 You know, the best "solution" is to go ask a postdoc. So what if they think you're an idiot - they will anyway.
CharonY Posted June 22, 2006 Posted June 22, 2006 That's not true. We had a postdoc from the MIT as a guest scientist. She was not able to make buffers. She told us that in her lab all buffers were made by the TAs. So probably asking TAs is a good start...
scicop Posted June 22, 2006 Posted June 22, 2006 You know, the best "solution" is to go ask a postdoc. So what if they think you're an idiot - they will anyway. NO!!!! , that is NOT the best "solution"!!! The best solution is to do calcalate exactly what how your going to make your solution and have it CHECKED/VALIDATED by a person who you TRUST/KNOW/ or experienced, And this is not limited to just buffers, also experiments. Communications folks is one of the best learning tools; and a good way establish a network, even if just starts with "hey, can you double check my calucations for making 0.05% SDS/4% BSA soln in 100mM Tris HCL pH 6.8?? with 10ug/ml of pSTAT3 mab?"
CharonY Posted June 22, 2006 Posted June 22, 2006 However, ideally you should let them check all your calculations together. It can be very disruptive if you go and ask every few minutes. Experiments (unless very simple ones) are ideally checked in groups.
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