# Another Dilution Problem... - Molar solutions... (Aug/27/2008 )

Hi,

I have asked similar questions related to dilution a million times but somehow i fail to see the logic behind it...

Well heres my doubt...

My stock concentration = 1mM (milli molar)

Required concentration = 10pM.

Now I know that I need to do 1:1000 dilution for 1microM.. Then a further 1:1000 dilution for 1nM... and then a 1:1000 dilution for 1pM... How much do I have to dilute the 1nM solution to get a 10pM???

Now I would like to know, if i need a concentration of 400nM, how do i go about doing that??? I mean the logic behind doin such dilutions...

Can you please explain to me in points.. please... I am so stuck in lab and takes ages for me to see the logic, I am a slow learner so please help me out..

I would also like to know that if I have a solution of 1mM concentration, would 1ul of it is 1mM and 30ul also have the same concentration???

Another thing is, how do I calculate the concentration having the final bath volume in mind?? For example, I have a dish with 100ul of Media and I need to add a drug of concentration 10pM.... If i make up a solution of 10pM and add 10ul to this then I would be diluting my drug.. So how do I calculate so that the drug I add is 10pM to the final volume?? Thanks a lot...

Also can u please show me, what 1mM is inmicromolar, nanomolar and picomolar..

And also, show me wat 0.5mM is in micro molar, nanomolar and picomolar...

If possible if anybody has the address of website or material that teaches me this basic calculations can u please forward it to me pleeeease. thanks a lot.... This would mean a lot to me...

You need to learn how to use the C1 X V1 = C2 X V2 equation. This says that the concentration of your stock (C1) multiplied by the volume of that stock (V1) will equal the concentration of your final product (C2) multiplied by the volume of the final product (V2). So, you have a stock of 1mM and want 10pM. This is a HUGE difference in concentrations and you would have to make multiple dilutions in order to accurately measure the volumes you need. So, let's turn your 1mM stock into a more dilute 1uM (this is how you write microMolar) stock. You said you know that this takes a 1:1000 dilution. How much of the 1uM stock do you need to make a 10pM working concentration? You first have to decide what volume of 10pM you want to make. Let's say you want to make 2mL of this 10pM solution. C1=your stock solution which is 1uM and V1 is what you are trying to figure out (how much of the 1uM do you need). C2=10pM but the units need to be the same as C1 so 10pM=0.000010uM and V2=2mL (this is whatever volume you are wanting to make). 1uM X V1 = 0.000010uM X 2mL Solve for V1 with simple algebra and you get 0.00002. This is in mL since V2 is in mL which means you need 0.02ul...unmeasurable as I mentioned earlier so...

If you dilute your stock down to 1nM you simply change the C1 (stock concentration) in the equation. C1= 1nM, V1 is what you are solving for. C2 is 10pM but again, the units must match so 10pM=0.010nM and again, say you want to make 2mL. The equation is: 1nM X V1 = 0.010nM X 2mL Solve for V1 and you get 0.02mL or 20ul. Voila! You need to add 20ul of your 1nM stock to 2mL (technically minus 20ul) to make your 10pM solution.

Ok, so now you want to make a 400nM solution. Obviously you can't use your 1nM stock since it is already more dilute than what you want to make so lets go back to the 1uM. C1=1uM, V1 is what you are trying to find, C2=400nM but the units don't match. 400nM=0.4uM V2=whatever volume you are trying to make, lets just say 8mL. The equation is: 1um X V1 = 0.4um X 8mL Solve for V1 and you get 3.2mL. You would need to add 3.2mL of your 1uM stock to 4.8mL of buffer/water/diluent to make the final volume 8mL. Now this is quite a bit and sometimes you don't want to use this much volume because it will throw off the buffer so how much of your 1mM stock would you use to make 8mL of a 400nM solution? Equation: 1mM X V1 = 0.0004mM (same as 400nM) X 8mL Solve for V1 and you get 0.0032mL. This is 3.2ul. See how the volume here is 1000X less than before? This is because your stock solution is 1000X more concentrated (1uM versus 1mM).

Next, if you have a solution of 1mM, any volume of that solution is going to have the same concentration. 1ul has the same concentration as 30ul. Think of it like this: You have milk that has 25% chocolate in it. You pour one cup, it's still 25% chocolate. You pour 5 cups, it's still 25% chocolate. You pour 1ul, it's still 25%. You pour 30ul, it's still 25%. It isn't until you add more milk (or somehow remove just milk..evaporation?) that the concentration changes.

Next, if you have a tissue culture dish with 100ul media and you need to add a drug so that your final concentration is 10pM, you need a stock of this drug that is more concentrated than 10pM. However, if you added 10ul of a 10pM stock into 100ul of media, what is your final concentration that your cells are experiencing? Well, your starting concentration (C1)=10pM. Your starting volume is 10ul as it's the amount of the stock drug you added to the dilution and this is V1. Now you are trying to figure out your final concentration (C2) and your V2 is the 100ul of media plus the 10ul of drug you added (110ul). The equation: 10pm X 10ul = C2 X 110ul. Solve for C2 and you will find that your final concentration is 0.9pM. Very low!

If you want your final concentration to be 10pM then you need to put 10pM in for C2, whatever volume you have your cells in as V2 (100ul here) and you must have a stock of the drug more concentrated than your desired final concentration (C1). Set up the equation and solve for V1. This is the amount of your concentrated stock drug you would need to add to your cells in order for them to be under 10pM concentrated drug. Granted, this is not exact (but very close) unless you actually remove the volume of media from the cells that you are going to add in drug in order to make your final volume 100ul (media+drug).

As for moving from mM, uM, nM, pM...it's just moving the decimal point three places for each step. It's knowing which way to move the decimal that's important.

1mM=1000uM=1000000nM=1000000000pM There are 1000uM in each mM. And 1000nM in each uM. And 1000pM in each nM. You multiply by 1000 each time.

1pM=0.001nM=0.000001uM=0.000000001mM One pM is only 1000th of a nM. And one nM is only 1000th of a uM and one uM is only 1000th of a mM. You divide by 1000 each time.

No matter what the number, you just move the decimal point three places. So a 0.5mM=500uM=500000nM=500000000pM Just multiply by 1000 for each step.

How about 23uM? How much is this in mM? There are 1000uM in one mM and you only have 23 so you know the number needs to get smaller. Move the decimal point three places to the left and you get 0.023mM. What about nM? There are 1000nM in every uM and you have 23uM so the number needs to get bigger. You move the decimal point three places to the right and you get 23000nM. It doesn't matter what the number is. When converting mM, uM, nM, pM, you just need to move the decimal point.

p.s. being able to do these calculations (and knowing how to make specific concentrated solutions from powder reagents) is one of the most critical skills to have in the lab. Without it, you will never be able to be an independent researcher!!! I HIGHLY suggest you get someone around you to help and you should be practicing these calculations. Just make up questions, answer them and have someone check your work.

Example:

you have a 1.3mM stock and need to make 50mL of a 0.4uM solution. How much 1.3mM stock do you need??? I just made this up. You can make these questions up and have someone check..heck, post them here and have us check. What's the answer to this one?

Oh My GOD!!! Thanks a lot.. I really understood that.. I know the formula, but point is I confuse myself a lot!!! About this decimal places etc, i need to keep working on that..

Answer to your ques:

V1 = 0.4uM x 50ml / 1300uM

=15.3ml

So I need to add 15.3ml form the stock to 34.7ml of water (or watever) to get a 0.4uM solution in 50ml...

Is it right????????

Yes, congratulations! I agree with the sentiment expressed by others here -- these are routine calculations you absolutely must become proficient at and comfortable with. The equation rkay447's given you (and the excellent explanation provided as well) seems to have cleared things up a bit for you...

This equation is one I (and likely many others here) use on an almost daily basis in the lab. No one requires that you're able to do such calculations in your head, but you must be able to do them -- I've been at this game for twenty-odd years, and I still write these dilution problems out...

Sorry the answer is wrong... The answer was 0.0153ml, so that is 15.3ul.... i think i am right this time!!!

And thanks a lot rkay..... really i understood it totally... muuah muuaah muuuah.....

And thanks a lot rkay..... really i understood it totally... muuah muuaah muuuah.....

Way to catch your own mistake!! I am SO glad that this was helpful. I posted it not sure if it would help or just confuse. Now..PRACTICE! The only way you are going to know for certain and have confidence in yourself is to practice until you know you can do this. How about calculating how many grams you need in order to make a specific molarity? How many grams of NaCl do you need to make 2L of 2.5M solution? How about 70mL of a 2.5M solution? These two calculations (dilution of stocks and making molar solutions) are the two most critical and most highly used calculations in the lab. Once you have these down you can make any buffer, solution or reagent you need! As HomeBrew said, you don't need to do these in your head. I still have to write it out and from time to time I still check with a labmate to make sure my math is correct, but you do need to know how to set the equations up. Keep up the great work!!!

Here Goes:

F.W of NaCl = 58.442

so, 58.44g in 1litre = 1M solution

therefore for 2l, 58.44g x 2l / 1l = 116.88g (still 1M solution)

For 2.5M solution,

if 116.88g in 2l gives 1M solution, then 2.5M x 116.88g / 1M = 292.2g (so 292.2g in 2l gives 2.5M solution)

To make 70ml of 2.5M solution,

if 292.2g in 2l gives 2.5M,

then,

70ml x 292.2g / 2000ml = 10.22g (so 10.22g in 70ml gives 2.5M solution)

WOOOOHOOOOOOOOOOO!!!!

RKay U r a GRRRRRRRRRRRRREAT TUTOR!!!!!

And thanks a lot rkay..... really i understood it totally... muuah muuaah muuuah.....

Way to catch your own mistake!! I am SO glad that this was helpful. I posted it not sure if it would help or just confuse. Now..PRACTICE! The only way you are going to know for certain and have confidence in yourself is to practice until you know you can do this. How about calculating how many grams you need in order to make a specific molarity? How many grams of NaCl do you need to make 2L of 2.5M solution? How about 70mL of a 2.5M solution? These two calculations (dilution of stocks and making molar solutions) are the two most critical and most highly used calculations in the lab. Once you have these down you can make any buffer, solution or reagent you need! As HomeBrew said, you don't need to do these in your head. I still have to write it out and from time to time I still check with a labmate to make sure my math is correct, but you do need to know how to set the equations up. Keep up the great work!!!

I have got another doubt (oooopsss...)

Stock=1mM ; Required=10pM ; Required in the volume of 10ml (ie i need to dilute the drug into this much ml of media)..

Volume of stock = 10pM x 10000ul / 1000000000pM

= 0.0001ul of stock..

= 0.1ml of stock???

Does this make any sense at all?????

Doesnt that contradict my answer of 0.0001ul??? GAWWDD!!!

THis is where i get stuck and dont know how to go about doing it after i end up like this.....

Now that I have got the knack of calculating things.. I have got a problem with moving the decimal places...

Now for example I was calculating,

2ug/ul x 5ul / 55ul = 0.18ug...

Now does that mean 180mg??? I know it doesnt but just to let u know that I am having problem in interpretiong the results...

Also I want to know a calculation for this..

Stock = 2.8ug/ul

Required = 10ug in a total of 10ml volume...

So do I have to use the c1V1=C2V2???

Cos if I do that I get a total of 50ml of stock!!!!!!!!!!!!!!!! Please help me out!!!! THanks...