# How do I prepare this solution? Dilution? CV=CV?

1. Apr 6, 2009

### x12179x

1. The problem statement, all variables and given/known data

In 50mL flask prepare sol 50 miM in SDS and 30 miM in Na C l"

MW of SDS = 288.38g/mol
MW of NaCl=58.44 g/mol

2. Relevant equations

N/A

3. The attempt at a solution

I think aqueous means I should dissolve it in water. So SDS in water, then NaCl in water...and then combine it to be 50mL? But then how much water do I dissolve in each? How many grams of each?

Do I prepare 25mL of 50mM SDS and then 25mL of 30mM NaCl and combine them?
Does it matter if it's 10mL of 50mM SDS and 40 mL of 30mM NaCl?

Last edited: Apr 7, 2009
2. Apr 6, 2009

### symbolipoint

The simpler approach is to first measure the volume of water which would fit into the flask, approximately - better, even somewhat less than that amount. You should not expect to use all of it. You want to dissolve each of the SDS and NaCl separately in enough water to dissolve ALL of each one. Pour and transfer all of each solution into the volumetric flask and rinse using a squirt water bottle from the first dissolved solutions of SDS and NaCl into the volumetric flask. You should not yet have reached the mark in the volumetric flask; mix gently, and then bring volume up to the mark (with pure water from squirt water bottle, maybe more carefully using a small dropper pipet as you get the volume nearer to the mark). Remember now to stopper the volumetric flask and invert a few times to mix thoroughly.

You can calculate for yourself how many grams of each material to weigh; you know the formula weights of each.

Note, you may be able to use a variation somewhat different than how I described, but in any case, you would not want to add all of the water all at once. Depending on the solubilities of those materials, you might be able to put them into the flask first without water, and then add most of the water and dissolve before adding the rest of the water.

3. Apr 6, 2009

### x12179x

I know to make 50mM of SDS it's 14.419 grams SDS per liter of water,
and for 30mM of NaCl its 1.7532 grams of it in 1L of water.

But does it matter how much of each is in the final solution?

Is it best to make 25mL of each and then combine it?

If I make 10mL of SDS and 40mL of NaCl, does that change anything?

4. Apr 6, 2009

### symbolipoint

Yes, if you mean the final solution in the volumetric flask, and yes if you mean avoiding exceeding the required amount of water, so that you still have some volume of water to use in transferring the materials to the volumetric flask.

No. Use less, but enough so that each material is dissolved; UNLESS you can transfer to the volumetric flask and add the water later. What do you know about the solubility of your "SDS" ?

You want to be sure that when you FINISH the solution in the volumetric flask, you must have 50 milliliters of volume. Think what this means!

Last edited: Apr 6, 2009
5. Apr 6, 2009

### x12179x

So I have a questions about 2 possible cases:

should I do this:

.72 grams of SDS and .0877 of NaCl and fill to 50mL

Is that last one 50mM of SDS and 30mM of NaCl?

or something else?

Last edited: Apr 7, 2009
6. Apr 7, 2009

### Staff: Mentor

Honestly, I have no idea how you can came to TWO different ways of thinking about the solution.

You want it it be 30 mM in NaCl. You want 50 mL of this solution.

How many moles of NaCl in 50 mL of 30 mM solution?

If you start with - say - 20 mL of 30 mM solution, do you have enough NaCl for the solution to be still 30 mM after it is diluted to 50 mL?

It is called mass conservation, you know. One of the most basic concepts in whole chemistry and physics.