Sciencemadness Discussion Board

Preparing a w/w% solution

Yttrium2 - 7-2-2015 at 07:33

The formula is grams of solute/grams of solvent x 100

Right?

gdflp - 7-2-2015 at 08:44

Yes, for example a 10% aqueous solution of sodium chloride contains 10g of NaCl and 90g of water per 100g of solution. Try posting these sorts of simple questions in the Short Questions Thread, or Beginnings not Chemistry in General.

Yttrium2 - 7-2-2015 at 08:55



[Edited on 7-2-2015 by Yttrium2]

blogfast25 - 7-2-2015 at 10:03

Quote: Originally posted by gdflp  
Yes, for example a 10% aqueous solution of sodium chloride contains 10g of NaCl and 90g of water per 100g of solution. Try posting these sorts of simple questions in the Short Questions Thread, or Beginnings not Chemistry in General.


So w% is weight of solute/(weight of solute + weight of solvent) x 100 %

10 g NaCl / (10 g NaCl + 90 g water) x 100 % = 10 w%

Edit: 10 % of course, NOT 90 %. Typo, my bad.


[Edited on 7-2-2015 by blogfast25]

gdflp - 7-2-2015 at 10:09

Quote: Originally posted by blogfast25  


So w% is weight of solute/(weight of solute + weight of solvent) x 100 %

10 g NaCl / (10 g NaCl + 90 g water) x 100 % = 90 w%


Wow, 90% is pretty concentrated for only adding 10g of NaCl.;)

Yttrium2 - 7-2-2015 at 10:19

Quote: Originally posted by gdflp  
Quote: Originally posted by blogfast25  


So w% is weight of solute/(weight of solute + weight of solvent) x 100 %

10 g NaCl / (10 g NaCl + 90 g water) x 100 % = 90 w%


Wow, 90% is pretty concentrated for only adding 10g of NaCl.;)


Hey now

Yttrium2 - 7-2-2015 at 10:28

So if one wanted 100mL of 10% NaCl w/w% how would this be done

Texium - 7-2-2015 at 10:34

Quote: Originally posted by Yttrium2  
So if one wanted 100mL of 10% NaCl w/w% how would this be done
I think blogfast must have had a typo when he said 90%. I think he meant 10%. 10g of solute in 90g of solvent is 10w/w%.

blogfast25 - 7-2-2015 at 10:36

Quote: Originally posted by Yttrium2  
So if one wanted 100mL of 10% NaCl w/w% how would this be done


If I tell you that the density of water is 1 g/mL, can you work it out from there?

[Edited on 7-2-2015 by blogfast25]

Yttrium2 - 7-2-2015 at 10:43

No, wouldn't the density of sodium come into play? Since it isn't 1g/ mL

[Edited on 7-2-2015 by Yttrium2]

Magpie - 7-2-2015 at 10:49

Quote: Originally posted by Yttrium2  
So if one wanted 100mL of 10% NaCl w/w% how would this be done


This is the way I would do it:

Let X = grams of NaCl to add to 100g of water to make a 10w% solution:

X/(X+100) = 0.10
X = 0.1X + 10
0.9X = 10
X = 10/0.9 = 11.1g

Mix 11.1g of NaCl with 100mL water. Pour off 100mL. :D


[Edited on 7-2-2015 by Magpie]

blogfast25 - 7-2-2015 at 10:50

Quote: Originally posted by Yttrium2  
No, wouldn't the density of sodium come into play? Since it isn't 1g/ mL

[Edited on 7-2-2015 by Yttrium2]


Strictly speaking yes, but in practice we don't do it that way.

Approx. 100 mL of 10 w% NaCl would mean dissolving 10 g of NaCl in 90g of water. 90 g of water is 90 mL of water. This would give you 100 g of solution. If the density of that solution is close to 1 g/mL, then 100 g of the solution will also be approx. 100 mL of solution. For most intents and purposes practical that is good enough.

If you did need to make exactly 100.0 mL of that solution you would have to take into account the density of pure NaCl.

[Edited on 7-2-2015 by blogfast25]

Yttrium2 - 7-2-2015 at 10:50



[Edited on 7-2-2015 by Yttrium2]

Yttrium2 - 7-2-2015 at 10:55

Quote: Originally posted by Magpie  
Quote: Originally posted by Yttrium2  
So if one wanted 100mL of 10% NaCl w/w% how would this be done


This is the way I would do it:

Let X = grams of NaCl to add to 100g of water to make a 10w% solution:

X/(X+100) = 0.10
X = 0.1X + 10
0.9X = 10
X = 10/0.9 = 11.1g

Mix 11.1g of NaCl with 100mL water. Pour off 100mL. :D


Could you explain each step

[Edited on 7-2-2015 by Magpie]

blogfast25 - 7-2-2015 at 10:57

Quote: Originally posted by Yttrium2  


[Edited on 7-2-2015 by Yttrium2]


Definitely NOT, you're a making a solution of NaCl in water, NOT Na in water.

Magpie - 7-2-2015 at 10:57

Do you understand basic algebra?

blogfast25 - 7-2-2015 at 10:58

Quote: Originally posted by Yttrium2  
Quote: Originally posted by Magpie  
Quote: Originally posted by Yttrium2  
So if one wanted 100mL of 10% NaCl w/w% how would this be done


This is the way I would do it:

Let X = grams of NaCl to add to 100g of water to make a 10w% solution:

X/(X+100) = 0.10
X = 0.1X + 10
0.9X = 10
X = 10/0.9 = 11.1g

Mix 11.1g of NaCl with 100mL water. Pour off 100mL. :D


Could you explain each step

[Edited on 7-2-2015 by Magpie]


Did you mean 'pour off 10 mL' ? :D

blogfast25 - 7-2-2015 at 11:00

Quote: Originally posted by Magpie  
Do you understand basic algebra?


Me? No. I skipped that bit and went straight to Calculus! ;)

Magpie - 7-2-2015 at 11:01

No. He wanted 100 ml. Realizing that my mixed solution will likely be slightly larger than 100 ml, I must "pour off" only 100ml.

Blogfast that comment about algebra was for the OP. not you! :D


[Edited on 7-2-2015 by Magpie]

Magpie - 7-2-2015 at 11:03

Blogfast that comment about algebra was for the OP. not you! :D

We are responding too fast. This is becoming a cluster-fuck. ;)

[Edited on 7-2-2015 by Magpie]

blogfast25 - 7-2-2015 at 11:25

Jokes aside, in some cases preparing mixtures to an exact volume is necessary of course. In the case of mixers with a precise volume for instance.

100.0 mL of 10 w% NaCl, assuming mixing NaCl with water does not result in contraction or expansion:

Take x g of NaCl and y g of water.

w% of NaCl = x / (x + y) 100 = 10 (eq.1)

Total volume is x/2.165 + y/1.000 = 100 => x + 2.165y = 216.5 (eq.2)

eq.1 => x = 0.1x + 0.1y => 0.9x = 0.1y => x= y/9

Insert into eq.2 => y/9 + 2.165y = 216.5 => 2.28y = 216.5 => y = 95.00 g water

x = 95.00/9 = 10.56 g NaCl



[Edited on 7-2-2015 by blogfast25]

DistractionGrating - 7-2-2015 at 12:07

Quote: Originally posted by blogfast25  
...assuming mixing NaCl with water does not result in contraction or expansion....


Well, that's a rather large and quite false assumption. http://bit.ly/1vxPnre

Yttrium2 - 7-2-2015 at 12:52

I understand basic algebra, though I do use the book for referencing some things that escape my memory.

just not getting the math! darnit!

will this type of math be included in general chemistry 2?

[Edited on 7-2-2015 by Yttrium2]

blogfast25 - 7-2-2015 at 12:53

Quote: Originally posted by DistractionGrating  
Quote: Originally posted by blogfast25  
...assuming mixing NaCl with water does not result in contraction or expansion....


Well, that's a rather large and quite false assumption. http://bit.ly/1vxPnre


Not sure how you reach that conclusion.

The link gives density of 1.1972 at 26 w% NaCl (solubility limit IIRW).

26/2.165 + 74 = 86 ml. 100 g/86 ml = 1.163 g/ml calculated, based on no contraction/expansion.


With the listed density of 1.1972, that's a discrepancy of 2.85 % with respect to 1.163. Hardly a "rather large and quite false assumption".

Care to explain?


[Edited on 7-2-2015 by blogfast25]

Magpie - 7-2-2015 at 12:59

Quote: Originally posted by Yttrium2  

will this type of math be included in general chemistry 2?


Yes. It should have been in general chemistry 1.

[Edited on 7-2-2015 by Magpie]

DistractionGrating - 7-2-2015 at 14:15

blogfast25: 1.1927 <> 1.163. You have proven that contraction has occurred. The correct value for the number of grams of NaCl for 100 mL of 10% solution at 20C is 10.71, not 10.53. Your answer would create a 9.845% solution.

Etaoin Shrdlu - 7-2-2015 at 14:58

Quote: Originally posted by Magpie  
Quote: Originally posted by Yttrium2  

will this type of math be included in general chemistry 2?


Yes. It should have been in general chemistry 1.

[Edited on 7-2-2015 by Magpie]

It should have been in high school.

DistractionGrating - 7-2-2015 at 15:22

Quote: Originally posted by Yttrium2  
So if one wanted 100mL of 10% NaCl w/w% how would this be done


One approach: Dissolve 11g NaCl in 99g H2O. The resulting solution will be a little more than 100mL. Transfer 100mL into a 100mL volumetric flask. Works as long as both the NaCl and the H2O are at ambient temperature.

If you want to create exactly 100mL of the solution, and need to know how much NaCl you will need, then you must first specify the temperature, and then determine the density of the solution at the target concentration and temperature. Assuming 20C, then the link I provided above should lead you to tables that show that a 10% w/w solution of NaCl at 20C has a density of 1.071. 100mL of a solution with this density will weigh 107.1g. 10% of 107.1g = 10.71g. Make sure you do this with everything at 20C.

Magpie - 7-2-2015 at 16:04

Quote: Originally posted by DistractionGrating  
Quote: Originally posted by Yttrium2  
So if one wanted 100mL of 10% NaCl w/w% how would this be done

One approach: Dissolve 11g NaCl in 99g H2O. The resulting solution will be a little more than 100mL. Transfer 100mL into a 100mL volumetric flask. Works as long as both the NaCl and the H2O are at ambient temperature.


This is the same way I proposed upthread. You have just done it more elegantly, disposing of any need for algebra.

This method eliminates any need to worry about volume changes with mixing, SpGs, temperature, etc. I use this method regularly when doing syntheses.

blogfast25 - 7-2-2015 at 18:11

Quote: Originally posted by DistractionGrating  
blogfast25: 1.1927 <> 1.163. You have proven that contraction has occurred. The correct value for the number of grams of NaCl for 100 mL of 10% solution at 20C is 10.71, not 10.53. Your answer would create a 9.845% solution.


While your point is correct it's also bit literalist: how often would one want to prepare a nominal 10 % and how often a 10.0 % solution? So for most purposes it's not a "rather large and quite false assumption".

I like your idea of using a VF but in most cases VFs would be used to create solutions of precise molarity rather than precise w%.

Using the densities as listed it's of course also possible to algebraically calculate the precise amounts for any really accurate w%.

DistractionGrating - 7-2-2015 at 18:49

Magpie: I agree.

blogfast25: The use of the VF was just to meet the 100mL requirement. I'll retract the words "rather large and".

Fulmen - 8-2-2015 at 01:12

Using densities is possible, but it does add extra margins of error. If you check tables of physical properties you shouldn't be surprised to find minute variations, and depending on the required accuracy it can make a difference. As a general rule weights should be weighed if accuracy is required. DistractionGratings approach is the best IMHO.