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I wish a couple of solutions to be calculated, and the calculations explained, please

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Delicieuxz
#1
Mar1-12, 07:04 AM
P: 5
Hi! I've got a beginner question. I'd like to know what quantities are required to make a couple solutions, but I'd also like to understand how the calculations are made.

I have 95-98% H2SO4 and 28-30% NH4OH, and I'm attempting to figure out the values to make 100ml of 5% H2SO4 w/w and also 100ml 10% NH4OH w/w. I previously didn't know what w/w meant, but I looked it up and saw that w/w stipulates that the ratio is to be based on mass.

If this means that I need to calculate based on the H2SO4 and NH4OH molarity's of 98.079 g/mol and 35.04 g/mol respectively, then I really am lost. But then, it might be as basic an issue as ensuring that my H2SO4 makes up 5% of 100ml, and then topping the remainder up with H2O until I hit the 100ml mark.

I'm guessing that my mind is making this out to be more complex that it really is.
Can someone please tell me the amounts of H2SO4, NH4OH, and H2O to use for both solutions, and also maybe crash course me on the method used to calculate solutions based on w/w? Thank you.
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DrDu
#2
Mar1-12, 03:16 PM
Sci Advisor
P: 3,567
I would rather consider using a balance. You also need the densities of the final solutions.
Borek
#3
Mar1-12, 03:44 PM
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As long as you are dealing with %w/w both before and after dilution, you don't need molarity. But you need masses.

Consider adding 35g of water to 20g of 95%w/w sulfuric acid. Final solution will have a mass of 55g (sum of masses), and it will contain 95%/100%*20g = 19g of sulfuric acid. Thus its concentration will be 19g/55g*100%=34.5%.

You can't do the same with volumes, as volumes are not additive. Mixing 20mL of the mentioned acid solution with 35mL of water yields 51.8mL of solution (checked with the - it has built in density tables, so it can convert volumes to masses, do calculations using masses, then convert them back to volumes).

Delicieuxz
#4
Mar1-12, 08:54 PM
P: 5
I wish a couple of solutions to be calculated, and the calculations explained, please

Ok, thanks. Is it standard that the concentrate solution, which is to be diluted, will be already calculated in w/w? It's 500ml ~18M.
If I take 5g of the concentrated HS2O4, count 0.175ml of it as being H2O content (averaging 95-98% as 96.5%), then add 91.5ml H2O, will I then have a ~5% HS2O4 w/w solution?
DrDu
#5
Mar2-12, 01:43 AM
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P: 3,567
To get 100g of a 5% solution (which is not exactly 100 ml) take 5/0.965 g=5.18 g of 96,5% H2SO4 and 100g-5,18g of water. Not so complicated.Mind to pour the acid into the water, not the other way round.
DrDu
#6
Mar2-12, 01:48 AM
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P: 3,567
Quote Quote by Delicieuxz View Post
Ok, thanks. Is it standard that the concentrate solution, which is to be diluted, will be already calculated in w/w? It's 500ml ~18M.
If I take 5g of the concentrated HS2O4, count 0.175ml of it as being H2O content (averaging 95-98% as 96.5%), then add 91.5ml H2O, will I then have a ~5% HS2O4 w/w solution?
Will work, but gives you less than 100g of solution.
Delicieuxz
#7
Mar2-12, 05:14 PM
P: 5
Quote Quote by DrDu View Post
To get 100g of a 5% solution (which is not exactly 100 ml) take 5/0.965 g=5.18 g of 96,5% H2SO4 and 100g-5,18g of water. Not so complicated.Mind to pour the acid into the water, not the other way round.
OK I get that now, it's pretty straight forward. Thanks.


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