Calculating Solutions w/w: Beginner Question

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Discussion Overview

The discussion revolves around calculating the quantities needed to prepare specific weight/weight (w/w) solutions of sulfuric acid (H2SO4) and ammonium hydroxide (NH4OH). Participants explore the methods for determining the necessary amounts of each substance and the role of density in these calculations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks clarification on the meaning of w/w and how to calculate the required amounts of H2SO4 and NH4OH for their desired solutions.
  • Another participant suggests using a balance and emphasizes the importance of knowing the densities of the final solutions.
  • It is noted that when working with %w/w, molarity is not necessary, but mass calculations are required.
  • A participant provides an example calculation for creating a solution, demonstrating how to determine the mass of sulfuric acid needed for a specific concentration.
  • There is a question about whether concentrated solutions are typically expressed in w/w, with a participant attempting to calculate the water content in their concentrated H2SO4 solution.
  • Another participant confirms that the proposed method for diluting sulfuric acid will yield a solution, but it may not result in exactly 100g of total solution.
  • One participant reiterates the calculation for creating a 5% solution, emphasizing the need to add acid to water, not the reverse.

Areas of Agreement / Disagreement

Participants express varying methods and calculations for preparing the solutions, indicating that multiple approaches exist. There is no consensus on a single method or outcome, as some calculations lead to slightly different results regarding total solution mass.

Contextual Notes

Participants mention the importance of density in calculations and the non-additive nature of volumes when mixing solutions, which may affect the final volume of the solution.

Who May Find This Useful

This discussion may be useful for beginners in chemistry or those interested in solution preparation, particularly in understanding weight/weight calculations and the implications of density in solution mixing.

Delicieuxz
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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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I would rather consider using a balance. You also need the densities of the final solutions.
 


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).
 


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?
 
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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.
 


Delicieuxz said:
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.
 


DrDu said:
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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