30,000 gals of a 25% solution of NaOH

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Discussion Overview

The discussion revolves around the calculation of the amount of dry product needed to create 30,000 gallons of a 25% NaOH solution. Participants explore different interpretations of what "25%" means in this context, as well as the implications of using different measurement standards and methods for preparing the solution.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes a calculation for the amount of dry NaOH needed based on a 25% solution, suggesting a figure of 62,625 pounds.
  • Another participant questions the meaning of "25%" and suggests it could refer to different concentration definitions (v/v, w/w, w/v, etc.).
  • A participant clarifies that their specification indicates "% by weight," which they interpret as w/w, and raises the question of whether the calculations differ between w/v and w/w, suggesting 83,500 pounds for w/w.
  • Concerns are raised about the accuracy of using solid NaOH due to its tendency to absorb moisture and CO2, and the exothermic reaction when mixed with water.
  • One participant emphasizes the importance of using the density of the solution rather than just water for accurate calculations, providing a link to a calculator for density.
  • Another participant notes that the density of a 25% NaOH solution is about 10.6 pounds per gallon, but questions the validity of this figure, suggesting it seems low.
  • A clarification is made regarding the use of US gallons and the conversion factors involved, highlighting the importance of understanding the units used in calculations.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the concentration percentage and the appropriate method for calculating the required amounts. There is no consensus on the correct approach or the figures involved, indicating an unresolved discussion.

Contextual Notes

Participants note the need for clarity on the definitions of concentration types and the importance of using accurate density values for solutions. There are references to potential variations in measurement standards (US vs. UK gallons) that could affect calculations.

obrien040362
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If I wanted to make 30,000 gals of a 25% solution of NaOH would the calculation of dry product needed be:

8.35 #/gal x 30,000 gal x .25 = 62,625 #

Thanks
 
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Depends what 25% means? v/v, w/w, w/v, v/w? by moles, mass, volume, etc..
 


I guess i thought it was w/v since it is a dry solute, but my spec says "% by weight".

So would "% by weight" be w/w?

So is w/v 62,625 and w/w 83,500 (33% more)
 
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obrien040362 said:
I guess i thought it was w/v since it is a dry solute, but my spec says "% by weight".

So would "% by weight" be w/w?

So is w/v 62,625 and w/w 83,500 (33% more)

I'm not sure where you are getting the "33% more" from. You may also need the identity of the solvent??
 
If you need 30,000 gallons of 25wt% NaOH you will have to use the density of the solution and not just water. This link has a calculator:
http://www.handymath.com/cgi-bin/spcfgrv.cgi
and from this we can calculate that 25% NaOH aq is about 10.6 lbs per gal. (temp unspecified but usually at 4C)

Use mass balance in your calculations and you should be able to get the mass of NaOH, H2O to make that mass of 25% aq NaOH
 
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Just a note: in industry one wouldn't try to make this solution from solid NaOH since it rapidly absorbs moisture and CO2 from the air and is hard to get exact weights. Also when it is dumped into water it reacts very exothermically and can superheat if not vigorously stirred and cooled during the additions. Instead one purchases the syrupy 50wt% aq NaOH and dilutes it by weight into the water with vigorous stirring and some cooling. This will give a solution that will not be cloudy from ppt of Na2CO3, and an aliquot can be titrated to give an exact concentration for use in calculations for its use as a reagent or reactant. The calculator gives the density of 50% aq NaOH as about 12.7 lbs per gal BTW...
 
MrSid said:
If you need 30,000 gallons of 25wt% NaOH you will have to use the density of the solution and not just water. This link has a calculator:
http://www.handymath.com/cgi-bin/spcfgrv.cgi
and from this we can calculate that 25% NaOH aq is about 10.6 lbs per gal. (temp unspecified but usually at 4C)

Use mass balance in your calculations and you should be able to get the mass of NaOH, H2O to make that mass of 25% aq NaOH

That figure sounds very low, especially as a gallon of water weighs 10 pounds itself. Maybe we need to figure out what gallon we're talking about as well.
 
I am talking about US gallons (1 gal =3.785 liters)- The beginning equation of Obrien's gave a clue that he was talking US gallons as pure water (spG 1.00, density at 4C 1.000 Kg/Liter) is about 8.33-8.35 pounds (#'s) per US gallon. The calculator in the link actually gives the density or spG in Kg/ Liter, so a unit transformation is in order anyway.
(10 #'s per gallon sounds like the UK or imperial gallon). The point is however that solutions are calculated from units solute per units solution; so that they can be used in making simple mass calculations for the solute as the reagent.
It is useful to evaluate the whole process using mass balances and stoichiometry, to avoid over or under purchases of reactants, and then leave room for Murphy ;-)
 

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