30,000 gals of a 25% solution of NaOH

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To prepare 30,000 gallons of a 25% NaOH solution, the calculation for the dry product needed is 83,500 pounds when considering weight/weight (w/w) and 62,625 pounds for weight/volume (w/v). The density of the 25% NaOH solution is approximately 10.6 pounds per gallon, which is crucial for accurate calculations. It is recommended to use a 50% aqueous NaOH solution for dilution instead of solid NaOH due to its hygroscopic nature and exothermic reaction when mixed with water. Proper mass balance and stoichiometry should be applied to ensure accurate reagent quantities.

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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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