• Baumé 2

    Posted by david08848 on April 18, 2020 at 11:22 pm

    I’ll make it brief:

    I’m analyzing another soap formula.  I need to know how much total water is used and how much KOH and NaOH is used so I can determine the ratio of KOH to NaOH that the author is using in this formula as a comparison to other formulas.

    KOH 38 Baume - 36

    NaOH 38 Baume - 19  

    I’m not good at this and have worked on this for a few days but have not succeeded.  I need help and would be grateful for it!  Thank you!

    david08848 replied 3 years, 10 months ago 3 Members · 10 Replies
  • 10 Replies
  • lmosca

    April 19, 2020 at 12:45 am

    Baume degrees are only an expression of specific gravity. You can find tables that relate percent of a certain solute to their Baume degrees. 
    Look for “Baume degrees and X” where X is NaOH or KOH. Or somewhere along these lines.

    But seriously, using Baume degrees should be fined and / or passable of jail time. Baumees are measured at 16 C. So you’ll also have to correct for the temperatures you’ll be using. 

  • lmosca

    April 19, 2020 at 12:57 am

    For KOH: 
    https://www.oxy.com/OurBusinesses/Chemicals/Products/Documents/CausticPotash/kohhandb.pdf (page 26)

    For NaOH:
    https://www.oxy.com/OurBusinesses/Chemicals/Products/Documents/CausticSoda/caustic.pdf (page 28)
    those also report the density of the respective solutions.

    Apart from that, I am not sure what you really need.

  • david08848

    April 19, 2020 at 1:51 am

    Imosca,  I appreciate your input.  I am usually a bit long-winded so I tried to make it brief but perhaps I was too brief.  I am trying to compare this shaving soap formula from 1933 with others from the same time period.

    I don’t have a cosmetic chemistry background but I have learned that it is a valuable tool to have and approach to use!  In making these comparisons with other formulas I need to be able to change it from a KOH “solution” with water and an NaOH “solution” with water to the actual amount of KOH, NaOH and H2O that are in this formula so that I can compare the RATIO of NaOH to KOH and know the size of the water phase as well.  I don’t have the knowledge to calculate it myself correctly so I just need some help from someone who knows all the details about calculating this part of the old formula that uses Baume.  I have found it the past that often there are similarities in various formulations from this era and that has been very helpful in finding the correct formula that will be easy to make and easy to perfect.  Thanks again!

  • lmosca

    April 19, 2020 at 4:26 am

    I see…
    That’s a lot less convoluted than you think. 
    From the tables I have linked, you can determine the percent % (w/w) of both the NaOH and KOH relative to the needed Baume concentrations reported in those formulas.
    Since you have the total weight of solution used you can calculate how much NaOH and KOH were used.
    W_base = W_solution × %_concentration / 100

    Those amounts will scale linearly, with the scaling of your formula, of course. To double check that the math is correct (and the formula) just plug in the results into a soap calculator, like soapcalc. 

    The difference between the weights of solution and the one of base will give you the amount of water. That one is only minimally important though, because eventually the water will evaporate upon curing of the soap. Water is there to provide a vehicle for the base and as a fluidifier during the saponification reaction.

  • david08848

    April 19, 2020 at 1:11 pm


    My problem is that I don’t understand the terminology you are using.  I am looking at the chart for KOH on page 26.  I followed the Degrees Baume down to 38.07 then over to the left and under %KOH I see the number 36.  The formula shows 38 Baume and the percentage used in the formula is 36.  Your calculation then says to multiply the weight of the solution X the % of concentration/100.  The formula is written in KG and I see Grams of KOH in Liter but I don’t know if that is the number I need for the %_Concentration because I am not sure what that term means.  Once I understand the terminology used in this chart and how one of these amounts is calculated I can then use these two documents to check out formula 1 AND Formula 2 and any other old formula I need to check and I can stop pestering you and everyone else here!  LOL
    I know all the soap calculators and fully understand the soapmaking process so fortunately that is not an issue and is part of my my business and knowing the relationship between the KOH and NaOH in any duel lye formula will help to show me the possible results compared to other formulas to know which is the best direction to take.  Thanks again!

  • pharma

    April 19, 2020 at 2:21 pm
    The 36 you read from the table means 36 % KOH and 64% water on a weight/weight basis.
    If your formula calls for 36 kg KOH of 38 Bé, multiply 36 kg by 0.36 (or 36%/100) to get 12.96 kg pure KOH and (obviously) 23.04 kg water.
    Ignore all the other columns, they’re just confusing without helping your cause ;) .
  • david08848

    April 19, 2020 at 3:34 pm

    Thanks Pharma!

    I took your second line in your post and filled in the numbers for NaOH because it made more sense for ME to try it this way but I was confused about where the 0.36 came from in your KOH calculation!  So here are two versions of it with different numbers:

    If your formula calls for 19 kg NaOH of 38 Bé, multiply 19
    kg by 0.32 (or 32%/100) to get 12.92
    kg pure NaOH and (obviously) 6.08 kg water.


    or should this read:


    If your formula calls for 19 kg NaOH of 38 Bé, multiply 19
    kg by 0.10(or 10%/100) to get 1.9 kg
    pure NaOH and (obviously) 17.1 kg water.

    I’m thinking that the second one is more likely correct!  What do you think?


  • pharma

    April 19, 2020 at 7:23 pm
    The first one is closer: 19 kg of 38 Bé has to be multiplied by 0.32 as you did but this results in 6.08 kg pure NaOH and 12.92 kg water ;) .
    The second one is ‘more likely correct’ regarding the maths itself (no mixing up alkali and water) but I have no clue from where you got those 10%…
  • lmosca

    April 19, 2020 at 7:57 pm


    KOH 38 Be corresponds to 36% KOH
    NaOH 38 Be corresponds to 32% NaOH

    I don’t know where you came up with 10% in your last calculation.

    If you are using formula 1 you need:
    36 kg of KOH solution (38 Be = 36%), that is 12.96 kg KOH (solid) dissolved in 23.04 kg H2O.
    Obtained from (kg of KOH) = 36 (kg of solution) x 0.36 (percent concentration or 36/100)
    Water is obtained by difference.


    19 kg of NaOH solution (38 Be = 32%), that is 6.08 kg NaOH (solid) dissolved in 12.92 kg H2O.
    Obtained from (kg of NaOH) = 19 (kg of solution) x 0.32 (percent concentration or 32/100).

  • david08848

    April 19, 2020 at 9:19 pm
    OK, Guys!  :)
    It appears that I did get the KOH solution correct with 12.96 kg of KOH dissolved in 23.04 kg H2O which totals their 36 kg. of 38 Baume KOH!
    For the NaOH either I misread the original formula or mistyped the 9 using a 0 which is right next to it on the keyboard!  Then I typed in 6.08 but listed it under H2O and listed the 12.98 under the NaOH - the calculations were correct but I put them both in the wrong places switching one for the other in my last reply!
    Pharma and Imosca, I would like to thank you from the bottom of my heart for your patience, for sharing your knowledge with me and assistance in this project.  I really, really appreciate your help today!  Thank you both 100X over! :)

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