Home Cosmetic Science Talk Formulating Hair Salt Curve Analysis - Sulfate Free Shampoo

  • Salt Curve Analysis - Sulfate Free Shampoo

    Posted by scharaya on August 3, 2016 at 3:18 pm

    Hi,

    I am working on sulfate free shampoo and I am using blend of following surfactants:

    8% Sodium Cocoyl Isethionate (78% Active)
    20% CAPB (30% Active)
    15% Stepan Mild PCL (25% blend of Sodium Methyl 2-Sulfolaurate (and) Disodium 2-Sulfolaurate (and) Sodium Lauryl Sulfoacetate)

    I have done Salt Curve Analysis on the finished product and I found at pH 6 to 6.75 viscosity drops at an addition of salt up to 1.5%. I adjusted pH to 7 and 7.25, I found viscosity starts building with the addition of salt which eventually drops after 1% of salt.

    This is really interesting and I was wondering what could be reason. Can any one explain to me what could be the reason for this?

    Let me know if you need more information. Thank you

    Dilfre replied 7 years, 9 months ago 3 Members · 9 Replies
  • 9 Replies
  • belassi

    Member
    August 3, 2016 at 4:02 pm

    You can not adjust pH without changing the percentage of ions in solution, that is what’s happening.

  • scharaya

    Member
    August 3, 2016 at 4:36 pm

    Thank you. My initial pH is 6 without adding any buffer and initial viscosity is 5000. I don’t have the associative thickener in the formula. If I add salt as it is, viscosity starts to drop. I adjusted the pH to 7 using NaOH and added salt, viscosity starts to build up. So, why at low pH salt doesn’t help in building viscosity whereas helps in neutral pH?

  • belassi

    Member
    August 3, 2016 at 5:55 pm

    6 isn’t low pH. It is all about how the ionisation works.

  • Dilfre

    Member
    August 3, 2016 at 8:49 pm

    Interesting. which compound are you using to adjust the pH? and for Mr. Belassi, Are all electrolytes viscosity modifiers?

  • scharaya

    Member
    August 3, 2016 at 9:01 pm

    I am using Sod Hydroxide to adjust and play on different pH.

  • belassi

    Member
    August 3, 2016 at 11:28 pm

    I don’t know if ALL electrolytes modify viscosity. Ionic compounds will.

  • Dilfre

    Member
    August 4, 2016 at 12:18 am

    thanks Belassi, sorry I’m refering to ionic electrolytes.

  • belassi

    Member
    August 4, 2016 at 4:20 am

    It’s a complicated topic and one that I doubt my own knowledge in. If I make any errors in this explanation will someone please correct me:
    Dilfre, ionic compounds, when added to water, differentiate into their (charged) ions. I’ll use sodium chloride and sodium hydroxide as examples but the same principle applies to acids as well as bases.
    So for instance when you add NaCl to water, the compound splits into its ions Na+ and Cl-. This causes the heads/tails of the surfactant to align into micelles, and the size and shape of the micelles determines the viscosity.
    If instead of NaCl you add NaOH, the ions are Na+ and OH-; however, adding NaOH, a strong base, generally causes a neutralisation reaction with one or more other components.
    (As an example, my primary surfactant in one shampoo is laureth-6 carboxylic acid. I have to neutralise it (otherwise the product would be horribly acidic) so I add NaOH. The output will be laureth-6 carboxylate + water. This is not a thickening reaction however)
    What makes things worse, and I am not sure from your post if you’re aware of this, is that different ionic compounds compete for solubility. For instance it is very difficult to dissolve even a small amount of salt in a solution of NaOH or KOH. The strong base binds so tightly to the water molecules that it doesn’t want to allow an interloper substance.
    If (you don’t specify) you are adding citric acid, then you will get citrates formed which are also salts that may or may not bind more tightly than NaCl.

  • Dilfre

    Member
    August 4, 2016 at 3:22 pm

    thanks Belassi, I guess L`Chatelier principle of common ion reduces the eficiency of NaCl as thickener due to common ion accumulation when adjust the pH with NaOH.

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