Sodium lauryl sulfate

Dear all!
We use sodium lauryl sulfate as a cleaning agent in cleaning equipments. But i don’t know how to decide the residue limit for the sodium lauryl sulfate?
Help me, please!

Dear Hang coi,

the limit for the cleaning agent is set in way a very similar to other analytical targets, only that you have to start from toxicological data instead of dosage data. In your case:

ADI = (1200 mg/kg day * 70 kg)/ 100000 = 0,84 mg/day for adults
ADI = (1200 mg/kg day * 18 kg)/ 100000 = 0,22 mg/day for a five year old girl

  • From there on the calculation is performed exactly as the calculation for contamination with an active ingredient, and depends on the approach taken at your site. In our case, we use the formula:

C= (ADI*U)/MDT, where:

U = number of therapeutic units per batch (e.g. tablets per batch)
MDT= Maximum daily takings (e.g. for Alprazolam the highest dosage in the literature is 4 mg/day, so the MDT for Alprazolam 0,5 mg is 4/0,5= 8 tablets/day)

The limit C is the amount of Sodium Lauryl Sulfate per batch of contaminated product, and so it is the highest amount of SLS accepted on the sum of the product contact surfaces in the equipment train. The limit L per surface unit is then:

L= C/surface in [mg/cm²]

To avoid a tedious calculation for each product and equipment train, we use as a worst case the product that has the lowest U/(MDT*surface) coefficient (usually the toxicity of the cleaning prodcts used is fairly low, so there is little risk to get unrealistically low limits).

As an example, with our figures (the product with that lowest coefficient is Alprazolam 0,5 mg, batch size 2000000 tablets/batch, equipment surface 364266 cm²/batch, MDT as above):

ADI = (1200 mg/kgday * 18 kg)/ 100000 = 0,22 mg/day

C= (0,22 mg/day * 2000000 tablets/batch)/(8 tablets/day) = 55000 mg/batch

L= 55000 mg/batch / 364266 cm²/batch= 0,15 mg/cm²

And this is your acceptance criterion

Best regards


Dear all!
We have a problem in determining the amount of Sodium Lauryl sulfate. Usually, we determine by titration method. But it’s no use to decide the residue limit for Sodium lauryl sulfate because its concentration is very low. Can you suggest me another method?
Thanks for your prompt reply!

Dear All,
SLS is used as Cleaning Agent and is highly water soluble.It is necessary to show the absence of the same.Since Residual Content of SLS is quite low the most appropriate metod is use of TOC Analyser.