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Solubility in organic solvents / fat solubility

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Endpoint:
solubility in organic solvents / fat solubility
Type of information:
experimental study
Adequacy of study:
key study
Study period:
28 Apr - 30 Aug 2017
Reliability:
1 (reliable without restriction)
Rationale for reliability incl. deficiencies:
guideline study
Qualifier:
equivalent or similar to
Guideline:
other: OECD 105 (Water Solubility)
Version / remarks:
adopted in 1995
Qualifier:
according to
Guideline:
other: CIPAC Handbooks as published by Collaborative International Pesticides Analytical Council available from: Marston Book Services Ltd, 160 Milton Park, Abindgon, Oxfordshire OX11 8LY, England
Medium:
n-octanol
Solubility:
ca. 127 g/L
Temp.:
20 °C

Solubility in n-octanol according to CIPAC MT 181
An amount of 100 mg of the test item formed a turbid solution in 0.5 mL n-octanol at 20 °C.

The preliminary solubility according to CIPAC MT 181 was > 200 g/L.

Following the further proceeding of the CIPAC method an amount of 2 x 500 mg was filled up to 2 mL volume with n-octanol in a graduated reagent tube. The mixture first was shaken at 20 °C, when it did not dissolve it was warmed in a drying cabinet (70 °C) until the test item was dissolved. After this the tubes were tempered at 20 °C overnight.

The test item was dissolved in a total volume of 2 mL. The liquid was light brown and turbid, a small white residue was visible on the ground of the tubes. The residue was considered to result from an impurity generally insoluble in organic solvent.

The solubility of the test item in n-octanol according to the evaluation scheme of CIPAC MT 181 was > 250 g/L for the test item ‘as is’, equivalent to > 225 g/L at 90.1% active. As no certificate of analysis was provided with the test item, the content of active of the test item was measured by 31P-NMR.

Solubility tests based on OECD 105
When the CIPAC method had shown high solubility of the test item in n-octanol, a mixture of 4.0 g of test item and 4.8 mL n-octanol was found to be manageable for saturation tests according to the flask method described in OECD 105.
The amount of test item originally proposed by the method for saturation of water is 5 times the amount required for saturation. In this case 5 x 250 g/L = 1250 g/L n-octanol. Such a mixture could not be stirred and would have most likely solidified after a warm/cool cycle.

The test was modified to dissolution at 80 °C on a tempered orbital shaker and subsequent cooling to 20 °C in a water bath for 1, 2 and 3 days. The original design is stirring at 30 °C for 1, 2, 3, 4, 5…days with subsequent equilibration to 20 °C for 24 hours. This however was not feasible with the high amounts of test item required as set out above.

The content of test item neat active (90.1% of test item) was determined by quantitative31P-NMR analysis. Two 200 µL fractions taken two independent samples each anaylsed and the mean of the 4 values obtained was issued as the result for the specific point of time.

Results n-octanol solubility at 20 °C:

Day 1: 133 g/L
Day 2: 129 g/L
Day 3: 120 g/L
Mean: 127 g/L
± 6.66 g/L (± 5.23% RSD, n= 3)

The concentrations were slightly decreasing from 133 g/L to 120 g/L over the 3 days of tempering at 20 °C, but were considered to represent a stable concentration level within ± 15%.

The n-octanol solubility of the test item neat active at 20 °C hence was determined to be 127 g/L ± 6.66 g/L. The correspondent result for the test item ‘as is’ (100% at 90.1% active) is 141 g/L ± 7.40 g/L.

With regard to the higher results obtained by the CIPAC method, the content of mineral oil in the test item may have had an effect here. At ca. 9% mineral oil, 0.28 g of mineral oil were introduced into the 4.8 mL n-octanol used for the test mixture by means of the 4.0 g of test item. At a density of 0.817 g/mL for n-octanol at 20 °C, the solvent mixture complied to 0.28 g/(0.28 g + 4.8 mL x 0.817 g/mL) * 100% = 6.7% mineral oil in n-octanol.

The correspondent value for the CIPAC method was 0.045 g mineral oil (9% of 0.5 g) in 1.6 mL n-octanol x 0.817 g/mL + 0.045 g, equivalent to 0.045 g/1.35 g * 100% = 3.3%. The higher concentration of mineral oil in the OECD test may have biased the result.

Description of key information

Solubility in n-octanol: 127 g/L +/- 6.66 g/L at 20 °C (modified OECD 105)

Key value for chemical safety assessment

At the temperature of:
20 °C
Solubility in organic solvents at 20°C:
127 g/L

Additional information