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Water solubility

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Description of key information

The test material has a water solubility of 0.0000128 g-mol/L (equating to 4.9 mg/L) at 29 °C and as such can be described as slightly soluble. 

Key value for chemical safety assessment

Water solubility:
4.9 mg/L
at the temperature of:
29 °C

Additional information

A value for the water solubility of the substance was obtained from a peer reviewed handbook. The data were awarded a reliability score of 2 in accordance with the criteria set forth by Klimisch (1997). The water solubility of the test material was reported to be 0.0000128 g-mol/L at 29 °C (equating to 4.9 mg/L) and, as such, the substance can be described as slightly soluble. Further clarification regarding the units for water solubility can be found below.

Supporting information is available in the form of a company Material Safety Data Sheet. The test material was reported to be insoluble in water. Since no information was provided on materials and methods the data were assigned a reliability score of 4.

Further discussion regarding the quoted units for water solubility:

The substance dierbium trioxide has a RMM of 382.5169 g/mol and a quoted water solubility (Merck) of 1.28 x 10-5 g-mol/L.

This equates to a water solubility of 4.9 mg/L according to the following rationale:

“g-mol/L” is a unit of molarity, i.e. the number of moles of a particular substance in a litre of solution.

Typically, molarity is written as mol/L, and it is understood that when converting to a mass-based solubility the calculation must be made using the molar mass in grams (as opposed to lb or any other non-metric unit). The “g” is not part of the units per se but merely confirms to the reader that this is a metric unit.

The use of “g” in the units in this case appears to be largely of historic relevance. Had the units been quoted as lb-mol/L, then in order to convert the mole-based solubility into a mass-based value, it would have been necessary to use the RMM for erbium oxide in the units of lb/mol.

The mass-based water solubility is therefore calculated by multiplying the molarity (1.28 x 10-5 g-mol/L) by the molar mass in grams (382.5169 g/mol) to give 0.00490 g/L (4.9 mg/L).