What is the freezing point of a 0.195 m aqueous solution of #K_2S#?

1 Answer
Jun 4, 2017




“Freezing point depression” depends on the solvent and the molar concentration of the solute.

Sulfide is highly basic, consequently #K_2S# completely and irreversibly hydrolyzes in water according to the following equation:

#K_2S + H_2O → KOH + KSH #

For many purposes, this reaction is inconsequential since the mixture of #SH^− and OH^−# behaves as a source of #S^(2−)#. Other alkali metal sulfides behave similarly. (Holleman, A. F.; Wiberg, E. "Inorganic Chemistry" Academic Press: San Diego, 2001. ISBN 0-12-352651-5).

Thus, in terms of the colligative properties, a 0.195M solution of #K_2S# results in three times the concentration of ions in solution, or 0.0.585 moles that need to be used in the freezing point depression calculation. The freezing point depression constant for water is 1.86’C/mol.

#3 * 0.195 mol * 1.86’C/mol = 1.088’C depression of the freezing point, or a change from 0’C for water to -1.09’C.