Assuming 100% dissociation, calculate the freezing point and boiling point of 2.50m K3PO4?
1 Answer
Jan 28, 2018
The freezing point depression and boiling point elevation are due to the same solute.
#DeltaT_f = T_f - T_f^"*" = -iK_fm#
#DeltaT_b = T_b - T_b^"*" = iK_bm# where
#T_(tr)# is the phase transition temperature.#"*"# indicates pure solvent.#i# is the van't Hoff factor, i.e. the effective number of dissociated particles per solute particle placed into the solvent.#K_f = 1.86^@ "C/m"# is the freezing point depression constant and#K_b = 0.512^@ "C/m"# is the boiling point elevation constant of water.#m# is the MOLALITY of water in units of MOLAL, i.e.#"m"# ,#"mol solute/kg solvent"# . It has nothing whatsoever to do with distance.
The dissociation is given as...
#"K"_3"PO"_4(aq) -> 3"K"^(+)(aq) + "PO"_4^(3-)(aq)#
...and so if we assume 100% dissociation,
Therefore,
#DeltaT_f = -4 cdot 1.86^@"C/m" cdot "2.50 m" = bbul(-18.6^@ "C")#
#DeltaT_b = 4 cdot 0.512^@ "C/m" cdot "2.50 m" = bbul(5.12^@ "C")#
Should the real value of
