Assuming 100% dissociation, calculate the freezing point and boiling point of 2.50m K3PO4?

1 Answer
Truong-Son N.
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, #i = 4#. (Why? See the definition above.)

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 #i# be #4#, or less than #4#?

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