What is the van't Hoff factor of #Na_3PO_4# in a 0.40 m solution whose freezing point is -2.6°C ?

Question

Ellie Andrews · Accepted Answer

#i = 3.5#

First, let's start by figuring out what you would expect the van't Hoff factor, #i#, to be for sodium phosphate, #"Na"_3"PO"_4#.

The ratio between the number of solute particles and the number of particles generated in solution* after the solute dissolves is known as the van't Hoff factor.

This relates to ionic compounds and how many ions will be generated per formula unit of solute.

Sodium phosphate will dissociate in aqueous solution to form sodium cations, #"Na"^(+)#, and phosphate anions, #"PO"_4^(3-)#

#"Na"_3"PO"_text(4(aq]) -> 3"Na"_text((aq])^(+) + "PO"_text(4(aq])^(3-)#

So, if every formula unit of sodium phosphate should produce #4# ions in solution, you should expect the van't Hoff factor to be equal to #4#.

Currently, the freezing-point depression equation looks like this:

#color(blue)(DeltaT_f = i * K_f * b)" "#, where

#DeltaT_f# - the freezing-point depression; #i# - the van't Hoff factor #K_f# - the cryoscopic constant of the solvent; #b# - the molality of the solution.

The cryoscopic constant of water is equal to #1.86 ""^@"C kg mol"^(-1)#

Tcriosebu.pdf https://tutor.hix.ai

The definition of the freezing-point depression is

#color(blue)(DeltaT_f = T_f^@ - T_f)" "#, where

#T_f^@# - the freezing point of the pure solvent #T_f# - the freezing point of the solution

If it were you, you would have

#DeltaT_f = 0^@"C" - (-2.6^@"C") = 2.6^@"C"#

Plug in your values into the equation for freezing-point depression and solve for #i#

#DeltaT_f = i * K_f * b implies i = (DeltaT_f)/(K_f * b)#

#i = (2.6 color(red)(cancel(color(black)(""^@"C"))))/(1.86 color(red)(cancel(color(black)(""^@"C"))) color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.40 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))) = color(green)(3.5)#

It is evident that the observed van't Hoff factor is lower than the predicted value.

The reason for this is that fewer ions are produced per formula unit than anticipated is because some of the ions will bind to form solvation cells.

In other words, some of the sodium cations will bind to the phosphate anions and exist either as #"Na"^(+)"PO"_4^(3-)#, or as #"Na"_2^(+)"PO"_4^(3-)# solvation cells.

This indicates that you've

#"Na"_3"PO"_text(4(aq]) -> 3"Na"_text((aq])^(+) + "PO"_text(4(aq])^(3-)#

and

#"Na"_3"PO"_text(4(aq]) -> 2"Na"_text((aq])^(+) + "Na"^(+) "PO"_text(4(aq])^(3-)#

#"Na"_3"PO"_text(4(aq]) -> "Na"_text((aq])^(+) + "Na"_2^(+) "PO"_text(4(aq])^(3-)#

Peyton Adams · Answer

undefined The van't Hoff factor (i) for Na₃PO₄ is 4, because it dissociates into four ions (3 Na⁺ ions and 1 PO₄³⁻ ion) in solution.