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

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#.

As you know, the van't Hoff factor tells you what the ratio between the number of particles of solute and the number of particles produced in solution* after dissolving the solute.

For ionic compounds, this comes down to how many ions will be produced 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#.

Now, the equation that describes freezing-point depression 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)#

http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf

The freezing-point depression is defined as

#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

In your case, 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)#

Notice that the observed van't Hoff factor is smaller than the expected van't Hoff factor.

This happens because some of the ions will actually bind to form solvation cells, which is why the number of ions produced per formula unit is smaller than the expected number.

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 means that you have

#"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-)#