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

##### 1 Answer

#### Answer:

#### Explanation:

First, let's start by figuring out what you would *expect* the van't Hoff factor, *sodium phosphate*,

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

So, if *every* formula unit of sodium phosphate should produce

Now, the equation that describes *freezing-point depression* looks like this

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

*van't Hoff factor*

*cryoscopic constant* of the solvent;

The cryoscopic constant of water is equal to

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

*pure solvent*

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

#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

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-)#