Question

Assuming 100% dissociation, calculate the freezing point and boiling point of 2.79 m K3PO4(aq). Constants may be found here.

Solvent

Formula

Kf value* (

Answer 1

Freezing Point Depression:
When a solute, like K3PO4, is added to a solvent, like water, the freezing point (fp) of the solution will be lower than that of the pure solvent (water= 0 C). To calculate the fp depression, we use the formula ?T= K m. ?T is the change in freezing point, K is the freezing point constant specific to each solvent, and m is the molality (moles of solute/ kg of solvent) of the solution.

The freezing point depression constant can be looked up using a book or a table. For water, K(fp) is 1.86 C/molal.

An important concept to take into account for this problem is that of the Van't Hoff factor (i). The Van't Hoff factor accounts for the dissociation of ions in solution, which affects molality. Basically, if the solute is ionic, you multiply the moles of solute by the number of particles that the solid will dissociate into. K3PO4 will dissociate into 4 particles, so we must multiply by four. Using this we can calculate the change in fp.

?T=(1.86C/molal)(1.23 molal)(4)
?T=9.15 C

Since the fp is dropping, we should subtract ?T from the freezing point of water which is 0 C. Our answer is -9.15 C.


Boiling Point Elevation:
Boiling Point Elevation is very similar to fp depression. The only difference is the constant K being used. The K(bp) for water is .512 C/molal. We can use this to calculate ?T. The molality of the solution does not change.

?T= K m
?T=(.512 C/molal)(1.23 molal)(4)
?T=2.52 C

In this problem, bp is being elevated, so we must remember to add our ?T to the normal boiling point of the solvent (water= 100 C). Our final answer is 102.52 C.