Question

Assuming 100% dissociation, calculate the freezing point and boiling point of 1.30 m k3po4(aq). constants may be found here.

Answer 1

Freezing point = -9.6°C
Boiling point = 102.7°C
Freezing point depression is modeled by the equation:
ΔTF = KF · b · i,
where
ΔTF = Change in freezing point
KF = Cryoscopic constant of solvent (water is 1.853 K*kg/mol)
b = molarity of solution
i = van't Hoff factor (number of ions per molecule of solute. Ideal for K3PO4 is 4)
So plugging in the given values and calculating gives:

ΔTF = KF · b · i,
ΔTF = 1.853 K*kg/mol · 1.30 mol/kg · 4
ΔTF = 9.6356 K
So the freezing point will be -9.6°C
For boiling point elevation the formula is
ΔTb = Kb · bB
where
ΔTb =Boiling point elevation
Kb = Ebullioscopic constant of solvent (water = 0.512 K*kg/mol)
bB = Molarity of solution after taking into account the dissociation.
So substitute the known values and calculate:
ΔTb = Kb · bB
ΔTb = 0.512 K*kg/mol · 1.30 m · 4
ΔTb = 2.6624 K
So the new boiling point will be 100°C + 2.7°C = 102.7°C