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3 years ago

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

1 answer:

4 0

Freezing point = -9.6Â°C
Boiling point = 102.7Â°C
Freezing point depression is modeled by the equation:
Î”TF = KF Â· b Â· i,
where
Î”TF = 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:

Î”TF = KF Â· b Â· i,
Î”TF = 1.853 K*kg/mol Â· 1.30 mol/kg Â· 4
Î”TF = 9.6356 K
So the freezing point will be -9.6Â°C
For boiling point elevation the formula is
Î”Tb = Kb Â· bB
where
Î”Tb =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:
Î”Tb = Kb Â· bB
Î”Tb = 0.512 K*kg/mol Â· 1.30 m Â· 4
Î”Tb = 2.6624 K
So the new boiling point will be 100Â°C + 2.7Â°C = 102.7Â°C

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