Water has a Kf of 1.86C/m and Kb of 0.51C/m. What will be the new freezing point of a 1.5 m solution of potassium phosphate?

Answer 1

Potassium phosphate = K3PO4. This will dissociate to 3K+ + PO43-. This produces 4 particles.

Formula for freezing point depression is ∆T = imK

∆T = change in freezing point = ?

i = vant Hoff factor = 4

m = molality = moles K3PO4 / kg water = 1.5 m

K = freezing point constant for water = 1.86ºC/m

Solving for ∆T, we have...

∆T = (4)(1.5)(1.86)

∆T = 11.2ºC

New freezing point of the solution = 0º - 11.2º = -11.2ºC