1. Mini-Unit: Colligative Properties Calculations with Colligative Properties Day 2 - Notes

2. After today you will be able to… Explain what a colligative property is Calculate b.p. elevation and f.p. depression

3. Recall, colligative properties are properties of a solution which are affected by the number of dissolved particles in the solvent. The chemical makeup of the solute does not matter. The effect is the same. Sugar has the same affect as salt on a solvent.The chemical makeup of the solute does not matter. The effect is the same. Sugar has the same affect as salt on a solvent.

4. Affected Colligative Properties Three colligative properties that are affected are: 1.Vapor pressure (lowering) 2.Boiling point (elevation) 3.Freezing point (depression) \. There are others, but these are the ones we will talk about in this class.

5. Boiling Point Elevation (ΔT B ) V.P. = atmospheric pressure. Recall, boiling point is the point at which V.P. = atmospheric pressure. depends on the number of particles the solute breaks into i= depends on the number of particles the solute breaks into a constant that depends on the solvent K B = a constant that depends on the solvent For H 2 O, molality of the solution (mol/kg) m= molality of the solution (mol/kg) ΔT B = (i)(K B )(m) K B = 0.512°C/m

6. Van’t Hoff Factor (i) To obtain the i-value: break apart into separate ions. When dissolved, ionic compounds will break apart into separate ions. Examples:NaCl  i= MgCl 2  i= Ca(SO 4 )  i= Ba 3 (PO 4 ) 2  i= Na +1 + Cl -1 Mg +2 + 2Cl -1 Ca +2 + (SO 4 ) -2 3Ba +2 + 2(PO 4 ) -3 2 3 2 5

7. Example a) What is the change in the boiling point of a 0.335m solution of KBr in H 2 O? b) What is the new boiling point? Δ ΔT B = i= m= K B = ΔT B =(i)(K B )(m) ? KBr  K +1 + Br -1 0.335m 0.512°C/m ΔT B =(2)(0.512°C/m)(0.335m) a) ΔT B =0.343°C Water boils at 100°C so… 100°C + 0.343°C = b) 100.343°C Add since this is a b.p. ELEVATION (b.p. is increasing!)

8. Freezing Point Depression (ΔT F ) the temperature at which a liquid  solid Recall, freezing point is the temperature at which a liquid  solid depends on the number of particles i= depends on the number of particles a constant K B = a constant For H 2 O, molality m= molality ΔT F = (i)(K F )(m) K F = 1.86°C/m

9. Example a) What is the change in the freezing point of a 2.5m solution of BaF 2 in H 2 O? b) What is the new freezing point? ΔT F = i = K F = m = Δ T F =(i)(K F )(m) ? BaF 2  Ba +2 + 2F -1 1.86°C/m 2.5m ΔT F = (3) (1.86°C/m)(2.5m) a) ΔT F = -14°C Water freezes at 0°C so… 0°C0°C0°C0°C- 14°C = b) -14°C Subtract since this is a f.p. DEPRESSION (f.p. is decreasing!)

10. Questions? Complete WS2