Mini-Unit: Colligative Properties

Calculations with Colligative Properties

Day 2 - Notes

After today you will be able to…

•Explain what a colligative property is

•Calculate b.p. elevation and f.p. depression

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 The chemical makeup of the solute does not matter. The effect solute does not matter. The effect is the same. Sugar has the same is the same. Sugar has the same affect as salt on a solvent.affect as salt on a solvent.

Affected Colligative Properties

Three colligative properties that are affected are:

1.1.Vapor pressure (Vapor pressure (loweringlowering))

2.2.Boiling point (Boiling point (elevationelevation))

3.3.Freezing point (Freezing point (depressiondepression))\

.

There are others, but these are the ones we will talk about in this class.

Boiling Point Elevation (ΔTB)

Recall, boiling point is the point at which V.P. = atmospheric pressure.V.P. = atmospheric pressure.

i= depends on the number of particles the depends on the number of particles the solute breaks intosolute breaks intoKB= a constant that depends on the solvent a constant that depends on the solvent

For H2O,

m= molality of the solution (mol/kg)molality of the solution (mol/kg)

ΔΔTTBB = (i)(K = (i)(KBB))(m)(m)

KKBB = = 0.512°C/m0.512°C/m

Van’t Hoff Factor (i)To obtain the i-value:When dissolved, ionic compounds will break break

apart into separate ions.apart into separate ions.

Examples: NaCl i=

MgCl2 i=

Ca(SO4) i=

Ba3(PO4)2 i=

NaNa+1+1 + + ClCl-1-1

MgMg+2+2 + + 2Cl2Cl-1-1

CaCa+2+2 + (SO + (SO44))--

22

3Ba3Ba+2+2 + 2(PO + 2(PO44))--

33

22

33

22

55

Examplea) What is the change in the boiling point of a 0.335m solution of KBr in H2O? b) What is the new boiling point?

ΔΔTB=

i=m=KB=

ΔΔTTBB=(i)(K=(i)(KBB)(m))(m)??

KBr KBr K K+1+1 + + BrBr-1-10.3350.335mm0.512°C/m0.512°C/m

ΔΔTTBB=(2)(=(2)(0.512°C/m0.512°C/m))(0.335m)(0.335m)a) a)

ΔΔTTBB=0.343°C=0.343°CWater boils at 100°C

so…100°C 100°C ++ 0.343°C0.343°C==

b) b) 100.343°C100.343°C

Add since this is a b.p. ELEVATION

(b.p. is increasing!)

Freezing Point Depression (ΔTF)

Recall, freezing point is the temperature the temperature at which a liquid at which a liquid solid solid

i= depends on the number of particlesdepends on the number of particlesKB= a constant a constant

For H2O,

m= molalitymolality

ΔΔTTFF = (i)(K = (i)(KFF))(m)(m)

KKFF = 1.86°C/m = 1.86°C/m

Examplea) What is the change in the freezing point of a 2.5m solution of BaF2 in H2O? b) What is the new freezing point?ΔTF =

i =KF =

m =

ΔΔTTFF=(i)(K=(i)(KFF)(m))(m)??

BaFBaF22 Ba Ba+2 +2 + + 2F2F-1-11.86°C/m1.86°C/m2.5m2.5m

ΔΔTTFF = (3) (1.86°C/m) = (3) (1.86°C/m)(2.5m)(2.5m) a) a) ΔΔTTFF = - = -

14°C14°CWater freezes at 0°C so…0°C 0°C -- 14°C14°C==

b) b) -14°C-14°C

Subtract since this is a f.p.

DEPRESSION (f.p. is decreasing!)

Questions?Complete

WS2