Transcript Unit 11: States of Matter

Mini-Unit:
Colligative Properties
Calculations with Colligative
Properties
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 solute
does not matter. The effect is the
same. Sugar has the same affect as
salt on a solvent.
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.
Boiling Point Elevation (ΔTB)
Recall, boiling point is the point at which V.P.
= atmospheric pressure.
ΔTB = (i)(KB)(m)
i= depends on the number of particles the
solute breaks into
KB= a constant that depends on the solvent
For H2O, KB = 0.512°C/m
m= molality of the solution (mol/kg)
Van’t Hoff Factor (i)
To obtain the i-value:
When dissolved, ionic compounds will break apart
into separate ions.
Examples: NaCl  Na+1 + Cl-1
i= 2
MgCl2  Mg+2 + 2Cl-1
i= 3
Ca(SO4)  Ca+2 + (SO4)-2
i= 2
Ba3(PO4)2  3Ba+2 + 2(PO4)-3 i= 5
Example
a) What is the change in the boiling point of a 0.335m
solution of KBr in H2O? b) What is the new boiling point?
ΔTB= ?
ΔTB=(i)(KB)(m)
+1
-1
i= KBr  K + Br
ΔTB=(2)(0.512°C/m)(0.335m)
m= 0.335m
a) ΔTB=0.343°C
KB= 0.512°C/m
Water boils at 100°C so…
100°C + 0.343°C =
Add since this is a b.p.
ELEVATION
(b.p. is increasing!)
b) 100.343°C
Freezing Point Depression (ΔTF)
Recall, freezing point is the temperature at
which a liquid  solid
ΔTF = (i)(KF)(m)
i= depends on the number of particles
KB= a constant
For H2O, KF = 1.86°C/m
m= molality
Example
a) 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)
ΔTF = ?
ΔTF = (3) (1.86°C/m)(2.5m)
i = BaF2 Ba+2 + 2F-1
a) ΔTF = -14°C
KF = 1.86°C/m
m = 2.5m
Water freezes at 0°C so…
Subtract since this is
0°C - 14°C =
a f.p. DEPRESSION
b) -14°C
(f.p. is decreasing!)
Questions?
Complete WS2