Solutions and Units
of Concentration
May 25, 2015
What is Solubility?
Solubility: the maximum amount of
solute that will dissolve in a certain
amount of solvent at a given
temperature
  Example: __ grams of salt (NaCl) in
100 g of water at __ ˚C.
 
Factors Affecting Solubility
1. Nature of Solute / Solvent: Like dissolves
like
2. Temperature:
a. Solids/Liquids- Solubility increases with
Temperature
- Increase K.E. increases motion and collision between
solute / solvent
b. Gases - Solubility decreases with Temperature
- Increase K.E. result in gas escaping to atmosphere
Factors Affecting Solubility
3. Pressure Factor:
a. Solids/Liquids - Very little effect (esp.
solids in liquids)
- Solids and Liquids are already close together, extra
pressure will not increase solubility
b. Gases - Solubility increases with
Pressure
- Increase pressure squeezes gas solute into solvent.
- HENRY’S LAW:
S = Solubility P = Press
Temperature & the Solubility of Gases
The solubility of gases DECREASES at higher temperatures
SOLUBILITY FORMULAS
amount of solute
amount of solvent
Given solubility
=
amount of solute
amount of solvent
unknown
SOLUBILITY PROBLEMS
1. The solubility of a solid is
15g / 100g of water. How many
grams of the solid must be dissolved
in 1 kg of water to make a saturated
solution? (1 kg = 1000 g)
SOLUBILITY PROBLEMS
amount of solute
amount of solute
=
amount of solvent
amount of solvent
15 g
X
------- = ------100g
1000g
X=150g
SOLUBILITY EXAMPLES
2. If you have 50g in 500g of water,
using solubility from problem #1 (15 g /
100 g), is the solution saturated?
15g
X
------ = -----100g 500g
X = 75g is saturated
No, 50 g is not saturated!
To read the
graph:
1. Find the
line for the
substance.
2. The
amount that
dissolves at a
given temp.
is on the yaxis.
How much KNO3
dissolves in 100g
H2O at 50oC?
1.  Find the line (red)
2.  Find the temperature
and follow up to the
line. (green)
3.  Read across to the yaxis and this is the
answer. (blue)
4.  Since it is above the
½-way between 80
and 90, 87g KNO3
will dissolve.
Types of Solutions:
 
Saturated solution: point on the line
–  Contains maximum amount of solute at given temp
–  Contains what it should hold
 
Supersaturated: above the line
–  Contains more solute than a saturated solution
–  Contains more than it should hold
 
Unsaturated: below the line is
–  Contains less solute than saturated solution
–  Contains less solute than it could hold
To do Calculations:
To calculate how much extra has been
dissolved:
Extra = Dissolved amt
saturated
in soln (given value)
line value
@ that temp.
 
To calculate how much more can be dissolved:
? Much more = saturated
- given
line value @ temp
value
 
Example 1:
 
 
 
How much less
KCl is dissolved
at 20oC than at
60oC in 100g
H2O?
Read the line
value:
32g at 20oC
Subtract it from
the given value:
•  45g – 32g =
13 g
Example 2:
 
 
 
How much more
KCl is required to
saturate the
solution if 25g are
dissolved at
40oC?
Read the line
value:
40g at 40oC
Subtract the given
value:
40g -25 g = 15 g
Your turn! Use your graph
 
 
 
 
1) What is the solubility of potassium nitrate at 300
C? 46 g/100 g
2) How many grams of ammonia can I dissolve in
100 grams of water at a temperature of 450 C? 32 g
3) At what temperature is the solubility of sodium
chloride the same as the solubility of potassium
chloride? 35 deg C
4) How many grams of ammonium chloride would I
need to make 100 grams of a saturated solution at
700 C? 60 g
 
5) What do all of the compounds that decreased in
solubility over the temperature range in the graph
have in common? All gases
 
6) What compound is least soluble at 400 C? SO2
Colligative Properties
When dissolving a solute in a solvent, the
properties of the solvent are modified.
  Vapor pressure
decreases
  Melting point
decreases
  Boiling point
increases
  Osmosis is possible (osmotic pressure)
Colligative Properties are properties of a liquid that
change when a solute is added
The magnitude of the change depends on the number
of solute particles in the solution, NOT on the identity
of the solute particles.
Vapor Pressure Lowering for a Solution
 
The diagram below shows how a phase diagram is affected
by dissolving a solute in a solvent.
Adding a solute increases the
boiling point.
Freezing Point Depression and
Boiling Point Elevation
 
Boiling Point Elevation
∆Tb =mkb
(for water kb=0.51 oC/m)
 
Freezing Point Depression
∆Tf=mkf
(for water kf=1.86 oC/m)
 
Note: m is the molality of the particles, so
if the solute is ionic, multiply by the #of
particles it dissociates to.
Example 1:
 
Find the new freezing point of 3m NaCl
in water.
Freezing Point Depression
  ∆Tf=mkf
(for water kf=1.86 oC/m)
  ∆Tf=(3m)(1.86 oC/m)
  5.58 oC
Example 2:
 
How many moles of ethylene glycol (C2H6O2)
should be added to 25 kg of water to lower
the freezing point by 45°C?
∆Tf=mkf (for water kf=1.86 oC/m)
45=(m)(1.86 oC/m)
= 24.2 m
m = moles
24.2 = moles
kg solvent
25
= 604 moles