UNIT XII
SOLUTIONS
CHAPTER 15; 4 PART 2
Solutions

Solution – a homogeneous mixture
 Not always a liquid- ex: air, brass,
glass
 Composed of:
Solute – the substance that gets
dissolved; the smaller
amount
Solvent – the substance that does
the dissolving; the larger
amount
Identify the solvent and the
solute:


50 g of NaCl in 100 mL of
water
50 mL of water in 100 mL
of alcohol
Rate of Solution


How fast a solute will
dissolve in a solvent
Increased by:
 Heating
 Stirring or Shaking
 Grinding up (a solid)
Solvation
The dissolving process
Example:
NaCl(s) + H2O ----> Na (aq) + Cl (aq)
Solvation
NaCl(s) + H2O ----> Na+(aq) + Cl- (aq)
1. Separate solute particles (endothermic)
Na
Cl
Cl
Na
Cl
+
Na
+
Na
Cl
Na
Cl
Cl
-
+
Na
-
Solvation
NaCl(s) + H2O ----> Na+(aq) + Cl- (aq)
2. Separate solvent particles (endothermic)
-Water is very polar and the hydrogen bonds must be broken.
Solvation
3.Solute and solvent particles are brought
together; this is called solvation (dissolving
process)
 If water is the solvent, it is also called
hydration.
Solvent molecules surround ions and
keep them apart
Cl-
H
H
H
Cl-
H
H
O
H
O
H
Na +
O
ClH
O
O
H
H
Cl-
Water molecules
surround the ions
and keep them
apart
Solvation of Ionic Compounds:

Dissociation – ionic compounds
(crystals) decompose into hydrated
ions (ions surrounded by water)
Solvation of Covalent
Compounds:
No dissociation occurs since covalent
compounds are not made of ions
 LIKE DISSOLVES LIKE!!!
 Polar dissolves polar (or ionic)
 Ex: water, acetone
 Both have charged particles or
molecules with charged poles
 Positive attracts negative

Like dissolves like

Non-polar dissolves non-polar
 No charges
 Nothing sticks together
Like dissolves like

Polar and non-polar molecules DO
NOT DISSOLVE each other


Polar molecules will stick together and will not
mix with non-polar molecules
EX:
water dissolves (I2 OR NaCl)
(WATER, ALCOHOL) dissolves ink (non-polar)
Exception: Sugar is non-polar and does dissolve
in water
Other Terms:
Miscible: two liquids that dissolve in each
other (such as water and alcohol)
Immiscible: two liquids that will not mix;
they form layers ( oil and water)
Heat of Solution
Heat released or absorbed as a solid
dissolves in a solvent
 Most heats of solution for solids are
endothermic (+ΔH)
 solid + H 0 + HEAT  aqueous solution
2



Adding something makes equilibrium shift to the
other side
Removing something makes equilibrium shift to
that side
Heat of solution
Heats of solutions for gases are usually
exothermic (-ΔH)
Gas + H2O  Aqueous solution + HEAT
Solubility

Amount of substance that dissolves
in a given quantity of a solvent at a
given temperature to produce a
saturated solution
Solubility
Types of solutions:
Supersaturated:
-contains more solute than a saturated solution
(contains more than it should)
 Saturated:
-contains maximum amount of solute for a
given amount of solvent at a constant
temperature (exactly what it should hold)

Unsaturated:
-contains less solute than a saturated solution
(contains less than it can hold)

SOLUBILITY
supersaturated
grams in
100 mL of
water
unsaturated
temperature
Factors affecting solubility

Temperature
Solids:
increase temperature, increases solubility
Gases:
increase temperature, decreases solubility

Pressure
Gases only:
increase pressure, increases solubility
SOLUBILITY GRAPHS
KNO3
140
120
KBr
Concentration (g/100 g water)
100
80
NaNO3
NH4Cl
60
Na2SO3
40
NaCl
20
10
20
30
40
50
60
temperature
70
80
SOLUBILITY
FORMULAS
amount of solute
amount of solvent
given
=
amount of solute
amount of solvent
unknown
Solubility Calculations


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 it
saturated solution?
Using the above solubility, is a solution of
50g/100g of water, unsaturated, saturated
or supersaturated?
Solubility Calculations

The solubility of a solute is 5g/100 g of
water at 20C and 7.5g/100g of water at
50 C.
 How much of the solute must be dissolved in
250g of water at 20C to prepare a saturated
solution?
 A saturated solution is prepared using 200 g of
water at 50 C, then allowed to cool to 20C.
Will it still be saturated? How much solute will
precipitate out?
Net Ionic Equations

A chemical equation that only
shows the ions involved in making
a precipitate or water


a precipitate is an insoluble solid
Can only be written for ionic
compounds since they dissociate
Steps in writing net ionic equations
1.Predict products and balance
equations
Pb(C2H3O2)2 (aq) + K2SO4 ----> 2K(C2H3O2)2 (aq) + PbSO4 (s)
Steps in writing net ionic equations
2. Change all soluble reactants
and products to separate ions
with oxidation numbers.
Pb
+2
+
2C2H3O2
2K +1+

-1
+
2K
+1
+
-2
SO4 ---->
2C2H3O2 -1 + PbSO4(s)
Precipitate (insoluble solid; will be a product), stays
a single compound.
Steps in writing net ionic equations

Eliminate ions that are not involved in
the formation of the precipitate or
water
+2
Pb
+
2C2H3O2
2K
+1
+
-1
+
2K
+1
2C2H3O2
-2
+
SO4 ---->
-1
+ PbSO4(s)
Steps in writing net ionic equations

Write the net ionic equation.
Pb+2+
SO4-2 ----> PbSO4(s)
The insoluble compound is the product; the ions
that form this compound are the reactants

Write the net ionic equation for:
Sodium hydroxide (aq) + hydrochloric acid (aq)
NaOH(aq) + HCl(aq) 
NaCl(aq) + H2O(l)
Na+ +OH- + H+ + Cl-  Na+ + Cl- + H20(l)
Net Ionic =
OH- + H+  H20(l)
Concentration

Amount of solute dissolved in a given
amount of solvent
Qualitative:
 Dilute-
solution containing a small amount of
solute
 Concentrated- solution containing a large
amount of solute
 Quantitative:
 Molarity (M) – number of moles of solute
dissolved in one liter of solution
M = moles of solute__
Liters of solution
 Molality
(m) – number of moles of solute
dissolved in one kilogram of solvent
m =
moles of solute__
kilograms of solvent
Concentration Calculations
What is the molarity of 2 moles of NaCl
in 5L of solution?
What is the molarity of 2000ml of water
containing 49g H3PO4?
How many grams of KBr are in one liter
of a 3M solution?
Concentration Calculations
What is the molality of a solution made
by dissolving 45g C6H12O6 in 500g of
water?
How many grams of water are required
to make a 0.5m solution containing
20g NaCl?
Dilution

Adding more solvent to a solution
spreads the same amount over a larger
volume
M1V1 =M2V2
(Molarity1)(Volume1) = (Molarity2)(Volume2)
Dilution Problems
100 mL of a 3M solution of HCl is diluted
to 375mL by adding water. What is the
new molarity?
How do you make 35mL of a 0.5M HCl
solution if all you have available is a
gallon of 12M HCl?
Colligative Properties

Properties that depend upon
concentration of the particles in a
solution and consequently affect the
boiling point and freezing points of
solutions
Colligative Properties

Adding solute particles:
 Raises boiling point
 Lowers freezing point
 Lowers vapor pressure
 Osmotic pressure
15.3

Colligative Properties of Solutions
Colligative Properties
Colligative properties (physical property) depend
on the concentration of the particles in the
solution
– Boiling point elevation: adding salt to water
for cooking
– Freezing point depression: salting the roads
before a freeze, antifreeze in cars, and
making homemade ice cream
– Osmotic pressure: responsible for plant’s cell
wall, sturdiness
15.3
Colligative Properties of Solutions
Colligative Properties

How does adding a solute change
physical properties?
– Solute particles get in the way of the
solvent molecules
 Makes
it harder for the solvent molecules to
boil (more energy needed – higher
temperature)
 Makes it harder for the solvent molecules to
freeze (need to release more energy –
lower temperature)
Boiling Point Elevation Calculations
DTb = (Kb) (m) (number of particles)
DTb = change in boiling point
Kb = (constant) 0.51C kgH2O
mol of solute
m = molality
# of particles = 1 if covalent
= count subscripts of ions
New BP = 100C +ΔTb
Freezing Point Depression Calculations
DTf = (Kf) (m) (number of particles)
DTf = change in freezing point (use C)
Kf = (constant) 1.86C kgH2O
mol of solute
m = molality
# of particles = 1 if covalent
= count subscripts of ions
New FP = 0C - ΔTf
EXAMPLE PROBLEMS
How many particles are in the following?
NaCl
AlCl3
C6H12O6
EXAMPLE PROBLEMS
Which of the following has more effect on
the boiling or freezing point?
NaF
K2SO4
C6H12O6
EXAMPLE PROBLEMS
Calculate the boiling point of a solution
containing 5.7g of sugar (C6H12O6)
dissolved in 50g of water.
Calculate the freezing point of a solution
containing 5.7g of sugar (C6H12O6)
dissolved in 50g of water.
EXAMPLE PROBLEMS
Calculate the freezing point of a
solution containing 1.2 mole of NaCl
dissolved in 800 g of water.
EXAMPLE PROBLEMS
10 g of covalent solute is dissolved in
500 g H2O. The boiling point is
100.85 C. What is the molecular
mass of the solute?