Transcript Solutions - 8th Grade Physical Science

Solubility Curve
RECALL
TYPES OF MIXTURES:
SUSPENSIONS
COLLOIDS
SOLUTIONS
All mixtures are physically combined and can be
physically separated.
DEFINITION
A solution is a homogeneous
mixture of two or more
substance in a single physical
state
Solution
 a mixture of two or more substances that is
identical throughout
 can be physically separated
 composed of solutes and solvents
the substance in the smallest
amount and the one that
dissolves in the solvent
Iced Tea Mix
(solute)
Salt water is
considered a
solution. How
can it be
physically
separated?
the substance in the larger
amount that dissolves the
solute
Iced Tea
(solution)
Water
(solvent)
Colloids (milk, fog, jello) are considered solutions
Solutes Change Solvents
 The amount of solute in a solution determines how
much the physical properties of the solvent are
changed
 Examples:
Lowering the Freezing Point
The freezing point of a liquid solvent
decreases when a solute is dissolved in it.
Ex. Pure water freezes at 320F (00C), but when salt is
dissolved in it, the freezing point is lowered.
This is why people use salt to melt ice.
Raising the Boiling Point
The boiling point of a solution is higher
than the boiling point of the solvent.
Therefore, a solution can remain a liquid at
a higher temperature than its pure solvent.
Ex. The boiling point of pure water is 2120F (1000C),
but when salt is dissolved in it, the boiling
point is higher. This is why it takes salt water
longer to boil than fresh water.
TYPES OF SOLUTIONS
SOLUTE
GAS
SOLVENT
GAS
EXAMPLE
Air
GAS
LIQUID
Seltzer (CO2)
LIQUID
LIQUID
Antifreeze (ethyl glycol in water)
SOLID
LIQUID
Sea water ( salt in water)
GAS
SOLID
Charcoal filter (poisonous gases in
carbon)
LIQUID
SOLID
Dental filling (mercury in silver)
SOLID
SOLID
Sterling silver (copper in silver)
SOLID SOLUTION
 Contain two or more metals called alloys
 Formed by melting the components and mixing them
together and allowing them to cool
 Properties of alloys are different from the original
component metals
TYPES OF ALLOYS
ALLOY
Babbitt
COMPONENT
Tin, antimony,
copper
USES
Bearings
Bell metal
Coinage metals
16 karat gold
Copper, tin
Bells
Copper, tin, zinc
Coins
Gold, copper, silver Jewelry
Sterling
Nichrome
Silver, copper
Nickel, iron,
chromium,
manganese
Jewelry, flatware
Heating elements
Important terminologies:
 Soluble – substance that dissolves another substance
 Insoluble – substance that does not dissolve another
substance
Solutes Change Solvents
 The amount of solute in a solution determines how
much the physical properties of the solvent are
changed
 Examples:
Lowering the Freezing Point
The freezing point of a liquid solvent
decreases when a solute is dissolved in it.
Ex. Pure water freezes at 320F (00C), but when salt is
dissolved in it, the freezing point is lowered.
This is why people use salt to melt ice.
Raising the Boiling Point
The boiling point of a solution is higher
than the boiling point of the solvent.
Therefore, a solution can remain a liquid at
a higher temperature than its pure solvent.
Ex. The boiling point of pure water is 2120F (1000C),
but when salt is dissolved in it, the boiling
point is higher. This is why it takes salt water
longer to boil than fresh water.
SOLUBILITY
Solubility
Solubility
maximum grams of solute that will dissolve
in 100 g of solvent at a given temperature
varies with temp
based on a saturated solution
Solubility
UNSATURATED
SOLUTION
more solute
dissolves
SATURATED
SOLUTION
no more solute
dissolves
SUPERSATURATED
SOLUTION
becomes unstable,
crystals form
increasing concentration
Concentration
 the amount of solute dissolved in a solvent at
a given temperature
•described as unsaturated if it
has a low concentration of
solute
•described as saturated if it
has a high concentration of
solute
•described as supersaturated if
contains more dissolved solute
than normally possible
C.
Determine if a solution is saturated,
unsaturated, or supersaturated.
• If the solubility for a given substance places
it anywhere on it's solubility curve it is
saturated.
• If it lies above the solubility curve, then it's
supersaturated,
• If it lies below the solubility curve it's an
unsaturated solution. Remember though, if
the volume of water isn't 100 cm3 to use a
proportion first as shown above.
Solubility  how much solute dissolves in a given amt.
of solvent at a given temp.
KNO3 (s)
SOLUBILITY
CURVE
KCl (s)
Solubility
(g/100 g H2O)
HCl (g)
Temp. (oC)
Unsaturated: solution could hold more solute; below line
Saturated: solution has “just right” amt. of solute; on line
Supersaturated: solution has “too much” solute dissolved in it;
above the line
Solubility vs. Temperature for Solids
140
KI
130
120
shows the dependence
of solubility on temperature
Solubility (grams of solute/100 g H2O)
Solubility
Table
110
NaNO3
gases
solids
100
KNO3
90
80
HCl
NH4Cl
70
60
NH3
KCl
50
40
30
NaCl
KClO3
20
10
SO2
0 10 20 30 40 50 60 70 80 90 100
LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 517
Solubility
• Solids are more soluble at...
– high temperatures.

Gases are more soluble at...
• low temperatures &
• high pressures
(Henry’s Law).
• EX: nitrogen narcosis,
the “bends,” soda
How to use a solubility graph?
A. IDENTIFYING A SUBSTANCE ( given the solubility
in g/100 cm3 of water and the temperature)
• Look for the intersection of the
solubility and temperature.
Example:What substance
has a solubility of 90 g/100
3
cm of water at a
temperature of 25ºC ?
Example:
What substance has a
3
solubility of 200 g/100 cm
of water at a temperature of
90ºC ?
B. Look for the temperature or solubility
•Locate the solubility curve needed and
see for a given temperature, which
solubility it lines up with and visa versa.
 What is the
solubility of
potassium
nitrate at 80ºC ?
At what
temperature
will sodium
nitrate have a
solubility of 95
g/100 cm3 ?
At what
temperature will
potassium iodide
have a solubility
of 230 g/100 cm3 ?
 What is the
solubility of
sodium chloride
at 25ºC in 150 cm3
of water ?
• From the
solubility
graph we see
that sodium
chlorides
solubility is 36
g.
Place this in the proportion below and solve for
the unknown solubility. Solve for the unknown
quantity by cross multiplying.
Solubility in grams = unknown solubility in grams
100 cm3 of water
other volume of water
___36 grams____ = unknown solubility in grams
100 cm3 of water
150 cm3 water
The unknown solubility is 54 grams. You can use this proportion
to solve for the other volume of water if you're given the other
solubility.
Solids and Gases dissolved in
Liquids
Solids dissolved in liquids
Sol.
Gases dissolved in liquids
Sol.
To
As To , solubility
To
As To , solubility
How many
additional grams
of solute must be
added in order to
make it
saturated?
From the graph
you can see that
the solubility for
potassium
nitrate at 50ºC is
84 grams
If there are already 30 grams
of solute in the solution, all you
need to get to 84 grams is 54
more grams ( 84g-30g )
Solubility vs. Temperature for Solids
140
KI
130
120
shows the dependence
of solubility on temperature
Solubility (grams of solute/100 g H2O)
Solubility
Table
110
NaNO3
gases
solids
100
KNO3
90
80
HCl
NH4Cl
70
60
NH3
KCl
50
40
30
NaCl
KClO3
20
10
SO2
0 10 20 30 40 50 60 70 80 90 100
LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 517
Solubility vs. Temperature for Solids
140
KI
Classify as unsaturated,
saturated, or supersaturated.
130
120
NaNO3
80 g NaNO3 @ 30oC =unsaturated
per
100
g
H2
O
60 g KCl @
60oC
=saturated
50 g NH3 @ 10oC = unsaturated
70 g NH4Cl @ 70oC =supersaturated
Solubility (grams of solute/100 g H2O)
110
gases
solids
100
KNO3
90
80
HCl
NH4Cl
70
60
NH3
KCl
50
40
30
NaCl
KClO3
20
10
SO2
0 10 20 30 40 50 60 70 80 90 100
Solubility vs. Temperature for Solids
140
KI
130
120
Per 500 g H2O, 120 g KNO3 @ 40oC
110
NaNO3
gases
saturation point @ 40oC for 100 g H2O = 66 g KNO3
So sat. pt. @
40oC
120 g < 330 g
for 500 g H2O = 5 x 66 g = 330 g
unsaturated
Solubility (grams of solute/100 g H2O)
solids
100
KNO3
90
80
HCl
NH4Cl
70
60
NH3
KCl
50
40
30
NaCl
KClO3
20
10
SO2
0 10 20 30 40 50 60 70 80 90 100
Describe each situation below.
(A) Per 100 g H2O, 100 g Unsaturated; all solute
NaNO3 @ 50oC.
dissolves; clear solution.
(B) Cool solution (A) very
Supersaturated; extra
slowly to 10oC.
solute remains in solution;
still clear.
(C) Quench solution (A) in
Saturated; extra solute
an ice bath to 10oC.
(20 g) can’t remain in
solution, becomes visible.