Transcript Chapter 13 Solutions

Chapter 13 Solutions
Mixtures
Mixtures can be either Heterogeneous
(particles are large enough to be seen
under a microscope….most of the time) or
Homogeneous (particles are moleculesized and the mixture appears uniform
throughout).
Colloids
Suspensions
Solutions
Homogeneous – uniform, consistent throughout
Heterogeneous - composed of 2 or more
regions with different compositions.
The 3 Types of Mixtures are classified by
there particle sizes.
1. Colloids
Heterogeneous Mixtures
Particles range in size between 2 – 500 nm
Particles are small enough
to remain suspended.
Colloids contain particles
with like charges, so they
repel one another, they
cannot clump together to
form large enough
molecules to settle out.
Milk is a colloid suspension
of protein, fat and whey.
Jello is a Colloid
Because its particles are too large to dissolve
completely into a solution.
2. Suspensions
Heterogeneous Mixtures
Particles larger than 500 nm
Large particles can
be suspended but
will eventually
settle out.
Has to be stirred to
re-suspend.
Starch Suspension
3. Solutions
Homogeneous Mixtures
Solute Particles less than 2 nm in size
Solutions are composed of
dissolved solute and a solvent.
Salt Water
In a solution the dipoles of the
solvent, usually water molecules,
are strong enough to separate and
attract the ions of the solute.
Solution
Homogeneous Liquid Mixture
Solvent – primary ingredient, usually water.
Solute – the substance to be dissolved
Solvent + Solute
H20
+
NaCl
Solution
NaCl (aq)
Solutions are Stable
NaCl + H20
NaCl(aq)
Once the salt is dissolved in the water, it will
never settle out…. The salt dissolved to
become a homogeneous mixture, the
particles are distributed throughout.
Aqueous Solutions
the symbol used in chemical equations is …..(aq)
Aqueous solutions use water as the solvent
Heats of Solutions
The Process of “Dissolving” involves breaking and
making Chemical Bonds… and that involves an
Energy Change.
Overall solution processes can be “Endothermic”
or “Exothermic”.
1- Breaking bonds of the solute
• endothermic
• requires energy
2- Forming bonds between H2O dipoles and solute
• exothermic
• gives off energy
Alloys
Mixture made up of Solids
Copper + Zinc
Copper + Tin
Iron + Carbon
Brass
Bronze
Steel
9 Types of Solutions based on Phase
(solvent/solute)
sol/liq
sol/sol
sol/gas
liq/liq
liq/sol
liq/gas
gas/liq
gas/sol
gas/gas
Separating Mixtures
Separation of Solution Components
1. Filtration
Separation by Size
2. Chromatography
Separation by differences in Attraction and Mass
Chromatography separates a
mixture because some
components have a greater
attraction to the paper… the
stronger the attraction, the
slower the compound
moves…. Mass can also
affect the speed at which a
particle moves… the larger
the mass, the slower it
moves.
3. Distillation
Separation by differences in Boiling Point
As one component reaches its boiling point, it evaporates and separates
from the other components…. As it cools, it condenses and can be
collected.
Decanting
Pouring off the liquid and leaving the solids behind.
Centrifuge
A centrifuge spins at a high rate of rotation and
separates solids by density… the denser material
will collect at the bottom of the collection tube.
Concentration and Molarity
Concentration
Concentration is the amount of a given substance in
a given quantity of a solution….
It’s the ratio of the solvent to the solute.
Concentration units
• mass % = g solute
g solute
g solution X 100 = g solute + g solvent X 100
•
molarity = M = mol solute
L solution
•
molality = m = mol solute
Kg solution
•
ppm = _____ grams solute________
–
1,000,000 grams of solution
Concentration by % Mass
(Percent Concentration)
Describes the amount of solute dissolved in 100
parts of solution.
g solute
mass %=
g solution
g solute
=
g solute + g solvent
Sample Problem
Determine the % mass of a solution containing 4.0 g
KCl and 46.0 g of H2O
Solute
+
Solvent
4.0 g KCl
46.0 g H2O
50.0 g KCl solution
Calculating Mass-Mass %
g of KCl
g of solvent
g of solution
=
=
=
4.0 g
46.0 g
50.0 g
% mass = 4.0 g KCl (solute) x 100 = 8.0% KCl
50.0 g KCl sol’n
Sample Problem
A solution contains 15 g Na2CO3 and 235 g of H2O?
What is the mass % of the solution?
1) 15% Na2CO3
2) 6.4% Na2CO3
3) 6.0% Na2CO3
Solution
mass solute
=
mass solution =
%
=
15 g Na2CO3
15 g + 235 g = 250 g
15 g Na2CO3
x 100
250 g solution
= 6.0% Na2CO3 solution
Sample Problem
An IV solution is prepared by
dissolving 25 g glucose (C6H12O6)
into 475 g water. What is the
percent mass of the glucose in the
IV solution?
1) 5.0%
2) 20.%
3) 50.%
Solution
1) 5.0%
% mass
= 25 g glucose x 100
500. g solution
= 5.0 % glucose solution
Sample Problem
• A solution of NaCl is needed. How many grams
of water must be added to 20.0 g of NaCl to
produced a 5.00% NaCl solution?
5.00% =
20.00 g NaCl x 100
(20.00 g NaCl + x g H2O)
5.00(20.00 + x) = 20.00 x 100
100. + 5x = 2000
x = 380 g H2O
Molarity
Molarity (M)
A concentration that expresses the
moles of solute in 1 L of solution
Molarity (M) = moles of solute
1 liter solution
Units of Molarity
2.0 M HCl
=
2.0 moles HCl
1 L HCl solution
6.0 M HCl
= 6.0 moles HCL
1 L HCl solution
Molarity Math Problems
3 types of Molarity problems
If you know the molarity
 1 - You can determine the number of moles (or grams) solute in a
given volume of solution.
 2 - You can calculate the volume of solution that contains a specific
number of moles solute if you know the molarity
If given the moles (grams) solute and volume of solution
 3 - determine the molarity
Calculating Molarity
Molarity Calculation
NaOH is used to open stopped sinks, to treat
cellulose in the making of nylon, and to
remove potato peels commercially.
 If 4.0 g NaOH (mm = 40.0 g/mol) are used to make 500.
mL of NaOH solution, what is the molarity (M) of the
solution?
 1 - calc the moles solute
 Mol = g/mm
 2 - Determine L of solution
 L = mL/1000
 3 - M = mol solute/L solution
Calculating Molarity
1. 4.0 g NaOH x 1 mol NaOH = 0.10 mol NaOH
40.0 g NaOH
2. 500. mL x 1 L
= 0.500 L
1000 mL
3. 0.10 mol NaOH = 0.20 mol NaOH
0.500 L
1L
= 0.20 M NaOH
Learning Check
A KOH solution with a volume of 400 mL contains 2
mole KOH. What is the molarity of the solution?
1) 8 M
2) 5 M
3) 2 M
Learning Check
A KOH solution with a volume of 400 mL contains
2 moles of KOH. What is the molarity of the
solution?
2) 5.0 M
M = 2 mole KOH = 5.0 M
0.4 L
Learning Check
 A glucose solution with a volume of 2.0 L contains 72
g glucose (C6H12O6). If glucose has a molar mass of
180. g/mole, what is the molarity of the glucose
solution?
 1 - Determine moles of solute (glucose)
 2 - Divide by L solution
1) 0.20 M
2) 5.0 M
3) 36 M
Solution
A glucose solution with a volume of 2.0 L contains
72 g glucose (C6H12O6). If glucose has a molar
mass of 180. g/mole, what is the molarity of the
glucose solution?
1) 72 g x 1 mole = 0.40 mol = 0.20 M
180. g
2.0 L
Calculating Moles from Molarity
Calculating Moles from Molarity
 Stomach acid is a 0.10 M HCl solution. How many moles of
HCl are in 1500 mL of stomach acid solution?
 1 - determine L of solution
 2 - convert from L to moles with molarity
1) 15 moles HCl
2) 1.5 moles HCl
3) 0.15 moles HCl
Solution
3) 1500 mL x
1 L = 1.5 L
1000 mL
1.5 L x 0.10 mole HCl = 0.15 mole HCl
1L
Calculating Mass from Molarity
Calculating Mass from Molarity
 How many grams of KCl are present in 2.5 L of 0.50
M KCl?
 1 - convert from L to moles solute, KCl
 2 - convert moles to grams of KCl
1) 1.3 g
2) 5.0 g
3) 93.0 g
Solution
3) 93.0 g
2.5 L x 0.50 mole x 74.6 g KCl = 93.0 g KCl
1L
1 mole KCl
Calculating Liters from Molarity
Calculating Liters from Molarity
 How many liters of stomach acid, which is 0.10 M
HCl, contain 0.15 mole HCl?
 1 - Convert from mol to L with molarity
1) 0.150 L
2) 1.500 L
3) 5.000 L
Solution
2) 1.500 HCl
0.15 mol HCl x 1 L soln = 1.500 L HCl
0.10 mol HCl
Molality (M)
Molality (m) = moles of solute
Kg of Solvent
Sample Problem
Example: A 1.00 molal solution of C12H22O11
(sucrose) is prepared by dissolving 1 mol of
C12H22O11 in 1.00 Kg of H2O.
What is the molality of a solution prepared by
dissolving 478 g of C12H22O11 (mm = 342 g/mol)
in 835 of H2O?
Plan: g sucrose
g H2O
g sucrose
Kg H2O
mol sucrose
Kg H2O
Dilutions
Dilutions Problems
Dilution means to add more solvent, usually water,
and reduce the solution’s concentration (M).
( Molarity x Volume (L) = moles solute)
Mole solute before = mole solute after
MbVb = MaVa
Sample Problem
If 750 mL of 0.50 M sodium chloride solution
evaporates until only 600 mL remains. What
will the new concentration (molarity) of the
sodium chloride solution?
Solution Stoichiometry
If M x vol (L) = moles
Then: mol A
mol B
gB
When 550.0 mL of a 0.600 M sol’n of NaCl is added to
AgNO3, solid AgCl is formed. Calc the g of AgCl
formed.
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
L NaCl  mol NaCl  mol AgCl  g AgCl
Molarity (mol/1.0 L)
Stoich ratio 1:1
mm (143.4g/1.0 mol)
0.5500 L x 0.600 mol x 1 mol AgCl x 143.4 g AgCl =
1.0 L
1 mol NaCl
1 mol AgCl
When 125.0 mL of a 1.50 M sol’n of CaCl2 is
added to Na2CO3, solid CaCO3 (mm = 100.1
g/mol) is formed. Calc the g of CaCO3 formed.
Na2CO3(aq) + CaCl2(aq)  CaCO3(s) + 2NaCl(aq)
Calculate the mass of water produced when 0.333
L of 0.500 M NaOH is added to acetic acid. The
equation is:
NaOH(aq) + HC2H3O2(aq)  NaC2H3O2(aq) + H2O(l)
Solubility
Solubility can be defined as ability of
one substance to dissolve in another at
a given temperature and pressure.
Expressed in the amount of solute that
will dissolve in a given amount of solute
to produce a saturated solution….
g/L, g/kg, mol/L, mol/kg, etc….
Solubility and Polarity
“Like dissolves Like”
Polar compounds dissolve Polar Compounds
Nonpolar compounds dissolve Nonpolar Compounds
Nonpolar compounds will not dissolve in Polar
compounds
Liquids in Liquids
Miscible – any two liquids such as ethyl alcohol and
water that are soluble in each other are called miscible.
Immiscible – any two liquids such as oil and water that
are insoluble in each other are called immiscible
Solubility of Solid Compounds
Saturated Solutions
Unsaturated Solutions
Supersaturated Solutions
know definitions
Saturations
An Unsaturated solution contains less solute than a
saturated solution and is able to dissolve
additional solute.
A Saturated solution cannot dissolve any more
solute under the given conditions.
A Supersaturated solution holds more solute than
what is required to reach equilibrium.
Saturation occurs at a point of
Solubility Equilibrium…. the rate of dissolution is
equal to the rate of recrystallization.
CuBr(s)
dissolution
recrystallization
Cu1+(aq) + Br1-(aq)
Solubility and Temperatures
How much solute
will dissolve into
100 g of solvent
(usually water) is
determined by
temperature and the
chemical properties
of the solute.
Temperature
Solubility
Temperature
Solubility
Direct relationship!
Solubility/Insolubility
“Like dissolves Like”
The more alike the IMFs holding
2 substances are, the more
likely they will dissolve in each
other.
Water, which is a dipole,
dissolves both ionic and
polar compounds
*Both have charged areas
Water won't dissolved
nonpolar compounds
Gases in Liquids
According to Henry's law if there is a constant
volume and temperature, an increase in
pressure will result in a proportional increase
in solubility.
Gas Pressure and Solubility
Pressure
Solubility
Pressure
Solubility
Electrolytes
aqueous solutions that conduct electrical currents
ionic compounds, salts,
are electrolytes if they dissolve in H2O
They dissociate into ions!
NaCl(s) --> Na+(aq) + Cl-(aq)
Nonelectrolytes
Molecular compounds
Molecular compounds dissolve as whole neutral molecules
C6H12O6(s) --> C6H12O6(aq)
No ions formed
Acids are the exception
Acids are “Molecular electrolytes”
They produce ions in water solution
They ionize!
HCl (g) --> H+(aq) + Cl-(aq)
Colligative Properties of Solutions
Colligative properties include freezing point
depression, boiling point elevation…..they
also include ….vapor pressure lowering,
and osmotic pressure.
Colligative Properties
These properties are dependent on the number
of solute particles in solution.
Based on “molality” – symbolized by “m”
Molality = mol solute
Kg solvent
Nonvolatile solutes, such as NaCl, depresses the freezing
point and elevates the boiling point.
Freezing Point Depression in Solutions
The freezing point depression (lowering) ΔTf is a colligative
property of the solution, and for dilute solutions is found to
be proportional to the molality of the solution:
ΔTf = Kf m
(Where Kf is called the freezing-point-depression constant… kf = 1.86°C/m)
The freezing point of pure water is 0°C, but that melting point
can be depressed by the adding of a solute such as a salt.
The use of ordinary salt (sodium chloride, NaCl) on icy roads
in the winter helps to melt the ice from the roads by
lowering the melting point of the ice. A solution typically has
a measurably lower melting point than the pure solvent.
Sample problems
If I add 45 grams of NaCl to 500 grams of water, what will the
boiling and freezing points be for the resulting solution?
(Since NaCl breaks into 2 ions when it dissociates in water, the
effective molality for the purpose of colligative property is
doubled….. In other words….after you calculate the
molality….”double” it!)
ΔTf = Kf m
(Where Kf is called the freezing-point-depression constant… kf = 1.86°C/m)
* Subtract your answer from 0 C to get the new freezing point!
ΔTb = Kb m
(Where Kb is called the freezing-point-depression constant… kb = 0.52°C/m)
* Add your answer from 100 C to get the new boiling point!
Solution
1. Find the concentration of the solution……Convert
grams of solute to moles and divide by the number
of kg of the solvent.
2. Because NaCl breaks into 2 ions you must multiply
your answer by 2.
3. Now just plug this molality (m) into the equation
and multiply by the constants.
You should get a 5.73 degree change for the melting
point, and a 1.60 degree change for the boiling
point.
I scream, you scream, we all
scream for ice cream.
An application of the freezing point depression is in the
making of homemade ice cream. The ice cream mix is
put into a metal container which is surrounded by
crushed ice. Then salt is put on the ice to lower its
melting point. The melting of the solutions lower the
temperature of the ice/water solution to the melting
point of the solution. This gives a temperature gradient
across the metal container into the saltwater-ice
solution which is lower than 0°C. The heat transfer
out of the ice cream mix allows it to freeze.
Boiling Point Elevation in Solutions
The boiling point elevation DTb is a colligative property of the solution, and
for dilute solutions is found to be directly proportional to the molality of
the solution:
DTb = Kb x m
(Where Kb is called the boiling-point-elevation constant… kb = 0.52°C/m)
The boiling point of pure water is 100°C, but that boiling point can be
elevated by the adding of a solute such as a salt. A solution typically has
a measurably higher boiling point than the pure solvent.
Solutions may be produced for the purpose of raising the boiling point and
lowering the freezing point, as in the use of ethylene glycol in
automobile cooling systems. The ethylene glycol (antifreeze) protects
against freezing by lowering the freezing point and permits a higher
operating temperature by raising the boiling point.
A Sample Problem
If I add 45 grams of NaCl to 500 grams of water, what
will the boiling point of the resulting solution?
DTb = Kb x m
(Where Kb is called the boiling-point-elevation constant… kb = 0.52°C/m)
* Add you answer to 100 C to get the new boiling point!
Surfactants
A Surfactant is a substance that concentrates at
the boundary surface between two immiscible
phases, such as solid-liquid, liquid-liquid, liquidgas….detergents are surfactants.
Surfactants allow oil to dissolve in water by
breaking the Hydrogen bonds between the
water molecules….. For the same reason,
surfactants lower the surface tension.
The surfactant’s lipophilic tails attach to the
fat molecules while its hydrophilic heads
attach to the surrounding water molecules.