SOLUTIONS
Chapter 14
MIXTURES
There are three basic types of mixtures: solutions, colloids, and suspensions. The particle size
determines the type of mixture. See the table below for differences between the three types.
Particle Type
Particle Size
Effect of Gravity
Effect of Light
Filtration
Examples
SOLUTIONS
COLLOIDS
ions, atoms, small
large molecules or
molecules
particles
0.1 – 1 nm
1 – 1000 nm
stable, will not settle stable, will not settle
out
out
no effect, light
passes through
cannot be filtered
scatters light;
Tyndall Effect
cannot be filtered
sugar water, salt
water, any ionic
substance in water
whipped cream,
milk, fog, jello,
blood, aerosol; (see
page 477)
SUSPENSIONS
large particles or
aggregates
>1000 nm
unstable, will
separate and settle
out upon standing
scatters light;
Tyndall Effect
particles can be
filtered out
oil and water; dirt
and water;
cornstarch and
water
TYNDALL EFFECT - particles dispersed in a mixture are big enough to scatter light.
SOLUTIONS
A solution is a homogeneous mixture of two or more substances in a single physical
state. (That means when a solid dissolves in a liquid, the solid is acting like a liquid – hence the
bit about the single phase.)
Solutions form when particles of solute are interspersed evenly throughout the solvent.
SOLVENT:
the substance doing the dissolving and in the greater amount. The most
common solvent is water. Called AQUEOUS.
SOLUTE:
the substance being dissolved and in less abundant.
SOLUBLE:
when a substance dissolves into another substance
INSOLUBLE: when a substance does not dissolve into another substance.
MISCIBLE:
two liquids that are soluble in each other.
IMMISCIBLE: two liquids that are insoluble in each other.
There are nine basic types of solutions based on the phases of the solvent. See page 454 for
other examples
Liquid Solutions:
1.
solid in liquid (salt water)
2.
liquid in liquid (vinegar)
3.
gas in liquid (carbonated drink)
Gaseous Solutions:
All gaseous mixtures are solutions
1.
solid in gas (soot in air)
2.
liquid in gas (humid air)
3.
gas in gas (air)
Solid Solutions
Homogeneous mixtures of solids are usually made from liquid solutions that have been
mixed and then solidified (frozen)
1.
solid in solid (alloys – brass, bronze)
2.
liquid in solid (dental fillings)
3.
gas in solid (charcoal gas mask)
SOLUTION EQUILIBRIUM
A solution is in DYNAMIC EQUILIBRIUM when the number of solute particles
returning to the crystal surface is equal to the number of solute particles leaving the crystal
surface. Saturation is dependent upon temperature and pressure.
A SATURATED solution is a solution with the maximum amount of solute dissolved at
a given temperature (Any more added goes to the bottom). It has reached dynamic equilibrium.
An UNSATURATED solution has the ability to dissolve more solute at a given
temperature
If a solution is SUPERSATURATED, then a hot solution is saturated and cooled. An
unstable condition results because the solution holds more solute that it normally does at a given
temperature.
Give an example of each type of solution:
Saturated:
Unsaturated:
Supersaturated:
======================================
VIDEO: Solutions
1. What are ion-dipole attractions?
2. During the process of solvation, how do ion-dipole attractions compare to the ionic bonds
inside an ionic crystal?
3. Describe the processes that are balanced when solubility equilibrium has been reached.
4. If more solute is added to a solution in equilibrium, will this always result in more dissolved
solute particles in the solution? Explain.
5. How does a solution become supersaturated?
6. After crystals precipitate out of a supersaturated solution, would you still call it
supersaturated? Explain.
THE DISSOLVING PROCESS:
How a solution forms
Dissolving occurs when the solute is pulled apart by the solvent. This takes place at the
surface of the solute. The solvent surrounds the solute. This process of surrounding the solute is
called SOLVATION. When the surrounding is done by water, this is called HYDRATION, a
particular type of solvation. When ionic compounds separate into their ions in a solvent,
DISSOCIATION occurs.
SOLUTE SOLVENT COMBINATIONS
Combinations of solute and solvent are based on the properties of each. There are four
main types:
1.
Polar Solvent – Polar Solute:
The polar solvent is attracted to the polar solute. The solvent gradually surrounds the
solute. The particles attach themselves due to polar attraction. Solvation occurs. Like
dissolves like.
2.
Polar Solvent – Nonpolar Solute:
Polar solvent particles are attracted to each other and not the solute. Solvation does not
occur and a solution is unlikely.
3.
Nonpolar Solvent – Polar Solute:
Nonpolar solvent particles have little attraction to the polar solute. Solvation does not
occur and a solution is unlikely.
4.
Nonpolar solvent – nonpolar solute:
Random motion of solute particles causes them to leave the surface of the solute and
become evenly dispersed in the nonpolar solvent. Solvation occurs. Like dissolves like.
VIDEO: Polar and Nonpolar Solvents (Disc chapter 30)
1. How would you predict whether carbon tetrachloride is a polar or nonpolar solvent? What
evidence have you observed that supports your prediction?
2. Based on chemical formula alone, can you tell whether iodine, I2, is a polar or nonpolar
solute? What evidence have you observed that supports your prediction?
3. Based on the chemical formula alone, can you tell whether copper (II) chloride, CuCl2, is a
polar or nonpolar solute? What evidence have you observed that supports your prediction?
SOLUBILITY
Solubility is how much solute can dissolve in a given amount of solvent. It is measured
in g/L or mol/L. It is usually the grams of solute per 100g of solvent. A CONCENTRATED
solution is said to have a high ratio of solute to solvent. A DILUTE solution is the opposite of
this.
SOLUBILITY CURVES




Solubility curves have
temperature and solubility
(usually in grams solute per
100g solvent) on their axes.
For solids, solubility generally
increases with increased
temperature.
For gases, solubility generally
decreases with increased
temperature.
Each line is a saturation line for
that substance indicating the
temperature and amount
dissolved.
FACTORS AFFECTING SOLUBILITY
1. Nature of the solvent and solute - “like dissolves like”. This means that polar solvent
dissolves a polar solute and nonpolar solvent dissolves nonpolar solute. But polar does not
dissolve nonpolar.
2. Temperature - increase the temperature and solubility increases (except gases).
3. Pressure - increase the pressure and you increase solubility (only with gases).
4. How much is already dissolved.
FACTORS AFFECTING THE RATE OF SOLUBILITY
(How fast something will dissolve)
1.
2.
3.
Agitation (shaking, stirring)
Increased temperature (except gases)
Smaller Particle size
These three allow more solvent to come into contact with the solute faster.
CONCENTRATION
Concentration is how many particles of a solute are dissolved. It is NOT dependent upon
the sample size and it can be measured. There are many units associated with concentration.
Below are a few:
Grams/100.0 grams
Parts per million (ppm)
Parts per billion (ppb)
Molarity
Molality
measures solubility
measures small concentrations
measures pollutants in small concentrations
used in lab chemistry
used for special calculation such as freezing point
depression
We will learn to calculate molarity, molality, mass percent, and mole fraction
MOLARITY, M
This is the ratio between the moles of dissolved substance and the volume of the solution
expressed in liters. It is the most useful measurement of concentration.
Molarity, M =
moles of solute
volume of solution in liters
 A one-molar (1M) solution of hydrochloric acid contains one mole of hydrochloric acid
in one liter of water.
 Which means it contains 36.46g of HCl in 1 liter of water.
Sample Problem:
Sandy dissolves 45.0 g of NaCl in 2.5 liters of solution. What is the concentration in
molarity of NaCl?
Mass = 45.0 g
V = 2.5 L
Mole Mass = 58.44 g/mol
Molarity = 45.0 g NaCl
1 mol NaCl
=
2.5 L
58.44 g NaCl
PRACTICE:
What is the molarity of 58.5g of NaCl dissolved in 2.0L of solution?
MOLALITY, m
This is concentration expressed in terms of moles of solute per kilogram of solvent.
Volume is not a factor.
Molality, m = moles solute
Kg solvent
A 1.0 molal aqueous sugar solution:
1 mole sugar
1 kg water
Sample Problem:
Calculate the molality of 98.0g RbBr in 0.824 Kg water.
m =
98.0 g RbBr
0.824Kg H2O
1 mol RbBr
165.38g RbBr
=
PRACTICE:
Calculate the molality of 85.2g SnBr2 in 140.0g water.
MASS PERCENT
Scientists frequently express the concentration of solutions in mass percent.
Mass % =
g of solute
x 100
g of solute + g of solvent
A 5% solution of sodium hydroxide contains 5g NaOH in each 100g of solution (95g
solvent and 5g of solute).
PRACTICE:
Calculate the mass percent of 98.0g RbBr in 824g water.
MOLE FRACTION
The concentration of solution can also be expressed in mole fractions.
Mole Fraction =
mole of solute
mole of solute + mole of solvent
PRACTICE:
Calculate the mole fraction of 98.0g RbBr in 824g water.
PREPARING AND DILUTING SOLUTIONS
A 3M solution of hydrochloric acid is not bought but made from 12M stock solutions. In
addition, a 1.0M solution of sodium hydroxide is made from a calculated amount of solid sodium
hydroxide added to water. It is important to know how to make different concentration of
solutions.
DISSOLVING A SOLID IN TO A LIQUID:
Prepare 1.0 liter of 1.0M NaOH
M = mol
L
1.0mol NaOH
1.0M = mol
1.0L
mol = (1.0M)(1.0L) = 1.0mol
40.00g NaOH = 40.00g NaOH
1 mol NaOH
So, you will put 40.00g
NaOH in a flask, add 1.0L
water, and mix well.
PRACTICE:
Prepare 1.0L of a 6.0M aqueous solution of KCl
DISSOLVING A LIQUID INTO A LIQUID:
To dilute a solution, you can form a ratio between molarity and volume.
M1V1 = M2V2
What volume would you use to make 0.500L of 6.0M HCl solution? You are essentially diluting
12.0M HCl to 6.0M HCl
V1 = M2V2 = (6.0M HCl)(0.500L HCl) = 0.250L HCl
M1
(12.0M)
Thus you would need 0.250L HCl and 0.250L water to make the solution.
PRACTICE:
What volume would you use of 12.0M HCl to make 1.0L of 0.10M HCl?
COLLIGATIVE PROPERTIES
“colligative” – depending upon the collection
These are properties that depend on the concentration (the number of) of the solute
particles. They are independent of the type of solute particle.
Three Factors:
1.
Vapor pressure reduction
2.
Boiling Point Elevation
3.
Freezing Point Depression
Vapor Pressure Reduction
A Review: Vapor pressure arises because some molecules of pure liquid leave the liquid surface
and enter the gas phase. At the same time, condensation occurs. If the rate of condensation and
the rate of vaporization are equal, then the system is in equilibrium. The gas pressure resulting
from the vapor molecules over the liquid is VAPOR PRESSURE.
The vapor pressure of a solvent containing a non-volatile solute is LOWER than the
vapor pressure of the pure solvent. It is NOT dependent upon the type (identity) of the solute.
WHY: The solute takes up space at the surface of the liquid and prevents some solvent
molecules from leaving the liquid. Now, more molecules leave the gas phase than enter it and
therefore, the vapor pressure is lower.
Boiling Point Elevation
A Review: The boiling point of a substance is the temperature at which the vapor pressure of a
liquid is equal to the external pressure on its surface (usually atmospheric pressure).
An addition of solute lowers that vapor pressure, therefore a higher temperature is
necessary to get the vapor pressure up to the external (atmospheric) pressure so that the solution
will boil. This is the boiling point elevation.
Freezing Point Depression
A Review: The freezing point of a substance is the temperature at which the vapor pressure of
the solid and liquid phases are the same.
An addition of solute lowers that vapor pressure, therefore a lower temperature is
necessary to get the vapor pressure of the liquid to equal the vapor pressure of the solid. When a
dissolved solute lowers the freezing point of its solution, you have freezing point depression.
CALCULATION
Dissociation factor
T = (K)(df)(m)
molality
constant (molal freezing pt or boiling pt constant)
 ∆T is directly proportional to the molality of the solution.
 Tbp is the difference between the boiling point of the solution and the boiling point of the
pure solvent. It is directly proportional to the number of solute particles per mole of solvent
particles.
 Tfp is the difference between the freezing point of the solution and the freezing point of the
pure solvent. It is directly proportional to the number of solute particles per mole of solvent
particles.
 The dissociation factor (df) is how many particle the solute breaks up into. For a non-ionic
compound (nonelectrolyte, organic compound), the df = 1.
 The molal constants for water are:
WATER
Freezing
Boiling
Normal BP and FP
0.00°C
100.00°C
Molal Constant
1.853°C/m
0.515°C/m
CALCULATING FOR A NON-IONIC SOLUTE
Determine the freezing and boiling point for 85.0g of sugar (C12H22011) in 392g water.
Need these things:
1.
2.
3.
Molality – need moles solute and kg solvent
number of particles (should be one)
Change in temp
1. 85.0 g C12H22011
1 mol C12H22011
0.392 g H2O
342.34 g C12H22011
= 0.633m
2. df = 1 (non-ionic solute)
3. Tfp = 1.853C 0.633 m =
m
0.00C –
100.000C +
Tbp = 0.515C 0.633 m
m
=
PRACTICE:
What is the freezing point for a solution of 210.0g of glycerol dissolved in 350.0g of
water? (The molecular mass of glycerol is 92.11 g/mol)
CALCULATING FOR AN IONIC SOLUTE
Determine the freezing and boiling point for 21.6 g NiSO4 in 100.0 g water.
Need these things:
1. 21.6 g NiSO4
0.1000g H2O
1.
2.
3.
molality
number of ions
change in temp
1 mol NiSO4
154.76 g NiSO4
=
2. There are two ions, Ni+2 and SO4-2, so the df is 2
3. Tfp = 1.853C 2 1.40 m
m
0.00C –
Tbp = 0.515C 2
m
1.40 m =
100.00C +
PRACTICE:
What is the boiling point of a solution containing 34.3 g of Mg(NO3)2 dissolved in 0.107
kg of water? (The formula mass of magnesium nitrate is 148.32 g/mol)
DETERMINING MOLAR MASS
Colligative properties provide a useful means to experimentally determine the
molar mass (molecular mass in one mole) of an unknown substance.
Steps:
1.
2.
3.
Solve for molality.
Solve for moles of solute.
Solve for molar mass
EXAMPLE:
A 10.0 g sample of an unknown compound is dissolved in 0.100 kg water. The
boiling point of the solution is elevated to 0.433oC above the normal boiling point of
water. What is the molar mass?
Step One:
Tbp = Kbp m
m=
Step Two:
m = mol solute
Kg solvent
mol solute = m x kg solvent
= 0.83 mol/kg x 0.100 kg
= 0.083 mol
Step Three:
mol solute =
= 0.433oC = 0.83 m (mol/kg)
0.52oC/m
mass solute
molar mass solute
molar mass = mass solute = 10.0 g = 120 g/mol
mol solute 0.083 mol
PRACTICE:
A solution of 3.39 g of an unknown in 10.00 g of water has a freezing point of –7.31oC.
The solution is a nonelectrolyte. What is the molar mass of the compound? (The molal
freezing point constant is 1.853C/m.)