7.1 Solutions
Name: _____________
What are solutions?
 a solution is a homogeneous mixture
 a homogeneous mixture is one where the particles are evenly mixed
 there are two components to a solution
What are the two components of a solution?
 a solution is made up of a solvent and a solute
 a solute is the component in a solution which exists in the smaller quantity
 a solvent is the component in a solution which exist in the greater quantity
 in general, in solution chemistry, the solid (solute) is dissolved in a liquid
(solvent)
Can we combine an infinite amount of solute with solvent?
 the amount of one substance that will dissolve in a certain amount of another at a
specific temperature refers to the solubility of a substance
 solubility varies with temperature  why?
 when no more solute will dissolve, the solution is saturated
Can liquids dissolve in a solution?
 yes  we refer to how liquids mix with one another as being miscible (soluble)
or being immiscible (insoluble)
 polar liquids will be miscible with another polar liquid e.g. water and alcohol
will mix in any proportions
 non-polar liquids will be miscible with another non-polar liquid e.g. salad oil and
motor oil will mix
 a non-polar liquid will not be miscible with a polar liquid e.g. oil and water will
not mix
 “like dissolves like”
What happens when an ionic solid is placed in water?
 ionic solids will dissociate when placed in water  why?
 water molecules can attach themselves to a surface ion and remove it from the
lattice
 when surrounded by attached water molecules, the ion is said to be hydrated
 a general term for this interaction between the solute and the solvent particles is
called solvation
 a typical dissociation equation:
 solvation can also happen to polar molecules (as opposed to ions) in water e.g.
MeOH/H2O
 solvation does not occur between polar and non-polar substances
 if both solvent and solute are non-polar, solvation may occur (through weaker
“Van der Waals” forces which are momentary dipoles caused by the nucleus of
one atom breifly attracting the electrons of another)
 e.g. benzene will dissolve moth balls (p-dichlorobenzene)
Converting between units of Solubility
 The solubility of Kbr is 1.3 x10-5 M (mol/L). What is this solubility in g/ml?
 Kbr = 119.00g/mol
 Complete problems on p. 366!
 Look at chart on p. 367 and complete problems on p. 368
 7.1 Review questions: 1-3, 6-8, 12 13
7.2 What Dissolves?
Name: _____________
Like Dissolves Like
 a solute will dissolve in a solvent if both contain similar types of intermolecular
forces of a similar magnitude
What are the forces involved in solutions?
 The forces between molecules are called intermolecular forces (between
molecules)
 There are 2 types:
o London forces (non polar molecules).
- Allows non-polar molecules to exist as liquids/solids bc there is a
force of attraction
- Induced dipole will induce more dipoles in surrounding molecules.
- Dipoles disperse throughout the sample which causes molecules to
attract each other = dispersion force, or London dispersion force
- Strength of London forces increase with as size of molecules
increase
o Dipole-dipole forces (polar molecules)
- (positive pole of one molecule is next to and attracts negative pole
of adjacent molecule)
- Network of dipole- dipole = high melting / boiling pts -> more
energy needed to overcome attraction
- The more polar the molecules are = stronger dipole – dipole force
What is the difference between the two?
 London forces are formed by inducing a dipole
 London forces exist between particles of all substances
 London forces have a temporary attraction between molecules
 Dipole-dipole forces are a result of molecules having a permanent dipole
 Molecules with dipoles have atoms with a difference in electronegativity
between the two atoms
Note: polar substances generally have higher boiling points than non-polar
substances
What is a hydrogen bond? (Special type of dipole – dipole force)
 a bond between polar molecules of a hydrogen atom of one molecule with
N, O, or F of another molecule
 The hydrogen bond is present in molecules containing a H-N, H-O, or H-F
bond
 The H-N, H-O, or H-F bonds are very polar bc of the large /\ En
In summary:
1. Hydrogen bonds are present when molecules contain H-F, O-H, or N-H
bonds.
2. When a permanent dipole is present, dipole-dipole forces and London forces
are present
3. When a permanent dipole is not present, only London forces are present (non
– polar molecules)
4. Londn forces are the weakest type of bonding force
Why does “like dissolve like”? (solute only dissolves in solvent if both have
similar intermolecular forces)
 Ionic solutes (e.g. NaCl) are held together by strong bonds
 Nonpolar solvents have only London forces between molecules – too weak
 Polar solvents have dipole-dipole forces between them – strong enough to
dissolve ionic solids
 Ions will attract opposite ends of polar water molecule = ion – dipole force
 Ion – dipole force stronger than attraction between ions in crystal lattice ->
crystal lattice breaks down
 This is also true for polar solutes
 Nonpolar solutes are only attracted by London forces
 Only nonpolar solvents have strong London forces
Three important steps for a substance to dissolve:
1. The particles in the solute must be separated to go into the solvent. The
attraction between solute particles must be overcome or replaced.
2. The particles in the solvent must be separated to allow space for the solute
particles. The attraction between solvent particles must be overcome or
replaced.
3. The solute and solvent particles must interact with each other.
Using your workbook take notes on the following cases (p.373-379)
NaCl (s) in Water
Ethanol in Water
For solute to dissolve, attraction bn Na+ and Cl- must
be overcome
Water molecules attracted to each other by H-bonds
Opposite ends of H2O and NaCl attract each other (
ion –dipole force)
Na+ and Cl- become hydrated
All ionic salts are soluble in water
Oxygen bonded to hydrogen at end of hydrocarbon
chain
Ionic compound OH= hydroxide, ethanol OH
=hydroxyl
Oh at end of alcohol = polar -> forms H-bonds with
nearby ethanol molecules
Hydrocarbon chain = non – polar end
Water molecules = polar = attracted to each other by
H-bonds
H-bonds b/n H2O and ethanol similar strength to Hbonds in ethanol – ethanol and H20-H20 = > water
and ethanol are miscible
Longer hydrocarbon chain = more non –polar = less
soluble (H-bonds weaker = less attraction to OH)
More OH groups = more soluble in water
Formaldehyde in Chloroform
Iodine in Carbon Tetrachloride
Formaldehyde = polar covalent, molecules attracted by
dipole-dipole force
Chloroform = polar covalent (dipole –dipole)
Similar strengths of dipole –dipole forces = soluble
Iodine is non – polar (only have weak London forces)
so only have temporary dipole
CCl4 = non – polar
Similar London forces b/n molecules of I2 and CCl4 =
soluble
Ionic and polar-covalent solutes dissolve in polar
covalent solvents
Non – polar covalent solutes only soluble in non –
polar solvents
Vitamins in Fat and or Water
Soaps
Vitamin A = long hydrocarbon with a polar end
Long hydrocarbon = non –polar = only london forces
(weak)
Fat molecules = non – polar molecues = weak London
forces
Fat and Vitamin A = soluble (can dissolve)
Vitamin A = non soluble in water (non – polar & polar
don’t mix
Organic molecules with pos. and neg. ions
Each neg. ion has long hydrocarbon ‘tail’ (non –polar)
and ‘head’ (polar)
COO- group = polar ->interacts with H20 (ion –
dipole, H-bonding)
Tail = non-polar -> dissolves other non-polar grease
p.381: 1-4, 6a, 8, 9
7.3 Dissociation and Solution Conductivity?
Dilution Calculations
It is often necessary to change the concentration of a solution e.g. you need a 2.5M
solution of HNO3 but you have an 8.0M solution. You have to be able to calculate
exactly how much you must dilute the original solution to get the concentration you
desire.
 in a dilution calculation, the number of moles before and after dilution will
remain the same, therefore, we can use the following equation to do dilution
calculations
M1V1 = M2V2
Where M = molarity of solution and V = volume of solution
Example 1
A student has 500 ml of a 0.40M NaCl solution. How much water is needed
to make it a 0.10M solution?
Solution
M1 = 0.40M
V1 = 0.500L
M2 = 0.10M
0.40M  0.500L = 0.10M  V2
V2 = (0.40M  0.500L)/0.10M
V2 = 2.0L
volume of water to be added = 2.0L – 0.5L = 1.5L
V2 = ?
Example 2
A chemist adds water to a 120 ml of a 6.0M solution of NaOH until the final
volume is 2.0L. What is the molarity of the resulting solution?
Solution
M1 = 6.0M
V1 = 0.120L
M2 = ?
V2 = 2.0L
6.0M  0.120L = M2  2.0L
M2 = (6.0M  0.120L)/2.0L =0.36M
Example 3
A chemistry teacher needs 5.0L of 0.10M HCl for a class experiment. A
supply of a 12M solution of concentrated hydrochloric acid is available. What
volume of the concentrated acid should be measured out to add to water to make up
the final volume of 5.0L?
Solution
M1 = 0.10M
V1 = 5.0L
M2 = 12M
V2 = ?
0.10M  5.0L = 12M  V2
V2 = (0.10M  5.0L)/12M = 0.042L
Example 4
What concentration results when 150 ml of a 0.36M solution of magnesium
sulfate, MgSO4, are added to 750 ml of water?
Solution
M1 = 0.36M
V1 = 0.150L
M2 = ?
0.36M  0.150L = M2  0.900L
M2 = (0.36M  0.150L)/0.900L = 0.060M
P. 384 Practice problems 1-4
V2 = 0.900L
Concentration of Ions
 in a 1.0M calcium nitrate solution the dissociation equation is given by the
following:
Ca(NO3)2 (s)  Ca+2(aq) + 2NO3-(aq)
 if the concentration of Ca+2 is 3.0M, then the concentration of NO3- is 2  3.0M =
6.0M
 we use square brackets [ ] to indicate the concentration of a substance in
moles/litre e.g. [NO3-] = 6.0M
 when two solutions are mixed, the volume increases and this reduces the
concentration of any dissolved particles i.e. same number of particles are in a
larger volume of solution
Example 1
300ml of 2.0M NaOH is mixed with 600ml of 5.0M MgI2. (a) What is the
concentration of each substance in the resulting solution? (b) What is the
concentration of each ion in the resulting solution?
Reactions in Solutions
 some substances have such a great attraction between the ions that they will not
dissociate in water
 two solutions may also contain ions that make up an insoluble substance when
combined  these ions attract each other and a precipitate is formed
 CaSO4 is considered insoluble  only 1.62 g will dissolve in 1L of hot water
(=0.01mol/L)
 anything with a solubility < 0.1mol/L is considered insoluble
 consider a reaction between Ca(NO3)2 and K2SO4
 Ca2+ and SO42- react to form solid CaSO4
 K+ and NO3- remain in solution  these are called spectator ions because they
do not take part in any reaction
 the ionic equation for the above reaction is:
Ca2+(aq) + 2NO3-(aq) + 2K+(aq) + SO42-(aq)  CaSO4(s) + 2K+(aq) + 2NO-(aq)
 to simplify, we may leave out the spectator ions to get the net ionic equation:
Acids and Bases
Acids and bases are two very large categories of compounds that can be
separated by their chemical behaviour. There are different definitions of what
makes a compound an acid and what makes a compound a base.
The Lewis Definition:
A Lewis acid is a compound that can accept a pair of electrons.
A Lewis base is a compound that can donate a pair of electrons.
e.g. H+ and OH- reacting to form H2O
The Bronsted-Lowry Definition:
A Bronsted-Lowry acid is a compound that can donate a proton.
A Bronsted-Lowry base is a compound that can accept a proton.
e.g. HNO3 is dissolved in H2O.
e.g. NH3 is dissolved in H2O.
Compounds that can both accept or donate a proton are called amphiprotic. Water
is an example of an amphiprotic compound.
The Arrhenius Definition:
An Arrhenius acid is a substance that can produce hydronium ions in solution.
An Arrhenius base is a substance that can produce hydroxide ions in solution.
e.g. A generic Arrhenius acid dissociates according to:
HA  H+ + Awhere the compound HA is the Arrhenius acid.
e.g. A generic Arrhenius base dissociates according to:
XOH  X+ + OHwhere the compound XOH is the Arrhenius base.
Properties of Acids and Bases:
Acids
Bases
 all taste sour
 turns blue litmus red
 acids destroy the chemical
properties of bases
 acids conduct electricity
 chemically reacts with active
metals to form H2
 all taste bitter
 turns red litmus blue
 bases destroy the chemical
properties of acids
 bases conduct electricity
 chemically reacts with fatty oils
in your skin to feel slippery
The Acid-Base Reaction:
When equal amounts of acids and bases react with each other, a neutralization
reaction occurs:
Acid + Base  Salt + Water
or
HA + XOH  XA + H2O
Indicators such as phenolphthalein and bromothymol blue will change colour to
indicate changes in the aciditiy of a solution. Red and blue litmus paper will
indicate changes in aciditiy as well. Acidity is measured on a scale called the pH
scale.
Electrical Conductivity of Solutions
 Some solutions can carry an electric charge thus conduct electricity
 Electric charge is carried through a solution by the movement of ions that
complete an electric circuit
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A solute that dissociates to form ions will conduct electricity
Salts with high solubility are will form solutions that are good conductors
A solution that conducts electricity well is a good electrolyte
Solutions that do not conduct electricity well are called weak electrolytes
Solutions that does NOT conduct electricity is called a nonelectrolyte
The positive ions move toward the negative electrode and the negative ions move
towards the positive electrode
Electrical Conductivity of Acids and Bases
 Acids and bases can conduct electricity
 Strong acids dissociate completely to form ionic solutions thus are good
electrolytes
 Weak acids do not dissociate completely; they stay mostly intact and only a
small number of molecules will dissociate to form ions
 The more ions the higher the electrical conductivity
 Covalent compounds do not from ions thus do not usually conduct electricity in
solution- the exception to this is some acids and bases
7.4 Titrations
When doing an acid-base reaction, often there is one reactant where the
concentration is unknown. With a known concentration of the other reactant, a
technique known as titration can be used to find the exact concentration of the
unknown.
 the solution whose concentration is known is called the standardized solution
 the equivalence point is the point where the amount of acid equals the amount of
base present in solution
 to indicate the equivalence point, an indicator is used  the indicator will
change colour at the equivalence point this is known as the endpoint
Solving Titration Problems
Mrs. Baker completely reacted 50.00mL of HCl with 90.00ml 0.300M NaOH.
Calculate the [HCl] (square brackets mean concentration).
- Write a balanced equation
- Calculate the moles of the solution you know (standardized solution)
o moles NaOH
- Use the balanced equation to convert from the standardized solution to the
unknown
o from moles NaOH to moles HCl
- Calculate the concentration by dividing the moles of unknown by litres of
unknown
o divide moles HCl by litres HCl from the question
p. 394 Practice Problems: 1-4
p. 399 Review Questions: 1-8,10