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Solutions are homogeneous (1 phase)
mixtures where 1 of the components
(solvent) is found in larger quantities
than the rest.
All other components are said to be
dissolved in the solvent. These
components are called solutes.
Both solutes and solvents can
be liquids, solids, or gases.
Create a chart with solutes
along the side and solvents at
the top which shows examples
of:
gas in gas, gas in solid, gas in
liquid, liquid in gas, etc.
SOLVENT
S Gas in
O
L
Liquid
U
in
T
E Solid in
Gas
Liquid
oxygen in
air
(nitrogen)
water in
air
oxygen in
water
Invisible
dust in air
Sugar in
water
(syrup)
Solid
air bubbles
in ice
alcohol in mercury in
water
silver
tin in
copper
(bronze)
When solutes are dissolved in solvents the
solutes formula is written followed by a
bracketed subscript which follows.
Examples:
magnesium chloride is dissolved in water
MgCl2(aq)
iodine is dissolved in alcohol
I2(al)
Aqueous solutions have water as the solvent.
They are always indicated by (aq) after the
formula.
Properties of Solutions
The ability to conduct electricity can be used to
classify solutions.
Electrolytes are substances which conduct
electricity when dissolved in water.
Ionic compounds are electrolytes and most
molecular compounds are non-electrolytes.
Solutions can also be categorized as acidic,
basic or neutral. Litmus paper can be used in
this determination.
Why does solid NaCl dissolve easily in water?
H and O atoms in water
molecules do not share electron
pairs equally.
H
O
H
Why does solid NaCl dissolve easily in water?
H and O atoms in water
molecules do not share electron
pairs equally.
-ve
H
O
+ve
Water molecules
+ve
H H
H
have oppositely
O
charged ends. -ve
They are polar molecules
Moving water molecules collide with the ions of
Na and Cl in solid NaCl crystals.
Why does solid NaCl dissolve easily in water?
H H
O
Here is a small crystal of NaCl Na1+ Cl1- Na1+
Drop the crystal in a
container of water
Cl1- Na1+ Cl1Na1+ Cl1- Na1+
Here is a small crystal of NaCl Na1+ Cl1- Na1+
Drop the crystal in a
container of water
Cl1- Na1+ Cl1Na1+ Cl1- Na1+
If the positive end of a water molecule strikes
a chloride ion with enough energy it pulls it
away. The same thing happens if the negative
oxygen end of a water molecule strikes a Na1+
ion.
Na1+ Cl1- Na1+
Na1+ Cl1Cl1- Na1+
These types of interactions are called
intermolecular and the NaCl crystal is
dissociating.
In reality each ion of Na and Cl become
surrounding by a number of water molecules.
Na1+ Cl1- Na1+
Na1+ Cl1Cl1- Na1+
These complexes are called hydrated ions.
All ions in water become hydrated.
Cl1- Na1+
Cl1-
Na1+ Cl1Na1+
Some substances do not
easily dissolve in water.
When air is exhaled in water it does
not easily dissolve.
Why?
Air is made up mostly of nitrogen and oxygen.
N2 and O2. Since they don't dissolve easily in
water they must be non-polar.
N N
Since all three pairs of
electrons are equally
shared this molecule is
non-polar.
Both N2 and O2 molecules are non-polar so they
are not strongly attracted by polar water
molecules.
In general like dissolves like.
Polar materials dissolve easily in polar solvents
and non-polar materials dissolve easily in nonpolar solvents.
Water is often called the universal solvent
because it dissolves so many different
substances.
This is due to the strong forces of attraction
water molecules have on each other and on
positive and negative particles in other
substances.
The H end of one water
molecule is strongly attracted
-ve to the O end of another water
H O
molecule.
+ve
H
The special force of attraction
is called a hydrogen bond and
-ve it occurs between molecules of
H O
substances with H and O, or H
+ve
H
and N, or H and F.
Any molecular substance containing O atoms
bonded to H atoms has polar regions which
exert these attractive H bonds.
For instance alcohols have OH groups. This
allows them to easily mix with water.
CH3OH
+ve
-ve
Attractive force
Any molecular substance containing O atoms
bonded to H atoms has polar regions which
exert these attractive H bonds.
For instance alcohols have OH groups. This
allows them to easily mix with water.
CH3OH
H bond
Alcohol will dissolve in water but this solution
does not conduct electricity. Why?
There are no mobile ions present.
CH3OH
H bond
Solutions are homogeneous mixtures.
Solutions involve 2 components. The substance
doing the dissolving (solvent) and the substance
being dissolved (solute).
Typically the amount of solute dissolved is
measured and compared to the total volume of
solution. This quantity is known as the
concentration of solution.
A 710 mL bottle of coke has 30 g of sugar.
What is the concentration in g/L (M/V)?
30 g / 0.710 L = 42 g / L
What mass of sugar is there in a 355 mL can of
coke?
42 g / L x 0.355 L = 15 g of sugar
Some solutions, like alcohol mixtures, list the
quantity of alcohol as a percentage by volume
since this number is bigger than the percentage
by mass for solutes with a density smaller than
water.
A can of regular beer is 5% (V/V) alcohol by
volume. What volume of alcohol is their in a
355 mL can of beer?
5/100 x 355 mL = 18 mL of alcohol
Which has more alcohol
45 mL of 40% (V/V) rye whiskey (typical shot)
310 mL of 7% (V/V) vodka cooler
341 mL of 4% (V/V) Coors light bottle of beer
180 mL of 12% (V/V) glass of red wine
45 mL x 40/100 = 18 mL in shot of rye
310 mL x 7/100 = 21.7 mL in the cooler
341 mL x 4/100 = 13.6 mL in the beer
180 mL x 12/100 = 21.6 mL in the wine
A 1.0 L sample of water is found to have 0.0012
g of lead.
The molar concentration works out to be a very
small number. To avoid using really small
numbers for concentrations of dilute solutions
another more practical scale is used. This scale
is called parts per million.
What is the ppm of lead for the example above?
1000 L would have 1.2 g of lead so it is 1.2 g in
1000 L or 1200 mg / 1000 L or 1.2 mg / 1.0 L
ppm can be expressed in a variety of ways
1 ppm = 1 g/1000 L or
1 ppm = 1 g / 1000 000 mL or
1 ppm = 1 g / 106 mL or
1 ppm = 1000 mg / 1000 L or
1 ppm = 1 mg / L
Calculating ppm
In a chemical analysis 3.4 mg of lead was
found in 100 mL of tap water. Find the ppm of
lead.
ppm = 1 mg/L = 3.4 mg / 0.1 L = 34 ppm
What fraction of a part per million (ppm) is a
part per billion (ppb)?
1/1000
So 1 ppm = ? ppb
1 ppm = 1000 ppb
An even smaller concentration unit is a part
per trillion (ppt)
1000 ppb = 1 ppt
1 mg in 1.0 L is 1 ppm
1 mg in 1000 L is a ppb
1 mg in 1 000 000 L is a ppt
Measuring Quantities of Solutes in Solutions
The quantity of solute can be measured in
grams or moles. The total volume of the
solution is measured in L. The amount of
solute in a given volume of solution is
measured using these units:
g
L
or
mol
L
or molL-1
=
kmol
m3
=
mol
dm3
kmolm-3 moldm-3
= M
This leads to the development of the
following equation:
Concentration
=
of a solution
C
# of moles of solute
Volume, in L, of solution
=
n
V
Preparing Solutions From Solid Reagents
Sample Problem
Describe how to prepare 500 mL of a
0.035 M solution of sodium thiosulfate.
Given: V = 500 mL = 0.500 L
C = 0.035 M
Asked to Find: Mass of Na2S2O3
Mass
Use n= m/MM
Mole
Use C = n/V
Concentration
Preparing Solutions From
Solutions
Determining Concentrations of
Concentrated Reagents
Concentrations of solutions, in
-1
molL , can be determined from
density and percentage
composition.
Sample Problem
A solution of concentrated
(conc.) HCl (hydrochloric
acid) has a density of 1.25
g/mL and it is 35% HCl by
mass. Find the concentration
of the HCl.
Given: density = 1.25 g/mL, 35% HCl
Change density into units of
mass and volume
m = 1.25 g, V = 1 mL =
Mass
0.00100 L
Mole
Concentration
Step 1 - Find mass of HCl
35% of 1.25 g = 0.4375 g
Step 2 - Find nHCl = m/mm
= 0.4375 g/ 36.45 g/mol = 0.01199 mol
Step 3 - Find C = n/V
= 0.01199 mol/ 0.00100 L
= 12 M
Describe how to prepare 1.5 L of
0.75 M HCl from this
concentrated reagent.
Solution: Find the volume of the
concentrated reagent needed to
prepare the solution.
Given:
Cd = 0.75 M , Vd = 1.5 L
Cc = 12 M, Vc = ?
# of moles
Concentrated
Reagent
=
# of moles
Diluted
Reagent
CcVc = CdVd
Cc Cc
Vc = 0.75 M x 1.5 L
12 M
= 0.094 L = 94 mL
1. Get a 1.5 L Volumetric Flask
2. Measure 94 mL of concentrated HCl
using gloves, apron, shield
3. Half fill the 1.5 L Volumetric flask
with distilled water
4. Add the 94 mL of conc. HCl
5. Top up with distilled water to the
calibration mark.
AW not WA
Describe how to prepare
2.0 L of a 1.5 M solution
of ammonium hydroxide
from a concentrated
reagent which is 14.5 M.
# of moles
Concentrated
Reagent
=
# of moles
Diluted
Reagent
CcVc = CdVd
Cc Cc
Vc = 1.5 M x 2.0 L
14.5 M
= 0.207 L = 210 mL
1. Get a 2.0 L Volumetric Flask
2. Measure 210 mL of concentrated
NH4OH using gloves, apron, shield
3. Half fill the 2.0 L Volumetric flask
with distilled water
4. Add the 210 mL of conc. NH4OH.
Top up with distilled water to the
calibration mark.