UNIT 3-SOLUTIONS & SOLUBILITY
SOLUTIONS – homogenous mixtures of substances composed of one solute and one solvent.
SOLUTE – a substance that gets dissolved in a solvent.
SOLVENT – the medium in which the solute gets dissolved in.
- often is a liquid e.g. water, paint thinner, etc.
HOMOGENOUS MIXTURE – a uniform mixture consisting of only one phase
- usually clear
HETEROGENOUS MIXTURE – a mixture where one can see distinct phases
usually opaque or translucent (cloudy)
e.g. oil in water, salad dressing, etc.
-
SOLUTE IN SOLVENT EXAMPLE OF SOLUTION
Gas in Gas
Gas in Liquid
Gas in Solid
Liquid in Gas
Liquid in Liquid
Liquid in Solid
Solid in Liquid
Solid in Solid
Oxygen in Nitrogen(air)
Oxygen in Water
Oxygen in Ice
Water in Air
Methanol in Water(anti-freeze)
Mercury in Silver(tooth fillings)
Sugar in Water(syrup)
Tin in Copper(bronze)

 a chemical formula representing a solution specifies the solute by using its chemical formula and shows the solvent by using a subscript. e.g. (1) NH
3(aq)
– ammonia gas (Solute) dissolved in
water(solvent)
(2) I
2(al)
– solid iodine(solute) dissolved in
alcohol(solvent)
 all aqueous solutions have water as the solvent and are clear
(transparent)
 compounds can be classified as either electrolytes or nonelectrolytes
 compounds are electrolytes if their aqueous solutions conduct
electricity and compounds are non-electrolytes if their aqueous solutions do not conduct electricity.
 electrolytes are mostly highly soluble ionic compounds
 most molecular compounds are non-electrolytes except for acids
PROPERTIES OF SOLUTES & THEIR SOLUTIONS
TYPE OF SOLUTE CONDUCTIVITY TEST
 light on conductivity apparatus glows, needle on ohmmeter moves
 light does not glow, needle does not move
TYPE OF SOLUTION LITMUS TEST
 blue litmus paper turns red
 red litmus paper turns blue
 no change in colour of litmus paper
HOMEWORK: P. 269 & 270 #’s 1-8.

 when we look at polar & non-polar solutes we generally find that
“like dissolves like” e.g. ethylene glycol is a polar molecule which has a high solubility(which also is polar)
 the theoretical explanation is that “polar molecules are surrounded and suspended in solution by polar solvent molecules”
 remember in Chapter 2…….H can be covalently bonded – N
- O
- F
 any of these compounds can “hydrogen bond” to their own molecules to increase the intermolecular forces more than either
London Dispersion forces or dipole-dipole forces. e.g.

 many more ionic compounds dissolve in water than any other solvent
 in Sweden, Arrhenius studying electrolytes proposed a hypothesis that when a substance dissolved, its particles separated from each other and dispersed into the solution
 he further explained that electrically charged particles must be present in their solutions e.g. NaCl
(s)
-- Na
(aq)
+ + Cl
(aq)
-
and (NH
4
)
2
SO
4(s)
- 2NH
4(aq)
+ + SO
4(aq)
-
 when an ionic compound dissolves we say that the compound
dissociates into individual aqueous ions
DO #’s 6 & 7 on p. 278 & 279.
 Why do polar solutes dissolve in non-polar solvents?
 they do not show evidence of polar molecules (dipoles) or hydrogen bonding
 we would then have to look at London Dispersion Forces which are weak intermolecular forces
WATER = “UNIVERSAL SOLVENT”
 due to its;
1.
small size
2.
highly polar nature
3.
considerable capacity for hydrogen bonding
Do #’s 9-12 on pages 279 & 280.

 because many solutions are colourless, we need to know the amount of solute that is present
 we use a ratio called the concentration
Concentration = Quantity of Solute
Quantity of Solvent
 if the concentration is high, we say that we have a concentrated solution
 if the concentration is low, we say that we have a dilute solution
–
1. c= V solute
x 100%
V solution
Concentration is expressed as
% V/V or percentage by volume
c= concentration
V solute
= volume of solute
V solution
= volume of the solution
Q: A photographic “Stop Bath” contains 140mL of pure acetic acid in a 500mL bottle of solution. What is the percentage by volume concentration of acetic acid?
A: c acetic acid
= 140mL x 100%
500mL
= 28.0% V/V
2.
 consumer products commonly use “percent weight by volume” or
% W/V e.g. Hydrogen Peroxide, H
2
O
2
, may be 3% W/V which means 3g of
H
2
O
2
is in every 100mL of solution c
H2O2
= 3g/100mL=3% W/V i.e. c=m solute
x 100%
v solution

3.
“percent weight by weight” or % W/W
c=m solute
x 100%
m solution
Q: A sterling silver ring has a mass of 12.0g and contains 11.1g of pure silver. What is the percentage weight by weight concentration of silver in the metal?
A: m
Ag
=11.1g
m alloy
=12.0g
c=11.1g x 100%
12.0g
c= 92.5% W/W
Therefore the ring is 92.5% silver.
 very low concentrations(dilute concentrations) are expressed using reasonable numbers for very small quantities of solute
 in the environment of swimming pools, etc. you often express concentration as parts per million (ppm) or 1:10 6
 we can express this depending on the units we are working with i.e. 1ppm=1g/10 6 mL
=1g/1000L
= 1mg/L
= 1mg/1kg
= 1μg/g
Q: In a chemical analysis of 250mL of water at SATP, 2.2mg of oxygen was measured. What is the concentration of oxygen in parts per million?

A: m
O2
=2.2mg
V
O2
= 250mL or 0.250L
c= 2.2mg
0.250L
= 8.8mg/L = 8.8ppm
Therefore the oxygen concentration is 8ppm.
 when we talk about chemical reactions the coefficients in a balanced equation refer to the number of moles
 concentration is communicated using Molar Concentration or C
 the molar concentration, C , is the amount of solute(in moles) dissolved in one litre of solution or
Molar Concentration = Amount of Solute (moles)
Volume of Solution (litres)
C=n/v C is expressed as moles/L
 Molar Concentration is sometimes indicated by the use of square brackets e.g. [NaOH
(aq)
]
Q: In a quantitative analysis, a stoichiometric calculation produced a
0.186mol of NaOH in 0.250L of solution. Calculate the molar concentration of NaOH?
A: n
NaOH
= 1.186mol
v
NaOH
= 0.250L
C
NaOH
= n/v
= 0.186mol/0.250L
= 0.744mol/L
Therefore the molar concentration of NaOH is 0.744mol/L

 in some cases, you might have to work with the equations and
manipulate them to find the answer
 in other cases, you might have to use the concentration ratio
(quantity of solute/quantity of solution) as a conversion factor.
Q: A box of apple juice has a fructose concentration of 12g/100mL
(12%W/V). What mass of fructose (C
6
H
12
O
6
) is present in a 175mL glass of juice?
A: c
C6H12O6
=12g/100mL c=m solute
v solution
V=175mL
m solute
=c x v solution
=12g/100mL x 175mL
=21g
Therefore there are 21g of fructose in a 175mL glass of apple juice.
Q: People with diabetes have to monitor and restrict their sugar intake. What volume of apple juice could a diabetic person drink, if their sugar allowance for that beverage was 9.0g? Assume that the apple juice has a sugar concentration of 12g/100mL (12% W/V) and that the sugar in apple juice is fructose.
A: c
C6H12O6
=12g/100mL m
C6H12O6
=9.0g
c=m solute
/v solution
--- v solution
=m solute
c
=9.0g
12g/100mL

=9.0g x 100ml/12g
=75mL
Therefore the person could drink 75mL of apple juice.
 when given the mass of a substance, volume of a substance and the molar concentration, we sometimes have to use 2 conversion factors;
1.
molar mass
2.
molar concentration to solve the problem
Q: To study part of the water treatment process in a lab, a student requires 1.50L of a 0.12 mol/L aluminum sulphate solution. What mass of aluminum sulphate must be measured for this solution?
A: v
Al2(SO4)3(aq)
=1.50L
C
Al2(SO4)3
=0.12mol/L
M
AL2(SO4)3
=342.14g/mol
Step 1: C=n/v, n
AL2(SO4)3
=C x v
=0.12mol/L x 1.50L
=0.180mol
Step 2: n=m/M, m
Al2(SO4)3
=n x M
=0.180mol x 342.14g/mol
=61.6g
Therefore the student will need 61.6g of aluminum sulphate.
Q: Sodium carbonate is a water softener that is a significant part of the detergent used in a washing machine.

A student dissolves 5.00g of solid sodium carbonate to make 250mL of a solution to study the properties of this component of detergent.
What is the molar concentration of the solution?
A: m
Na2CO3
=5.00g
V
Na2CO3
=250mL
M
Na2CO3
=105.99g/mol
Step 1: n=m/M, n
Na2CO3
=5.00g/105.99g/mol
=0.0472mol
Step 2: C=n/v, C
Na2CO3
=0.0472mol/0.250L
=0.189mol/L
Therefore the molar concentration of the solution is 0.189mol/L.
Answer questions 19-22 on page 290.
 70% of the Earth is covered with a dilute aqueous solution averaging about 4km deep
 very little of this is useful for drinking.
 Only about 2% of Earth’s water is freshwater from lakes and rivers
 about 0.06% more is ground water which has soaked into the soil and porous rock of the Earth’s crust
 in Canada we have a very abundant supply of freshwater, which can be divided into 2 types;
1.
Surface water-Great Lakes, rivers, streams, etc
2.
Ground water-made up of thousands of aquifers*
*an aquifer is an underground formation of loose material or permeable rock that can produce useful quantities of water when tapped by a well.

 sometimes water will naturally contain substances like living organisms, suspended particles and naturally occurring dissolved chemicals from surrounding rocks and minerals
e.g. Ca 2+ , Mg 2+ , Na 1+ ……common cations
HCO
3
-1 , SO
4
2, Cl -1 ….common anions
 true contaminants are classified into 3 categories;
1.
Biological-viruses, bacteria & algae
2.
Physical-suspended particles (make water look dirty)
3.
Chemical-all dissolved substances including metals,
manufactured chemicals, etc
 sometimes damage from chemicals can be reversed but many like
PCB’s, dioxins, heavy metals and some pesticides cause damage that CAN NOT be reversed!

 TABLE 1 on P. 293 lists common sources of water contamination.
 All levels of Government have some degree of responsibility in ensuring the safety of our drinking water!
 Federal-Provincial Governments have set-up MAC (Maximum
Acceptable Concentrations) lists of chemicals in our water.
Read Section 6.4 “Drinking Water” P. 291-298 & answer questions 1-4, & 6 on pages 294/295. Also questions 11 & 12 on page 299.
 Solutions with precisely known concentrations are called
Standard Solutions
 these are in many cases made by dissolving a solid solute into a liquid or aqueous solution
 done by diluting an existing solution
 commonly done by taking a standard solution or stock solution & adding solvent to decrease the concentration to the level we need
 Solutions with a molar concentration of < 0.1mol/L are often described as dilute, while those with a molar concentration
>0.1mol/L are referred to as concentrated.
 As the mass (or amount) of solute is not changed by adding more solvent…we can say that m i
=m f
of n i
=n
f m i
=initial mass of solute m i
=final mass of solute

n i
=initial amount of solute (in moles) n f
=final amount of solute (in moles) we know that C=n/v -- n=v x C
& that c=m/v -- m=c x v therefore we can compare 2 solutions and express the constant quantity of solute in terms of v and C, using the formula: v i
C i
=v f
C
f
Q: Water is added to 0.200L of 2.40mol/L of NH
3(aq)
cleaning solution, until the final volume is 1.000L. Find the molar concentration of the final diluted solution.
A: v i
=0.200L
V f
=1.000L
C i
=2.40mol/L
Find C f
V
I
C i
=v f
C f
C f
=v i
C
i
- 0.200L x 2.40mol/L
v f
1.000L
=0.480mol/L
Therefore the final concentration is 0.480mol/L.

Q: A student is instructed to dilute some concentrated HCL
(aq)
(36%) to make 4.00L of a 10% solution. What volume of HCl should the student measure to do this?
A: c i
=36%
V f
=4.00L
C f
=10%
Find v i v i c i
= v f c f
--- v i
= v f c f
c i
= 4.00L x 10%
36%
= 1.1L
Therefore the initial volume was 1.1L.
Answer Questions 1-5 on Pages 306-307.