Solution Chemistry
(Chp. 7)
Chemistry 2202
Topics
Molar Concentration (mol/L)
 Dilutions
 % Concentration (pp. 255 – 263)
 Solution Process
 Solution Preparation
 Solution Stoichiometry
 Dissociation

Terms
solution
solvent
solute
concentrated
dilute
aqueous
miscible
Immiscible
alloy
solubility
molar solubility
saturated
unsaturated
supersaturated
dissociation
electrolyte
non-electrolyte
filtrate
precipitate
x
limiting reagent
excess reagent
actual yield
theoretical yield
decanting
pipetting
dynamic equilibrium

Define the terms in bold and italics from
pp. 237 – 240.

Solids, liquids, and gases can combine
to produce 9 different types of solution.
Give an example of each type.

p. 242 #’s 5, 7, 9, & 10
Terms
solution
solvent
solute
concentrated
dilute
aqueous
miscible
immiscible
x
alloy
solubility
molar solubility
saturated
unsaturated
supersaturated
dynamic equilibrium
Factors Affecting Solubility (pp.243 – 254)
List 3 factors that affect the rate of
dissolving.
How does each of the following affect
solubility?
1.
2.



particle size
temperature
pressure
Factors Affecting Solubility
What type of solvent will dissolve:
 polar solutes
 nonpolar solutes
 ionic solutes
4. Why do some ionic compounds have
low solubility in water?
3.
p. 254 #’s 1, 2, 4 - 6
Section 7.2 (pp. 243 – 252)
State the generalizations regarding
solubility and solutions (in italics)
 Define terms (in bold)

rate of dissolving
dipole
ion-dipole attractions
hydrated
electrolyte.
non-electrolytes
Rate of Dissolving
for most solids, the rate of dissolving is
greater at higher temperatures
 stirring a mixture or by shaking the
container increases the rate of dissolving.
 decreasing the size of the particles
increases the rate of dissolving.

“Like Dissolves Like”
ionic solutes and polar covalent
solutes both dissolve in polar solvents
 non-polar solutes dissolve in nonpolar solvents.

Solubility
small molecules are often more
soluble than larger molecules.
 the solubility of most solids increases
with temperature.
 the solubility of most liquids is not
greatly affected by temperature.
 the solubility of gases decreases as
temperature increases

Solubility

An increase in pressure increases the
solubility of a gas in a liquid.
Applications
1. An opened soft drink goes ‘flat’ faster if
not refrigerated.
2. Warming of pond water may not be
healthy for the fish living in it.
3. After pouring 5 glasses of pop from a
2 litre container, Jonny stoppered the
bottle and crushed it to prevent the
remaining pop from going flat.
Molar Concentration
Review:
- Find the molar mass of Ca(OH)2
- How many moles in 45.67 g of Ca(OH)2?
- Find the mass of 0.987 mol of Ca(OH)2.
Molar Concentration
The terms concentrated and dilute are
qualitative descriptions of solubility.
A quantitative measure of solubility
uses numbers to describe the
concentration of a solution.
Molar Concentration
The MOLAR CONCENTRATION of a
solution is the number of moles of solute
(n) per litre of solution (v).
Molar Concentration
FORMULA:
Molar Concentration = number of moles
volume in litres
C= n
V
eg. Calculate the molar concentration of:
 4.65 mol of NaOH is dissolved to
prepare 2.83 L of solution.

15.50 g of NaOH is dissolved to
prepare 475 mL of solution.
p. 268 - # 19
Eg. Calculate the following:
a) the number of moles in 4.68 L of 0.100 mol/L
KCl solution.
b) the mass of KCl in 268 mL of 2.50 mol/L KCl
solution.
c)
d)
the volume of 6.00 mol/L HCl(aq) that can be
made using 0.500 mol of HCl.
the volume of 1.60 mol/L HCl(aq) that can be
made using 20.0 g of HCl.
p. 268 #’s20-24
Dilution (p. 272)
Number of moles
before dilution
When a solution is diluted:
- The concentration decreases
- The volume increases
-
The number of moles remains
the same
ni = nf
Number of moles
after dilution
Dilution (p. 272)
ni = nf
Ci Vi = C f Vf
eg. Calculate the molar concentration of a
vinegar solution prepared by diluting 10.0
mL of a 17.4 mol/L solution to a final
volume of 3.50 L.
p. 273 #’s 25 – 27
p. 276
#’s 1, 2, 4, & 5
DON’T SHOW UP UNLESS
THIS IS DONE!!
Solution Preparation & Dilution




standard solution – a solution of known
concentration
volumetric flask – a flat-bottomed glass vessel
that is used to prepare a standard solution
delivery pipet – pipets that accurately measure
one volume
graduated pipet – pipets that have a series of
lines that can be use to measure many different
volumes
To prepare a standard solution:
1. calculate the mass of solute needed
2. weigh out the desired mass
3. dissolve the solute in a beaker using less
than the desired volume
4. transfer the solution to a volumetric flask
(rinse the beaker into the flask)
5. add water until the bottom of the
meniscus is at the etched line
To dilute a standard solution:
1. Rinse the pipet several times with
deionized water.
 2. Rinse the pipet twice with the standard
solution.
 3. Use the pipet to transfer the required
volume.
 4. Add enough water to bring the solution
to its final volume.

Percent Concentration
Concentration may also be given as a %.
 The amount of solute is a percentage of
the total volume/mass of solution.
 liquids in liquids - % v/v
 solids in liquids - % m/v
 solids in solids - % m/m

Percent Concentration
mass of solute (g)
Percent (m/v) 
x 100
volume of solution (mL)
p. 258 #’s 1 – 3
DSUUTID!!
mass of solute (g)
Percent (m/m) 
x 100
mass of solution (g)
p. 261 #’s 5 – 9
DSUUTID!!
volume of solute (mL)
Percent (v/v) 
x 100
volume of solution (mL)
p. 263 #’s 10 – 13
DSUUTID!!
Concentration in ppm and ppb
Parts per million (ppm) and parts per
billion (ppb) are used for extremely
small concentrations
ppm x msolution
msolute 
6
10
ppb x msolution
msolute 
9
10
eg. 5.00 mg of NaF is dissolved in
100.0 kg of solution. Calculate the
concentration in:
a) ppm
b) ppb
ppm = 0.005 g x 106
100,000 g
= 0.05 ppm
ppb = 0.005 g x 109
100,000 g
= 50.0 ppb

p. 265 #’s 15 – 17

pp. 277, 278
#’s 11, 13, 15 – 18, 20
DON’T SHOW UP UNLESS
THIS IS DONE!!
Solution Stoichiometry
1. Write a balanced equation
2. Calculate moles given
n=m/M
OR
n=CV
3. Mole ratios
4. Calculate required quantity
n
V
C
OR
n
C
V
OR
m  nM
Solution Stoichiometry
eg. 45.0 mL of a HCl(aq) solution is used
to neutralize 30.0 mL of a 2.48 mol/L
NaOH solution.
Calculate the molar concentration of the
HCl(aq) solution.
p. 304: #’s 16, 17, & 18
Worksheet
Sample Problems
1. What mass of copper metal is needed
to react with 250.0 mL of 0.100 mol/L
silver nitrate solution?
2. Calculate the volume of 2.00 M HCl(aq)
needed to neutralize 1.20 g of dissolved
NaOH.
3. What volume of 3.00 mol/L HNO3(aq) is
needed to neutralize 450.0 mL of 0.100
mol/L Sr(OH)2(aq)?
Sample Problem Solutions
Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq)
Step 2
n = 0.02500 mol AgNO3
Step 3
n = 0.01250 mol Cu
Step 4
m = 0.794 g Cu
Sample Problem Solutions
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Step 2
n = 0.0300 mol NaOH
Step 3
n = 0.0300 mol HCl
Step 4
V = 0.0150 L HCl
Sample Problem Solutions
2 HNO3(aq) + Sr(OH)2(aq) →
2 H2O(l) + Sr(NO3)2(aq)
Step 2 n = 0.04500 mol Sr(OH)2
Step 3
n = 0.0900 mol HNO3
Step 4
V = 0.0300 mol/L HNO3
The Solution Process (p. 299)
Dissociation occurs when an ionic
compound breaks into ions as it
dissolves in water.
 A dissociation equation shows what
happens to an ionic compound in water.

eg. NaCl(s) → Na+(aq) + Cl-(aq)
K2SO4(s) → 2 K+(aq) + SO42-(aq)
The Solution Process (p. 299)
Solutions of ionic compounds conduct
electric current.
 A solute that conducts an electric
current in an aqueous solution is
called an electrolyte.

The Solution Process (p. 299)
Acids are also electrolytes.
 Acids form ions when dissolved in
water.
eg. H2SO4(aq) → 2 H+(aq) + SO42-(aq)

HCl(s) → H+(aq) + Cl-(aq)
The Solution Process (p. 299)
Molecular Compounds DO NOT
dissociate in water.
eg. C12H22O11(s) → C12H22O11(aq)


Because they DO NOT conduct electric
current in solution, molecular compounds
are non-electrolytes.
The Solution Process (p. 299)
The molar concentration of any dissolved
ion is calculated using the ratio from the
dissociation equation.
eq. What is the molar concentration of
each ion in a 5.00 mol/L MgCl2(aq)
solution:
5.00 mol/L
5.00 mol/L
10.00 mol/L
p. 300 #’s 7 – 9
What mass of calcium chloride is
required to prepare 2.00 L of 0.120
mol/L Cl-(aq) solution?
p. 302 # 14
p. 311 #’s 11, 12, 16, & 18