Solution Chemistry
(Chp. 7)
Chemistry 2202
Solutions
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Terms
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
soluble
insoluble
limiting reagent
excess reagent
actual yield
theoretical yield
decanting
pipetting
filtrate
precipitate
dynamic equilibrium
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Define the terms in bold and italics from
pp. 237 – 240.
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Solids, liquids, and gases can combine
to produce 9 different types of solution.
Give an example of each type.
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p. 242 #’s 5, 9, & 10
Terms
solution - is a homogeneous mixture
solute - the substance that dissolves OR the
substance in lesser quantity
solvent - the substance which dissolves the
solute OR the substance in greater quantity
concentrated - a large amount of solute relative
to the amount of solvent
dilute - a small amount of solute relative to the
amount of solvent
Terms
saturated – contains the maximum amount of
dissolved solute at a given temperature and
pressure
unsaturated – contains less than the maximum
amount of dissolved solute at a given
temperature and pressure
supersaturated – contains more than the
maximum amount of dissolved solute for a
given temperature and pressure
Terms
miscible – liquids that dissolve in each other
immiscible – liquids that do not dissolve in
each other
aqueous - the solvent is water
alloy - a solid solution of two or more metals
Terms
Solubility - the maximum amount of solute that
can be dissolved under specific temperature
and pressure conditions
eg. the solubility of HCl at 25 °C is 12.4 mol/L
eg. 100.0 mL of water at 25°C dissolves 36.2 g
of sodium chloride
Terms
soluble – solubility is greater than 1 g per
100 mL of solvent.
insoluble - solubility is less than 0.1 g per
100 mL of solvent.
9 Types of Solution
Factors Affecting Solubility (pp.243 – 254)
1.
2.
List 3 factors that affect the rate of dissolving.
How does each of the following affect
solubility?
 particle size
 temperature
 pressure
Factors Affecting Solubility
3.
4.
What type of solvent will dissolve:
 polar solutes and ionic solutes
 nonpolar solutes
Why do some ionic compounds have low
solubility in water?
p. 254 #’s 1, 2, 4 - 6
Rate of Dissolving
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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.
Solubility
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small molecules are often more soluble than
larger molecules.
solubility of solids increases with temperature.
the solubility of most liquids is not affected by
temperature.
the solubility of gases decreases as
temperature increases
an increase in pressure increases the
solubility of a gas in a liquid.
“Like Dissolves Like”
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ionic solutes and polar covalent solutes
both dissolve in polar solvents
non-polar solutes dissolve in non-polar
solvents.
compounds with very strong ionic bonds,
such as AgCl, tend to be less soluble in
water than compounds with weak ionic
bonds, such as NaCl.
Applications
1. An opened soft drink goes ‘flat’ faster if not
refrigerated.
2. Thermal pollution (warming lake water) is
not 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 how much
solute is dissolved or 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
mol/L OR
M (molar)
eg. Calculate the molar concentration of:
1. 4.65 mol of NaOH is dissolved to
prepare 2.83 L of solution.
(1.64 mol/L)
1.
15.50 g of NaOH is dissolved to
prepare 475 mL of solution.
( 0.3875 mol → 0.816 mol/L)
p. 268 - # 19
Rearranged
Formulas
n CxV
or
n
V 
C
eg. Calculate the following:
a) the number of moles in 4.68 L of 0.100
mol/L KCl solution. (0.468 mol)
b) the mass of KCl in 268 mL of 2.50 mol/L
KCl solution. (0.670 mol → 49.9 g)
p. 268 #’s 20 - 24
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.
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!!
June 2010 # 12.
A lab technician prepares a dilute solution of
hydrochloric acid. If 50.0 mL of 2.50 mol/L
hydrochloric acid is added to 450.0 mL of
water, what is the new concentration?
(A) 0.250 mol/L
(C) 3.60 mol/L
(B) 0.278 mol/L
(D) 4.00 mol/L
Solution Preparation & Dilution
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standard solution – a solution of known
concentration
volumetric flask – a flat-bottomed glass vessel
that is used to prepare a standard solution
delivery pipet – a pipet that accurately
measures one volume
graduated pipet – pipets that have a series of
lines that can be used 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 final 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.
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Percent Concentration
Concentration may also be given as a %.
 The amount of solute is a percentage of
the total volume or mass of solution.
 liquids in liquids - % v/v
 solids in liquids - % m/v
 solids in solids - % m/m
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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
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p. 265 #’s 15 – 17
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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
OR
n = CV
n=m/M
3. Mole ratios
4. Calculate required quantity
m = n x M OR
n
V
C
OR
n
C
V
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)
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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)
Ca(NO3)2(s) → Ca2+(aq) + 2 NO3-(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.
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The Solution Process (p. 299)
Acids are also electrolytes.
 Acids form ions when dissolved in
water.
eg. H2SO4(aq) → 2 H+(aq) + SO42-(aq)
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HCl(s) → H+(aq) + Cl-(aq)
The Solution Process (p. 299)
Molecular Compounds DO NOT
dissociate in water.
eg. C12H22O11(s) → C12H22O11(aq)
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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 a solution
in which the Cl-(aq) concentration is
0.120 mol/L ?
p. 302 # 14
p. 311 #’s 11, 12, 16, & 18