Solutions

High School Chemistry Rapid Learning Series - 19

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Chemistry :: Biology :: Physics :: Math

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Solutions

HS Ch i t R id Learning Series
Chemistry Rapid L
i
S i
Wayne Huang, PhD
Kelly Deters, PhD
Russell Dahl, PhD
Elizabeth James, PhD

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© Rapid Learning Inc. All rights reserved.

© Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com

1


Learning Objectives
By completing this tutorial you will learn…
How solutions form.
Factors affecting solubility.
F t
ff ti
l bilit
Concentrations of solutions.
Calculations with solution
concentrations.
Electrolyte solutions.
Colloids.

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Concept Map
Previous content

Chemistry
New content
Studies

Matter
Solution
Formation

Solute & Solvent
properties govern

One type of

Solutions
Solute & Solvent
ratios described by

Various factors
affect

Concentration

Solubility

Used in calculations

Dilutions

Stoichiometry

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2


Forming Solutions

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Definition: Solution, Solute & Solvent
Solution – Homogeneous mixture, solution =
solute + solvent.
Solute Substance being dissolved, present in a
S l t –S b t
b i di
l d
ti
smaller quantity (solute dissolves into solvent).
Solvent – Substance (in a greater quantity) doing
the dissolving.
Water is called the “Universal
Solvent
Solvent” because it s used for so
it’s
many solutions.
Solutions can be any combination
of solids, liquids and gases!
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Solute vs Solvent Mnemonic: Dissolve solute into solvent = “The
police came, the thief hide the lute in the vent.”


3


Steps for Solution Formation
In order for a solution to form:

1

Intermolecular forces in the solvent are broken
(
(“expanding the solvent”): Solvent - - X- - Solvent
p
g
)

2

Intermolecular forces in the solute are broken
(“expanding the solute”): Solute - - X - - Solute

3

Solute and solvent particles form new
intermolecular forces together: Solute –– Solvent

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Energy of Solution Formation
1

Intermolecular forces in the solvent are broken.

2

Intermolecular forces in the solute are broken.

3

Solute and solvent particles form new
intermolecular forces together.
IMF – Intermolecular Force

Breaking IMF’s requires energy

Forming IMF’s releases energy

If energy required >> energy released…
gy q
gy
the solution will not form.
If energy required similar or < energy released…
The solution will form.
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4


Why Don’t Oil and Water Mix?
Let’s look at the IMF’s broken and formed.
IMF’s Broken

IMF’s Formed

Water—London Dispersion

Oil & Water together—
W t t
th
London Dispersion

Water—Dipole-Dipole
Water—Hydrogen bonding

The two are more stable
separate than together.

Oil—London Dispersion
4 IMF’s are broken.
Only 1 IMF can be formed (and it’s the weakest kind).
Much more energy would be required than released.
Therefore, the solution doesn’t form.
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“Like Dissolves Like”
This general rule that can be followed for solution
formation stems from:
Compounds that
have similar
properties and
bond
characteristics
form similar
intermolecular
forces.

If similar
intermolecular
forces can be
formed between
2 compounds as
within each
compound, a
solution is likely
to form.

Things with
“Like” bond
types will
dissolve other
things with
“Like” bond
types.

Things dissolves other things with “Like” bond
properties (“Like dissolves Like”):
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Examples: Polar compounds (HCl) dissolve polar compounds
(H2O). Non-polar (C5H12) dissolves non-polar (C6H6), etc.


5


Factor’s Affecting
Solubility

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Pressure - 1
How does pressure affect solubility of a gas in a
liquid?
Gas molecules above the solvent
cause pressure.
When a gas molecules comes into
contact with the surface of the solvent,
it can be “dissolved”.

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6


Pressure - 2
How does pressure affect solubility of a gas in a
liquid?
If there are more gas
th
particles above the
solution (a higher
pressure)…

More will come in
contact with the
t t ith th
solvent and
“dissolve”.
Higher pressure
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Higher solubility for gases

(Pressure has no affect on solids or liquids dissolving!)

Temperature-Dissolving Gases - 1
When dissolving gases:

As
A gas particles move around, when
ti l
d h
they reach the surface of the solution,
they can “escape” if they have
enough energy.

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7


Temperature-Dissolving Gases - 2
When dissolving gases:

If gas particles have
ti l h
more energy (higher
temperature)…
They will move faster,
reach the surface
more often, and have
more energy to
“escape” when they
do reach the surface.
Higher temperature

Lower solubility for gases

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Dissolving Gases
Gases dissolve in a liquid best at:

High Pressure

Low Temperature

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8


Temperature—Dissolving Solids
For most solids dissolving, increasing the
temperature of the solvent:

Increase in
solvent
temperature
increases energy
of solvent
particles.

Increased energy
allows for more
IMF’s of the
solvent to be
broken (the
solvent is more
“expanded”).

There is more
“room” for
solute particles
in between the
solvent particles.

For most solids, as temperature increases, solubility increases.

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Solubility Curves
The solubility of a solid at different temperatures is
shown in a solubility curve:
Solubility
changes a lot as
temperature is
increased.
Solubility is
barely affected
by change in
temperature.

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9


Saturation of
Solutions

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Definitions: Saturation
Unsaturated Solution – The
solvent can still hold more
solute particles.
Saturated Solution – The solvent
is holding as many solute
particles as it can possibly hold.
Supersaturated Solution – The
solvent is holding more solute
particles (crystals) than it
should be able to at that
temperature.

More
solute can
dissolve.

No more
solute can
dissolve.

More
solute can
crystallize.

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10


How are Supersaturated Solutions Formed?
How can a solution hold more solute particles than
it usually can at that temperature?
Solution is
saturated.

Temperature of
T
t
f
solution is
raised.
Solution is no
longer saturated
at the new
temperature.
More solute
particles are
added.

Solution is slowly
and carefully
cooled back
down.
Solute particles
remain dissolved.
Solution is now
supersaturated to
form crystals.

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Concentrations of
Solutions

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11


Definition: Concentration
Concentration – Measure of the
qua t ty o so ute per quantity
quantity of solute pe qua t ty
of solvent or total solution.
Common Units:
Percent Composition by Mass (%)
Molarity M (Moles of solute per volume
solution L)
Molality m (Moles of solute in 1kg solvent)
y (
g
)
Mole Fraction (ratio of moles in one
component over moles of all components)

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Definition: Concentrated & Dilute
Concentrated – Large ratio of solute : solvent.
Dilute Small ti f l t
Dil t – S ll ratio of solute : solvent.
l
t
Do not use the terms “weak” or “strong” to
describe concentrations -those words have
very specific meanings in chemistry.

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12


Percent by Mass
To determine concentration in % by mass:
% mass =
Example:

mass solute
×100
mass solution

The mass units must
match!

A sample contains 1.25 g NaCl in 100 mL of water.
(1 mL water = 1 g water). Determine the % by
mass of the solution.

Mass solute = 1.25 g NaCl
Mass solution = 1.25 g NaCl + 100 g water = 101.25 g
% by mass = ?
% mass =

1.25 g NaCl
×100 = 1.23% NaCl
101.25 g solution

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Percent by Volume
To determine concentration in % by volume:
% volume =
Example:

volume solute
×100
volume solution

The volume units
must match!

What volume of water is needed to make a 15% by
volume solution of alcohol if you have 5 mL of
alcohol?

Volume solute = 5 mL
Volume solution = 5 mL solute + x mL solvent = (5+x) mL
% by volume = 15%

15% =

5mL
× 100
(5 +x)mL
x mL =

(5 + x)mL =

5mL
×100
15%

5mL
×100 − 5mL
15%
x = 28 mL solvent

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13


Percent Mass/Volume
To determine concentration in % mass/volume:
% mass / volume =

Example:

mass solute
×100
volume solution

Use mass in grams.
Use volume in milliliters.

Vinegar is a 5% solution of acetic acid in water (by
mass/volume). What mass of acetic acid
(CH3COOH) is in 45 mL of vinegar?

Mass solute = ? g
Volume solution = 45 mL
% mass/volume = 5%

5% =

mass acetic acid
×100
45mL

(45mL) × 5%
= mass acetic acid
100
2.25 g CH3COOH

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Molarity
One of the most commonly used concentration
units is Molarity (M):
M = Molarity
n
n = moles solute
M=
V
V = volume solution (in Liters)
Example:

Find the molarity if 12.5 g NaCl is dissolved in 500
mL of water.

Solute = 12.5 g NaCl = 0.214 mole NaCl
Solution = 500 mL = 0.500 L
0.214mole NaCl
Molarity =
Molarity = ? M

0.500 L

12.5 g NaCl

1 mole NaCl
= 0.214 mole NaCl
58.44 g NaCl

500 mL
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0.428 M NaCl

0.001 L
1 mL

= 0.500 L


14


Molality
To determine molality (m):
n
m=
kg solvent
Example:

m = molality
n = moles solute

How many moles NaNO3 are needed to make a 0.24
m solution with 1.5 L (1.5 kg) of water?

Solute = ? mole NaNO3
Solvent = 1.5 kg
molality = 0.24 m
0 24

0.24m =

mole NaNO3
1.5 kg

1.5kg × 0.24m = mole NaNO3
0.36 mole NaNO3

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Molarity vs Molality Mnemonic: MolaRity (moles to Liter solution) and
MolaLity (moles to kg solvent) = “Rose to Lover and Loyal to King!”

Calculations
with
Concentrations

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15


Dilutions - 1
Often, you are supplied with a more concentrated
solution than you need.
In order to dilute a solution to a lower concentration, more solvent is
added.
dd d
The moles of solute did not change:
moles solute before = moles solute after

Solvent

Solute

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Dilutions - 2
Often, you are supplied with a more concentrated
solution than you need.
In order to dilute a solution to a lower concentration, more solvent is
added.
added
The moles of solute did not change:
moles solute before = moles solute after
And if

M=

n
V

Then

n1 = n2

n = M ×V

M 1 × V1 = M 2 × V2

The original molarity times the volume of the original solution = the
new molarity times the new volume of the solution.
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16


Dilution Example
A dilution calculations example:
M 1 × V1 = M 2 × V2

Example:

2 volume units must match!

You need 55 mL of 0.10 M HCl solution. You
currently have 12 M HCl solution. What volume of
the concentrated solution will you dilute to 55 mL?

M1 = 12 M HCl
V1 = ? mL
M2 = 0 10 M HCl
0.10
V2 = 55 mL

12 M × V1 = 0.10 M × 55mL

V1 =

0.10 M × 55mL
12 M
V1 = 0.46 mL

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Using Molarity in Conversions
Molarity is used to convert between moles and liters.
Example: If 0.85 moles NaOH are needed and you have a 1.5 M
solution, how many liters of the solution do you need?
From concentration:
1.5 mole NaOH = 1 L

0.85 mole NaOH

1

L

0.57
= ________ L

1.5 mole NaOH

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17


Solution Stoichiometry
Example: If you need 15.7 g Ba(OH)2 to precipitate, how many
liters of 2.5 M NaOH solution is needed?
2 NaOH + BaCl2
Ba(OH)2 + 2 NaCl
From balanced equation:
q
2 mole NaOH
1 mole Ba(OH)2
Concentration of NaOH:
2.5 mole NaOH = 1 L
Molar Mass of Ba(OH)2:
1 mole Ba(OH)2 = 171.35 g
15.7 g Ba(OH)2
( )

1

mole Ba(OH)2

171.35 g Ba(OH)2

2

mole NaOH

1

mole Ba(OH)2

1

L NaOH

2.5 mole NaOH

0.0733
= ________ L NaOH

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Electrolyte Solutions

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18


Definition: Electrolyte

Electrolyte – Ionic compound
which dissolves in water,
producing free-floating ions.
Free-floating ions can conduct
electricity (hence “electro”).
e.g.

When dissolved in water:
NaCl
Na+ + ClCa(NO3)2
Ca2+ + 2 NO3-

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Definition: Strong, Weak & Non-Electrolytes
Strong Electrolyte – Most of the ions
dissociate and are free floating.
g
Weak Electrolyte – Only some of the ions
dissociate and are free floating (weakly
conducts electricity).
Non-Electrolyte – Dissolved substance
does not produce ions at all (does not
conduct electricity).
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19


Breaking Compounds into Electrolytes
How do you break up a compound when forming
electrolytes?
1

Do not break up polyatomic ions.

2

Use subscripts that are not a part of a polyatomic
ion as coefficients.
e.g. CaCl2 doesn’t have “Cl2” ions, it has 2 “Cl” ions.

Example:

Break up the following strong electrolytes:

Na3PO4
(NH4)2CO3

3 Na+ + PO432 NH4+ + CO32-

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Colloids

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20


Definition: Colloid

Colloid Solution with
C ll id – S l ti
ith
solute particles large
enough to deflect light
as it travels through the
solution.

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Definition: Tyndall Effect
Tyndall Effect – Property exhibited by
colloids. The scattering light is visible
through the solution.

Light coming in

Light going out

The light is not
scattered, and is not
seen traveling through
the solution.

Solution

The light is scattered,
and is seen traveling
through the colloid.
Light coming in
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Light going out
Colloid


21


Learning Summary

Solutions are composed
p
of solute and solvents.

The l ti
Th solution
process is governed by
energetics of solution
formation and factors
affecting solubility.

Colloids are mixtures
which have large enough
g
g
solute particles to scatter
light (exhibit the Tyndall
Effect).

Concentration is an
expression of the ratio of
i
f th
ti
f
solute to solvent particles.
Concentrations are used
in dilution and
stoichiometry
calculations.

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Congratulations
You have successfully completed
the core tutorial

Solutions


22


Chemistry :: Biology :: Physics :: Math

What’s N t
Wh t’ Next …

Step 1: Concepts – Core Tutorial (Just Completed)
Step 2: Practice – Interactive Problem Drill
Step 3: Recap – Super Review Cheat Sheet

Go for it!

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