Chemistry-Unit 8 Notes HC (Solution Chemistry).pdf

Essential Question
What are the different ways to
determine the composition of a
solution?
By the end of the period…
I will be able to understand the
different relationships to determine
the composition of solutions
1

The Nature of Solutions
■ What is a solution?
• Solution:
▪ A homogenous mixture of two or more substances
uniformly spread out
• Solute:
▪ The substance that dissolves in a solution
• Solvent:
▪ The substance that dissolves the solute
2

■ Types of Mixtures:
• Homogenous Mixture:
▪ Same throughout entire mixture; consistent
▪ Example ➔ A pitcher of lemonade
• Heterogeneous Mixture:
▪ Not the same throughout; not well mixed
▪ Example ➔ A fruit salad
3

■ Miscible vs. Immiscible:
▪Miscible:
– Two or more liquids that can be mixed together
▪ Example ➔ Coffee and cream
▪Immiscible:
– Two or more liquids that cannot be mixed
▪ Example ➔ Oil and vinegar
4

Heterogeneous mixtures that
are liquids can be separated
with a spout as seen in the
picture. The heavier (more
dense) particles sink to the
bottom of the cup, leaving
the less dense liquid on the
top.
5

• Predict what you think is going to happen…
➢ When 3 drops of food coloring is placed into
two different beakers containing equal
amounts of different temperature water.
➢ Which beaker do you think contains the
warmer/colder water? Be ready to support
your answer with specific evidence
7

The Kinetic Theory
• All substances are made up of tiny particles
• All particles are in motion
• Temperature has an impact on the movement
of particles
8

Stop and Think
■ Imagine that you place the sugar
cube (in the diagram) in a glass of
water (solution)…
▪ What can you do to the sugar cube
to make it dissolve quicker?
▪ What can you do to the solution to
make the sugar cube dissolve
quicker?
▪ Will the sugar cube dissolve?
9

■ Dissolving a sugar cube…
▪ The water molecules strike the
sides of the sugar cube
▪ Energy is transferred from the
water molecules to the sugar
particles
▪ Sugar particles leave the sugar
cube into the water
▪ More sugar particles are uncovered
▪ Repeat process
10

■ The Rate of Dissolving:
➢ Three major factors for dissolving:
▪Temperature
▪Surface Area
▪Stirring or shaking
11

Rates of Dissolving
■ Temperature (of the solvent)
▪ As temperature increases ➔ INCREASES
▪ As temperature decreases ➔ DECREASES
▪ It’s all about collision of particles…
▪ High temp ➔ particles move faster
▪ Low temp ➔ particles move slower
▪ Faster moving particles = higher frequency of
collisions
➔ More solute can be dissolved
12

Rates of Dissolving
■ Surface Area (of the solute)
▪ More surface area ➔ INCREASES
▪ Less surface area ➔ DECREASES
▪ More surface for particles of solvent to
interact with particles of solute
▪ Greater rate of collisions
13

Rates of Dissolving
■ Agitation:
▪ More agitation ➔ INCREASES
▪ Less agitation ➔ DECREASES
▪ Affects both the solute AND solvent
▪ Agitation (stirring or shaking) of a solution
causes the particles in solution to collide more
often
14

Molarity and Concentration
■ Concentration:
▪ The measure of the amount of solute that is
dissolved in a given quantity of solvent
▪ Dilute Solution:
➢ Low number of solute particles in a given
solution
▪ Concentrated Solution:
➢ Large number of solute particles in a given
solution
18

■ Molarity:
▪ The number of moles of a solute that is dissolved
in a given solution (in Liters)
▪ Unit = M ➔ pronounced “molar”
Molarity (M) =
moles of solute
Liters of solution
=
mol
V
19

■ Molarity Equations and Algebra
M =
mol
V
moles = (M) x (V)
V =
moles
M
M = molarity
mol = moles of solute
V = Liters of solution
20

■ Example #1:
▪ Calculate the molarity of a solution prepared by
dissolving 0.0428 mol of HCl into 26.8 mL of
solution.
Moles of solute =
Volume of solution = = 0.0268 L
0.0428 mol HCl
26.8 mL
Equation ➔ Molarity =
moles
volume
21

■ Example #1:
▪ Calculate the molarity of a solution prepared by
dissolving 0.0428 mol of HCl into 26.8 mL of
solution.
Molarity (M) =
mol HCl
L solution
= 1.60 M
0.0428 mol HCl
0.0268 L solution
=
22

■ Example #2:
▪ A salt water solution contains 0.90 grams of
NaCl per 100 mL of solution. What is its
molarity?
Moles of solute =
Volume of solution = = 0.1 L
0.90 grams NaCl
100 mL
Equation ➔ Molarity =
moles
volume
0.90 g NaCl
58.4 g NaCl
1 mol NaCl
= 0.015 mol NaCl
23

■ Example #2:
▪ A salt water solution contains 0.90 grams of
NaCl per 100 mL of solution. What is its
molarity?
Molarity (M) =
mol NaCl
L solution
= 0.15 M
0.015 mol NaCl
0.1 L solution
=
24

■ Example #3:
▪ How many moles of sulfuric acid (H2
SO4
) would
be in solution if you have 1.5 L of 0.10 M of
H2
SO4
solution
Moles of solute =
Volume of solution =
???
1.5 L
Equation ➔
Molarity = 0.10 M
moles =(molarity) x (L of solution)
25

■ Example #3:
▪ How many moles of sulfuric acid (H2
SO4
) would
be in solution if you have 1.5 L of 0.10 M of
H2
SO4
solution
mol = (molarity) x (L of solution)
mol = (0.10 M) x (1.5 L) = 0.15 mol H2
SO4
26

Dilutions
■ A lot of chemicals are purchased in the
concentrated form
▪ Large # of mol of solute in a given amount
of solution
▪ When making a dilution…the number of
moles before the dilution EQUALS the
number of moles of solute after the dilution
28

Dilutions
■ Need to rearrange the Molarity equation to
solve for moles of solute
Molarity (M) x
moles of solute
Liters of solution
# of moles of solute = # of L of solution
Molarity (M) =
29

Dilutions
■ Since the total number of moles of solute
remains unchanged upon dilution, you can write
the following relationship.
■ The M1
and V1
values represent the initial
solution’s Molarity and volume and M2
V2
are
the final solution’s molarity and volume
M1
x V1
= moles of solute = M2
x V2
30

Dilutions
■ KEY POINTS:
▪ The volumes can be measured in either mL or
Liters…but BOTH volumes must be the same
▪ When using this equation, you are adding (or
removing solvent) to make a new concentration
M1
V1
= M2
V2
31

Dilutions
■ Dilution Equation and Algebra
M1
V1
= M2
V2
M2
V2
M1
=
V1
M2
V2
V1
=
M1
M1
V1
M2
=
V2
M1
V1
V2
=
M2
32

Dilutions
■ What concentration of HCl acid would you produce
if you mixed 2.5 L of a 2.0 M solution of HCl to
make a TOTAL volume of 4.0 L?
M1
V1
M2
=
V2
(2.0 M) (2.5 L)
M2
=
4.0 L
= 1.25 M
33

Dilutions
■ What volume of 16 M sulfuric acid (H2
SO4
) must be
used to prepare 1.5 L of a 4 M sulfuric acid solution?
How much water must be added to create 1.5 L of
solution?
M2
V2
V1
=
M1
(4 M) (1.5 L)
V1
=
16 M
= 0.375 L
H2
SO4
34

Dilutions
■ What volume of 16 M sulfuric acid (H2
SO4
) must be
used to prepare 1.5 L of a 4 M sulfuric acid solution?
How much water must be added to create 1.5 L of
solution?
How much H2
SO4
acid is needed?
0.375 L H2
SO4
How much H2
O is needed to make 1.5 L of solution?
1.5 L sol’n - 0.375 L H2
SO4 = 1.125 L H2
O
35

Molality
■ Molality:
▪ The number of moles of a solute dissolved in
1 kilogram (1000 grams) of solvent
▪ Unit = m ➔ pronounced “molal”
Molality (m) =
moles of solute
kilograms of solvent
37

Molality
■ Molality Equations and Algebra
m =
moles
kg
moles = (m) x (kg)
kg =
moles
m
m = molality
mol = moles of solute
kg = kg of solvent
38

Molality
■ Significance of Molality:
▪ Used when both the solute and solvent need to
be weighed (accurately)
▪ Density variations due to temperature and
pressure differences
▪ Mass of a pure substance is more relevant
than volume (in some cases)
▪ More useful in experiments involving
significant temperature changes
39

Molality
■ Example #1:
▪ Glucose (C6
H12
O6
) is a naturally occurring sugar. What
is the molality of a solution containing 5.67 g glucose
dissolved in 25.2 g of water?
Solute =
Solvent =
5.67 g glucose
25.2 g H2
O
Molality = ???
5.67 g C6
H12
O6
180.0 g C6
H12
O6
= 0.0315 mol C6
H12
O6
1 mol C6
H12
O6
= 0.025 kg H2
O
➔ convert to moles
40

Molality
■ Example #1:
▪ Glucose (C6
H12
O6
) is a naturally occurring sugar.
What is the molality of a solution containing
5.67 g glucose dissolved in 25.2 g of water?
m =
0.0315 mol
0.0252 kg
= 1.25 m C6
H12
O6
m =
mol
kg
41

Molality
■ Example #2:
▪ How many grams of potassium iodide (KI) must
you dissolve in 500 g of water to produce 0.060
molal KI solution?
Solute =
Solvent =
Potassium iodide (KI)
➔ 500 g H2
O
Molality = 0.060 m
= 0.500 kg H2
O
= ???
H2
O
42

Molality
■ Example #2:
▪ How many grams of potassium iodide must you
dissolve in 500 g of water to produce 0.060 molal
KI solution?
moles = (m) x (kg)
moles = (0.060) x (0.500) = 0.03 mol KI
0.030 mol KI
1 mol KI
= 5.0 g KI
166.0 g KI
43

Mole Fraction
■ Mole Fraction:
▪ The ratio of the moles of solute in solution to
the total number of moles of BOTH solvent and
solute ⇒ (solute + solvent = solution)
Mole fraction
(ΧA
)
mol substance A
total moles of solution
=
Mole fraction
(XA
)
mol substance A
molsolute
+ molsolvent
=
45

Mole Fraction
■ Significance of Mole Fraction:
▪ It is NOT temperature dependent
▪ Do not need to know density values
▪ Mole fractions can be determined from the
masses of solute and solvent
▪ It is a unitless property
46

Mole Fraction
■ What are the mole fractions of glucose and water in a
solution containing 5.67 grams of glucose (C6
H12
O6
)
dissolve in 25.2 grams of water?
5.67 g C6
H12
O6
180.0 g C6
H12
O6
= 0.0315 mol C6
H12
O6
1 mol C6
H12
O6
25.2 g H2
O
18.0 g H2
O
= 1.40 mol H2
O
1 mol H2
O
1.40 mol H2
O + 0.0315 mol C6
H12
O6
= 1.432 mol
(solvent) (solute) (total moles)
47

Mole Fraction
Mole fraction
(C6
H12
O6
)
= 0.0220
1.432
=
0.0315
Mole fraction
(H2
O)
= 0.978
1.432
=
1.40
= 1.000
The sum of the mole fraction is
ALWAYS 1.000
Mole fraction
(XA
)
mol substance A
molsolute
+ molsolvent
=
48

Mass Percent
■ Mass Percent (or Percent by Mass):
▪ The ratio of the mass of a solute divided by the
mass of the solution (solute + solvent = solution)
Mass Percent
mass of substance
mass of solution
= X 100
Mass Percent
mass of substance
masssolute
+ masssolvent
= X 100
50

Mass Percent
■ Significance of Mass Percent:
▪Used to measure…
▪ The amount of solute compared to the
TOTAL solution
▪ The amount of solvent compared to the
TOTAL solution
▪Independent of lab conditions
▪ Pressure, temperature and volume
▪It is a unitless property
51

Mass Percent
■ Example:
▪ What is the percent by mass of 15.5 g NaCl in
80.0 g H2
O?
Mass Percent
mass of substance
masssolute
+ masssolvent
= X 100
Mass Percent
mass of NaCl
massNaCl
+ massH2O
= X 100
52

Mass Percent
■ Example:
▪ What is the percent by mass of 15.5 g NaCl in
80.0 g H2
O?
Mass Percent
15.5
15.5 + 80.0
= X 100
Mass Percent = 16.2 % NaCl
53

Mass Percent
■ Example:
▪ How many grams of NaCl are found in 60.0 g of a
4.0% solution?
Mass Percent
mass of solute
mass of solution
= X 100
4.0 %
mass of NaCl
60.0 g
=
Convert to decimal…
divide by 100
X 100
54

Mass Percent
■ Example:
▪ How many grams of NaCl are found in 60.0 g of a
4.0% solution?
0.04
mass of NaCl
60.0 g
=
mass of solute = (0.04) (60.0)
= 2.4 g NaCl
55

• Solubility:
▪ The amount of a solute that dissolves in a given
amount of grams of a solvent at a given
temperature to produce a saturated solution
▪ A graph of solubility (called a solubility curve) is
created by comparing the temperature of the
solvent to the amount of solute
per 100 grams of solvent
57

■ Saturated Solution
▪ The maximum amount (mass) of solute that can be
dissolved into solution at a certain temperature
▪ Interpreting the Solubility Curve
▪ All points located ON the “solubility curve”
59

Saturated Solution
60
For any point
ON the solubilitycurve
the solution is SATURATED

■ Unsaturated Solution
▪ A solution where more solute can dissolve into
solution
▪ Any point BELOW the “solubility curve” of a
substance at a certain temperature.
61

For any point
BELOW the solubility curve
the solution is UNSATURATED
Unsaturated Solution
62

■ Supersaturated Solution
▪ A solution holding more dissolved solute than is
“allowed” at a given temperature
▪ Any point ABOVE the “solubility curve” of a
substance at a certain temperature.
63

For any point ABOVE the
solubility curve the
solution is
SUPERSATURATED
Supersaturated Solution
64

1. At 40°C, 100g of
water can dissolve
how much solute?
2. At 40°C, is the
solution saturated,
unsaturated or
supersaturated?
3. If 30 grams of solute
is dissolved at 80°C,
is the solution
saturated,
unsaturated or
supersaturated?
Saturation at 40°C
with 45 grams of
solute
65

■ Factors that affect solubility
▪ Temperature
▪ Physical state of solute
▪ Partial pressure above solution
67

■ Temperature of the solution
▪ Predicting temperature dependence on solubility
can be difficult due to chemical properties
▪ Solubility of solids tend to increase with
increasing temperatures
▪ Solubility of gases tends to decrease with
increasing temperatures
▪ The type of substance is a determining factor
on the solubility
68

■ Physical state of the solute
▪ The solubilities of gases are greater in cold
water than in warmer water
▪ Example ➔ Boiling water
▪ When you first begin to heat water to bring it to
its boiling point, bubbles begin to form
▪ These bubbles are dissolved gases that are
escaping from the solution
69

■ Partial Pressure
▪ The solubility of a gas increases as the partial
pressure of the gas above the solution is
increased
▪ Example ➔ Carbonated Beverages
▪ Most containers of soda are under high
pressure…this keeps the CO2
in solution
▪ When the bottle is opened, the partial pressure
above the solution decreases, causing the
dissolved CO2
to be released from the solution
71

Colligative Properties
■ Colligative Properties:
▪ Properties of a solution that depend on the
concentration (number of particles) of the
solute in solution
▪ The properties of a solution are different from
pure solvents
▪ Three key colligative properties:
▪ Lowering vapor pressure
▪ Increasing boiling point
▪ Lower freezing point
73

■ Lowering Vapor Pressure
▪ Vapor Pressure
▪ Pressure that is exerted by a vapor in a closed
system
▪ A solution containing a non-volatile solute
ALWAYS has a lower vapor pressure than its
pure solvent.
▪ Non-volatile = solute does not vaporize
▪ More particles of solute = lower vapor pressure
74

■ Example ➔ Adding salt to water
▪ When placed in H2
O, sodium and chloride ions
dissociate in solution
▪ The ions become surrounded by H2
O (dissolved)
▪ The water “shell” around each ion creates a
barrier for water molecules to gain enough KE
to escape the solution as a vapor
75

■ Increasing Boiling Point
▪ Boiling Point
▪ The temperature of a liquid at which its vapor
pressure equals external pressure (usually
atmospheric pressure)
▪ Adding a solute to a liquid reduces the overall
vapor pressure, therefore increasing the overall
temperature required to reach boiling point
77

■ Increasing Boiling Point
▪ Attractive forces exist between the solvent and
particles of solute
▪ To overcome these attractive forces, the
particles need to have greater KE
▪ Higher KE = higher boiling point temperature
78

■ Decreasing Melting/Freezing Point
▪ Melting/Freezing Point
▪ The temperature at which a substance changes
from a solid to a liquid (or liquid to solid)
▪ When a substance freezes, the particles tend
to organize in an orderly pattern
▪ Presence of solute particles disrupts this
order…decreasing the overall temperature at
which the solid will melt
79

■ Examples:
▪ The amount of solute and type of solute has
different effects
▪ The addition of 1 mol of NaCl to 1000 g of H2
O
lowers the freezing point of H2
O to -3.72°C
▪ The addition of 1 mol of glucose to 1000 g of H2
O
lowers the freezing point of H2
O to
-1.86°C
▪ Adding salt to icy pavement will lower the
freezing point of water…causing the ice to melt
80

Essential Question
What are acids and bases?
I will be able to understand how solutes
dissolved in solution create different
solution properties
84

Acids and Bases
■ Common Properties of Acids
▪ React with most metals to form hydrogen gas
▪ Taste sour
▪ Can feel sticky when on skin
▪ Usually liquids or gases
85

Acids and Bases
■ Common Properties of Bases
▪ React with oils and greases
(non polar molecules)
▪ Taste bitter
▪ Feels slippery (reacts with your skin)
▪ Frequently solids or liquids
86

Acids and Bases
■ Acids
▪ Arrhenius Concept
▪ An acid is a substance that produces a H+
ion when
dissolved in water
▪ Bronsted-Lowry Model
▪ A proton donor
▪ Donates proton (H+
) ion to another compound
87

Acids and Bases
■ Bases
▪ Arrhenius Concept
▪ A base is a substance that produces a hydroxide (OH–
)
ion when dissolved in water
▪ Bronsted-Lowry Model
▪ A proton acceptor
▪ Receives proton (H+
) ion from the other
compound (acid)
88

Acids and Bases
■ Conjugate Acid
▪ A substance formed by receiving a proton lost by
the acid
▪ Proton is transferred to the base (H2
O)
▪ Forms a hydronium ion ➔ H3
O+
■ Conjugate Base
▪ Is formed by the removal of a proton from an acid
HA (aq) H3
O+
(aq) A-
(aq)
+
+ H2
O (ℓ)
Acid Base Conjugate
Acid
Conjugate
Base 89

Acids and Bases
■ Conjugate Acid-Base Pair
▪ Consists of two substances related to each other by
donating and accepting a single proton
HCl (aq) H3
O+
(aq) Cl-
(aq)
+
+ H2
O (ℓ)
Acid Base Conjugate
Acid
Conjugate
Base
HCl (aq) and Cl-
(aq)
and H3
O+
(aq)
H2
O (ℓ)
➔ HA donates a proton to form A-
➔ H2
O receives a proton to form H3
O+
90

■ Electrolytes:
▪ A substance that when dissolved in water produces a
solution that can conduct electricity
■ Nonelectrolytes
▪ Solutions that DO NOT conduct electricity
91

▪ The extend at which a solution can conduct
electricity depends on the concentration of ions in
solution
▪ Cations ➔ positive charges
▪ Anions ➔ negative charges
92

■ Strong Electrolytes
▪ Substances that are completed ionized (dissociate)
when they are dissolved in water
▪ Can create a basic chemical equation by splitting up the
ions
▪ Use the reverse criss-cross method ➔ from Chapter 5
CaCl2
Ca2+
(aq) Cl-
(aq)
+
H2
O
MgSO4
Mg2+
(aq) SO4
2-
(aq)
+
H2
O
93

■ Strong Acids
▪ Substances that completely ionize (dissociate)
in water…producing H+
ions in solution
▪ Strong Acids:
▪ HCl
▪ HBr
▪ HI
▪ HNO3
▪ HClO4
▪ H2
SO4
94

■ Strong Acids
▪ Monoprotic Acids
▪ Produces one H+
ion when in aqueous solution
▪ Hydrochloric acid (HCl); Nitric acid (HNO3
)
▪ Diprotic Acids
▪ Produces two H+
ions when in aqueous solution
▪ Sulfuric acid (H2
SO4
)
HCl H+
(aq) Cl-
(aq)
+
H2
O
H2
SO4
2 H+
(aq) SO4
2-
(aq)
+
H2
O
95

■ Strong Bases
▪ Substances that completely ionize (dissociate)
in water…producing OH-
ions in solution
▪ Strong Bases:
▪ NaOH
▪ KOH
NaOH Na+
(aq) OH-
(aq)
+
H2
O
KOH K+
(aq) OH-
(aq)
+
H2
O
96

■ Weak Electrolytes
▪ Substances that exhibit a small degree of ionization
when dissolved in water
▪ Produce very few ions in solution
▪ Weak Acids
▪ Any acid that dissociates and produces very few H+
ions in water (solution)
▪ Weak Bases
▪ Any base that dissociates and produces very few OH-
ions in water (solution)
97

Essential Question
What is pH and how is it calculated?
I will be able to understand how pH and
pOH is used to measure the level of
acidity in a solution
98

■ What is pH?
▪ Scale used to measure level of acidity in a solution
▪ Acid vs. Base
▪ Determined based on the amount of H+
ions in
solution
▪ Measured on a scale ➔ 0 – 14
▪ 0 = more acidic
▪ 7 = neutral
▪ 14 = more basic
99

■ What is pH?
▪ Developed on a log scale based on powers of 10
▪ Since pH is a log scale, pH changes by 1 for every
power of 10
▪ Example:
▪ A solution with a pH of 3 has an H+
concentration 10x
that of a solution with a pH of 4
100

■ Ion Product Constant (Kw
)
▪ Autoionization of water
▪ Involves the transfer of a proton (hydrogen ion) from
water molecule to another
▪ Produces a hydroxide (OH-
) ion and a hydronium
(H3
O+
) ion
▪
101

)
▪ The total amount of H+
ions and OH-
ions in pure
water at 25°C MUST be equal
[H+
] = [OH-
] = 1.0 x 10-7
M
Kw
= [H+
] [OH-
] = (1.0 x 10-7
) (1.0 x 10-7
)
Kw
= 1.0 x 10-14
= [H+
] [OH-
]
102

)
▪ A neutral solution where [H+
] = [OH-
]
▪ An acidic solution where [H+
] > [OH-
]
▪ A basic solution where [H+
] < [OH-
]
[H+
] and [OH-
] ions will ALWAYS
= 1.0 x 10-14
103

Example
If [H+
] = 1.0 x 10-5
M, what is the [OH-
] of this
solution?
[H+
] = 1.0 x 10-5
M
What do we know???
What are we solving for?
[OH-
] of solution
What equation are we going to use?
Kw
= [H+
] [OH-
] 104

Example
If [H+
] = 1.0 x 10-5
M, what is the [OH-
] of this
solution?
Kw
= [H+
] [OH-
]
[OH-
] = 1.0 x 10-9
M
[OH-
] = Kw [OH-
] = (1.0 x 10-14
)
[H+
] (1.0 x 10-5
)
105

Example
If [H+
] = 1.0 x 10-5
M, what is the [OH-
] of this
solution?
[OH-
] = [1.0 x 10-9
]
Is this solution basic, neutral or acidic?
[H+
] = [1.0 x 10-5
]
[H+
] > [OH-
]
Acidic
106

Example
If [OH-
] = 5.20 x 10-7
M, what is the [H+
] of this
solution?
[OH-
] = 5.20 x 10-7
M
What do we know???
[H+
] of solution
Kw
= [H+
] [OH-
] 107

Example
If [OH-
] = 5.20 x 10-7
M, what is the [H+
] of this
solution?
[H+
] = 1.92 x 10-8
M
Kw
= [H+
] [OH-
]
[H+
] = Kw [H+
] = (1.0 x 10-14
)
[OH-
] (5.20 x 10-7
)
108

Example
If [OH-
] = 5.20 x 10-7
M, what is the [H+
] of this
solution?
[H+
] = [1.92 x 10-8
]
Is this solution basic, neutral or acidic?
[OH-
] = [5.20 x 10-7
]
[OH-
] > [H+
]
Basic
109

■ Calculating pH ➔ Acids
▪Calculating pH from the [H+
] concentration
pH = -log [H+
]
▪ [H+
] concentration is equivalent to the hydrogen ions
in solution
▪ To determine the pH of a solution, need to know the
ratio of major species (ions)
111

Example
What is the pH of a solution that has a hydrogen-ion
concentration [H+
] of 1.0 x 10-10
M?
Hydrogen-ion concentration = 1.0 x 10-10
M
What do we know???
pH of solution
pH = -log [H+
]
112

Example
What is the pH of a solution that has a hydrogen-ion
concentration [H+
] of 1.0 x 10-10
M?
Calculate pH of solution
pH = -log [H+
] = -log [1.0 x 10-10
]
pH = 10.0
113

Example
What is the pH of a solution if the [OH-
] = 4.0 x 10-11
M?
[OH-] = 4.0 x 10-11
M
What do we know???
pH of solution
Kw
= [H+
] [OH-
] pH = -log [H+
]
114

Example
What is the pH of a solution if the [OH-
] = 4.0 x 10-11
M?
Calculate [H+
] ion concentration
Kw
= [H+
] [OH-
] [H+
] = Kw
[H+
] = (1.0 x 10-14
)
[H+
] = 2.5 x 10-4
M
[OH-
]
(4.0 x 10-11
)
pH = -log [H+
] = -log [2.5 x 10-4
]
pH = 3.60 115

Essential Question
What is pOH and how is it
calculated?
I will be able to understand how pOH is
used to measure the level of acidity in
a solution
116

■ What is pOH
▪pOH is the measured level of OH-
ions in a particular
solution
▪ Measured on a scale of 1-14 (like pH)
▪Opposite scale of pH
▪ LOW pH = acid ➔ LOW pOH = base
▪ HIGH pH = base ➔ HIGH pOH = acid
117

■ Calculating pOH ➔ Bases
▪Calculating pOH from the [OH-
] concentration
pOH = -log [OH-
]
▪ [OH-
] concentration is equivalent to the hydroxide
ions in solution
118

Example
What is the pOH of a solution that has a hydroxide-ion
concentration [OH-
] of 4.10 x 10-9
M?
[OH-
] = 4.10 x 10-9
M
What do we know???
pOH of solution
pOH = -log [OH-
]
119

Example
What is the pOH of a solution that has a hydroxide-ion
concentration [OH-
] of 4.10 x 10-9
M?
pOH = -log [OH-
] = -log [4.10 x 10-9
]
pOH = 8.4
120

■ Calculating [H+
] concentration
▪When do I use this equation???
▪When I need to calculate [H+
] ion concentration from pH
▪ You need to make sure that you change the pH to a
negative value to determine the correct [H+
]
concentration
[H+
] = antilog (pH)
[H+
] = 10(-pH)
121

Example
What is the hydrogen-ion concentration [H+
] of a
solution with a pH of 6.0?
pH = 6.00
What do we know???
Hydrogen-ion concentration [H+
]
[H+
] = 10(-pH)
122

Example
What is the hydrogen-ion concentration [H+
] of a
[H+
] = 10(-pH)
[H+
] = 1.0 x 10-6
➔ 10(-6.0)
123

■ Calculating [OH-
] concentration
▪When do I use this equation???
▪When I need to calculate [OH-
] ion concentration from pOH
▪ You need to make sure that you change the pOH to a
negative value to determine the correct [OH-
]
concentration
[OH-
] = antilog (pOH)
[OH-
] = 10(-pOH)
124

Example
What is the hydroxide-ion concentration [OH-
] of a
solution with a pOH of 12.5?
pOH = 12.5
What do we know???
Hydroxide-ion concentration [OH-
]
[OH-
] = 10(-pOH)
125

Example
] of a
solution with a pOH of 12.5?
[OH-
] = 3.16 x 10-13
M
[OH-
] = 10(-pOH)
➔ 10(-12.5)
126

■ Understanding pH and pOH
▪ If you are determining the pH of a basic solution
(OR the pOH of an acidic solution), you can use the
following formula:
pH + pOH = 14.00
▪ This provides another option for determining pH,
pOH, [H+
] or [OH-
]
127

Example
What is the pH of a solution has a pOH of 5.25?
pOH = 5.25
What do we know???
pH
What equation(s) are we going to use?
pH + pOH = 14.0 128

Example
What is the pH of a solution has a pOH of 5.25?
pH + pOH = 14.0
pH = 14.0 - pOH
pH = 14.0 – 5.25
pH = 8.75
- pOH - pOH
129

Example
What is the [OH-
] of a solution has a pH of 4.75?
pH = 4.75
What do we know???
[OH-
] of solution
What equation(s) are we going to use?
pH + pOH = 14.0 [OH-
] = 10(-pOH)
130

Example
What is the [OH-
] of a solution has a pH of 4.75?
pOH = 14.0 - pH
pOH = 14.0 – 4.75
pOH = 9.25
pH + pOH = 14.0
- pOH - pOH [OH-
] = 10(-pOH)
[OH-
] = 10(-9.25)
[OH-
] = 5.62 x 10-10
M
131

Example
] of a
pH = 3.35
What do we know???
Hydroxide-ion concentration [OH-
]
[H+
] = 10(-pH) Kw
= [H+
] [OH-
]
132

Example
] of a
[H+
] = 4.47 x 10-4
M
[H+
] = 10(-pH)
[H+
] = 10(-3.35)
Kw
= [H+
] [OH-
]
[OH-
] = Kw
[H+
]
[H+
]
[H+
]
[OH-
] = (1.0 x 10-14
)
(4.47 x 10-4
)
[OH-
] = 2.23 x 10-11
M 133

Essential Question
What is pH and pOH and how is it
calculated?
I will be able to understand how pOH is
used to measure the level of acidity in
a solution
135

■ Calculating pH ➔ Strong Acids and Bases
▪Need to know the following:
▪Major species (ions) dissociated in solution
▪ Need to know the number of H+
or OH-
ions in
solution
▪ The number of ions EQUALS the [H+
] or [OH-
]
concentration
▪Concentration (molarity) of the solution
▪Moles of solute
▪Volume of solution
136

Example
What is the pH of a solution that contains 25.0 grams
of HCl acid dissolved in 22.5 Liters of water
HCl acid and H2
O
What are the major species (ions)?
H+
; Cl-
; H2
O
137

Example
Mass of HCl acid = 25.0 g
Volume of solution = 22.5 L H2
O
What do we know???
pH of solution
138

Example
Convert mass HCl to moles HCl
25.0 g HCl
36.45 g HCl
= 0.686 mol HCl
1 mol HCl
Calculate molarity of HCl
Molarity =
moles
volume
= 0.0305 M
0.686 mol
22.5 L
=
139

Example
pH = -log [H+
] = -log [0.0305]
pH = 1.52
140

Example
of H2
SO4
acid dissolved in 40.25 Liters of water
H2
SO4
acid and H2
O
2 H+
; SO4
2-
; H2
O
141

Example
of H2
SO4
Mass of H2
SO4
acid = 42.0 g
Volume of solvent = 40.25 L H2
O
What do we know???
pH of solution
142

Example
of H2
SO4
Convert mass H2
SO4
to moles H2
SO4
42.0 g H2
SO4
98.1 g H2
SO4
= 0.428 mol H2
SO4
1 mol H2
SO4
Calculate molarity of H2
SO4
Molarity =
moles
volume
= 0.0106 M
0.428 mol
40.25 L
=
143

Example
of H2
SO4
Account for the 2 H+
ions per molecule
Molarity = = 0.0212 M H2
SO4
0.0106 x 2
pH = -log [H+
] = -log [0.0212]
pH = 1.67 144

Example
What is the pOH of a solution that contains 1.77 mol
NaOH dissolved in 2.5 Liters of water?
NaOH (base) and H2
O
Na+
; OH-
; H2
O
145

Example
Moles of base = 1.77 mol NaOH
Volume of solution = 2.5 L H2
O
What do we know???
pOH of solution
146

Example
Calculate molarity of NaOH
Molarity =
moles
volume
= 0.708 M
1.77 mol
2.5 L
=
Calculate pOH of solution
pOH = -log [OH-
] = -log [0.708]
pOH = 0.150
147

Example
What is the pH of a solution that has a pOH of 0.150?
Determine the pH of the solution
pH + pOH = 14.00
pH + (0.150) = 14.00
- (0.150) - (0.150)
pH = 13.85
148

Example
What is the pOH of a solution that has a pH of 1.67?
Determine the pH of the solution
pH + pOH = 14.00
(1.67) + pOH = 14.00
- (1.67) - (1.67)
pH = 12.33
149

■ Acid-Base Neutralization Reactions
▪ A chemical reaction involving an acid and a
base
▪ Products formed with ALWAYS
be a salt (ionic compound) AND
water (H2
O)
▪ GOAL:
▪ Determine how much acid (or base) is
needed to create a solution that
has a pH near 7
151

■ Acid-Base Neutralization Calculations
▪ Need to know the balanced CHEMICAL equation
▪ Know the mole ratio between acid and base
(reactants)
▪ Calculate moles of the REACTANTS
▪ Use the volume provided in the problem
▪ Use a conversion table…volume ➔ mol
▪ Concentration (molarity) = mol/L
▪ You can convert mL to Liters prior to conversion
table
152

■ Acid-Base Neutralization Calculations
▪ Use the “molarity equation” to…
▪ Determine moles, volume or concentration (molarity)
of acid
▪ Determine moles, volume or concentration (molarity)
of base
▪ Use algebra to determine variables
▪ Remember…molarity is moles/Liters (or M)
Molarity =
moles of solute
Liters of solution
153

■ How many moles of sulfuric acid (H2
SO4
) would you
require to neutralize 0.50 mol of sodium hydroxide
(NaOH)
Use mole ratio to determine moles of NaOH
H2
SO4
(aq) + 2 NaOH (aq) ➔ 2 H2
0 (ℓ) + Na2
SO4
(aq)
0.50 mol NaOH
2 mol NaOH
= 0.25 mol H2
SO4
1 mol H2
SO4
155

■ What is the concentration of a HCl solution if 26.0 mL
of it is neutralized by 15.0 mL of 0.0300 M sodium
hydroxide (NaOH) solution?
__ HCl (aq) + __ NaOH (aq) ➔ __ H2
0 (ℓ) + __ NaCl (aq)
1 1 1 1
Moles HCl to Moles NaOH ➔ 1:1
What is the mole ratio between the acid and base?
156

Volume of HCl acid = 26.0 mL
Volume of NaOH = 15.0 mL
What do we know???
Concentration of base ➔ NaOH = 0.0300 M
Concentration (molarity) of HCl acid
__ HCl (aq) + __ NaOH (aq) ➔ __ H2
0 (ℓ) + __ NaCl (aq)
1 1 1 1
157

Need to determine moles of NaOH
mol = (0.0300 M) x (0.0150 L)
= 0.00045 mol NaOH
158

Determine molarity of HCl from volume and moles:
0.00045 mol NaOH
1 mol NaOH
= 0.00045 mol HCl
1 mol HCl
Molarity =
moles
volume
= 0.0173 M
0.00045 mol
0.026 L
=
159

You can solve this problem another way…
Modify the “dilution” equation ➔ M1
V1
= M2
V2
Macid
Vacid
= Mbase
Vbase
160

Macid
Vacid
= Mbase
Vbase
Macid
(26.0 mL) = (0.0300 M) (15.0 mL)
(26.0 mL) (26.0 mL)
= 0.0173 M 161

■ What is the concentration of a sodium hydroxide
(NaOH) solution if it takes 25.0 mL of 0.05 M HCl to
neutralize 345.0 mL of NaOH solution?
■
__ HCl (aq) + __ NaOH (aq) ➔ __ H2
0 (ℓ) + __ NaCl (aq)
1 1 1 1
Moles HCl to Moles NaOH ➔ 1:1
What is the mole ratio between the acid and base?
163

Volume of HCl acid = 25.0 mL
Volume of NaOH = 345.0 mL
What do we know???
Concentration of acid ➔ HCl = 0.05 M
Concentration (molarity) of NaOH 164

Need to determine moles of HCl (acid)
mol = (0.05 M) x (0.0250 L)
= 0.00125 mol HCl
165

Determine molarity of NaOH from volume and moles:
0.00125 mol HCl
1 mol HCl
= 0.00125 mol
NaOH
1 mol NaOH
Molarity =
moles
volume
= 0.0036 M
0.00125 mol
0.345 L
=
166

Macid
Vacid
= Mbase
Vbase
(0.05 M) (25.0 mL) = (Mbase
) (345.0 mL)
(345.0 mL) (345.0 mL)
= 0.0036 M 167

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