lectures 17-20, Trimester 2, AY 22-23.pptx

Chemistry for Health Sciences
Two credits
Trimester 2, AY 2022/2023
King Saud bin Abdulaziz University for Health Sciences

1. Acids and Bases: Definitions
• Acids: Taste sour, dissolve some metals,
cause plant dye to change color
• Bases: Taste bitter, are slippery, are
corrosive
• Three theories that help us to understand
the chemistry of acids and bases
1. Arrhenius Theory
2. Brønsted-Lowry Theory
3. Lewis Theory

• Acid - a substance, when dissolved in
water, dissociates to produce hydrogen
ions:
– Hydrogen ion: H+ also called “proton”
HCl is an acid:
HCl(aq)  H+(aq) + Cl-(aq)
Arrhenius Theory

• Base - a substance, when dissolved in
water, dissociates to produce hydroxide
ions
NaOH is a base
NaOH(aq)  Na+(aq) + OH-(aq)
Arrhenius Theory

Brønsted-Lowry Theory
• Acid – proton (H+) donor
• Base - proton (H+) acceptor
– Notice that acids and bases are not defined
using water
– When writing the reactions, both accepting
and donating are evident

HCl(aq) + H2O(l)  Cl-(aq) + H3O+(aq)
What donated the proton? HCl
Is it an acid or base? Acid
What accepted the proton? H2O
Is it an acid or base? Base
base
acid

base acid
NH3(aq) + H2O(l) NH4
+(aq) + OH-(aq)
Now let us look at NH3 and see why it is a
base
Did NH3 donate or accept a proton? Accept
Is it an acid or base? Base
What is water in this reaction? Acid

Lewis Theory
• Developed in 1923 by G.N. Lewis.
– This is the most general and comprehensive definition of
acids and bases.
– Emphasis on what the electrons are doing rather than what
the protons are doing.
• Acids are defined as electron pair acceptors.
• Bases are defined as electron pair donors.
• Acid-base reactions are accompanied by coordinate
covalent bond formation.

• One Lewis acid-base example is the ionization of ammonia.
• Look at this reaction in more detail paying attention to the
electrons.
N
H
H
H +
O
H H
N
H
H
H
H +
+ O
H -
Base - it donates
the electron pair
Acid - it accepts
the electron pair Notice that a coordinate
covalent bond is formed
on the ammonium ion.
Lewis Definition
Lewis Theory

• The reaction of lithium fluoride (LiF) and boron
trifluoride (BF3) provides an example of a reaction
that is only a Lewis acid-base reaction.
– It does not involve H+ at all, thus it cannot be an
Arrhenius nor a Brønsted-Lowry acid-base reaction.
+ BF3  Li+ + BF4
-
• You must draw the detailed picture of this reaction
to determine which is the acid and which is the
base.
Lewis Theory

• A base must have a lone pair of electrons to accept H+
from an acid.
• A base can be either neutral (B:) or negatively charged
(B:–).
Summary: Acid-Base Definitions

Summary: Acid-Base Definitions
• Arrhenius, Brønsted-Lowry, and Lewis Acid-Base
theories expand on one another.

Example
• Look at the reaction of ammonia and hydrobromic
acid.
NH3 + HBr NH4
++ Br-
• Is this reaction an example of:
1. Arrhenius acid-base reaction
2. Brønsted-Lowry acid base reaction
3. Lewis acid-base reaction
4. or a combination of these?
You do it!

• The acid base reaction can be written in
the general form:
• Notice the reversible arrows
• The products are also an acid and base
called the conjugate acid and base
HA + B A- + BH+
2. Conjugate Acids and Bases
acid base

• Conjugate Acid – what the base becomes after it
accepts a proton
- BH+ is the conjugate acid of the base B
• Conjugate Base – what the acid becomes after it
donates its proton
- A- is the conjugate base of the acid HA
HA + B A- + BH+
acid acid
base base
Conjugate Acids and Conjugate Bases
• Two species that differ by a proton are called acid-base
conjugate pairs.

• Conjugate Acid-Base Pair – the acid and
base on the opposite sides of the equation
- BH+ / B constitute a conjugate acid-base
pair
- HA / A- constitute a conjugate acid-base
pair
HA + B A- + BH+
acid acid
base base
Conjugate Acid- Base Pair

• An important part of BrØnsted-Lowry acid-base theory
is the idea of conjugate acid-base pairs.
• For example we can use this reaction:
• HNO3 + H2O  H3O+ + NO3
-
1. Identify the reactant acid and base.
HNO3 is the acid, conjugate base is NO3
-
2. Find the species that differs from the acid by a proton,
that is the conjugate base.
H2O is the base, conjugate acid is H3O+
Conjugate Acid- Base Pair

Acid-Base Properties of Water
• Water possesses both acid and base properties
– Amphoteric – a substance possessing both acid
and base properties
– Water is the most commonly used solvent for both
acids and bases
– Solute-solvent interactions between water and both
acids and bases promote solubility and dissociation

Conjugate Acid-Base Practice
Write the chemical reaction for the following acids
or bases in water.
Identify the conjugate acid base pairs.
1. HF (a weak acid)
2. H2S (a weak acid)
3. HNO3 (a strong acid)
4. CH3NH2 (a weak base)
Note: The degree of dissociation also defines weak
and strong bases

3. Acid and Base Strength
• Acid and base strength – degree of
dissociation
– Not a measure of concentration
– Strong acids and bases – reaction with water is
virtually 100% (Strong electrolytes)
– Example:
HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq)
KOH(aq) K+(aq) + OH-(aq)
strong acid
strong base

Strong Acids and Bases
• Strong Acids:
– HCl, HBr, HI Hydrochloric Acid, etc.
– HNO3 Nitric Acid
– H2SO4 Sulfuric Acid
– HClO4 Perchloric Acid
• Strong Bases:
– NaOH, KOH, Ba(OH)2
– All metal hydroxides
Memorize

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)
H2CO3(aq) + H2O(l) HCO3
-(aq) + H3O+(aq)
Weak Acids
• Weak acids and bases – only a small
percent dissociates (Weak electrolytes)
• Weak acid examples:
– Acetic acid:
– Carbonic Acid:

• Weak base examples:
– Ammonia:
– Pyridine:
– Aniline:
C6H5NH2(aq) + H2O(l) C6H5NH3
+(aq) + OH-(aq)
C5H5NH2(aq) + H2O(l) C5H5NH3
+(aq) + OH-(aq)
NH3(aq) + H2O(l) NH4
+(aq) + OH-(aq)
Weak Bases

HA + B A- + HB+
Acid-Base Dissociation
• The reversible arrow isn’t always written
– Some acids or bases essentially dissociate 100%
– One way arrow is used
• HCl + H2O  Cl- + H3O+
– All of the HCl is converted to Cl-
– HCl is called a strong acid – an acid that dissociates
100%
• Weak acid - one which does not dissociate 100%

• The weaker the acid or base, the stronger the
conjugate partner.
• The relative strength of any acid depends on the
weakness (or stability) of its conjugate base:
Acid strength order HCN < CH3COOH < HF
Conjugate base order CN- > CH3COO- > F-
Conjugate Acid-Base Strength

• An acid-base equilibrium always favors reaction
of the stronger acid with the stronger base, and
formation of the weaker acid and base.

• Determine in which direction the equilibrium is
favored in these examples

• Pure water is virtually 100% molecular
• Very small number of molecules dissociate
– Dissociation of acids and bases is often called
ionization
• Dissociation of water is self-ionization
• Very weak electrolyte
H2O(l) + H2O(l) H3O+(aq) + OH-(aq)
4. Self-ionization of Water

• H3O+ is called the hydronium ion
• In pure water at room temperature:
– [H3O+] = 1  10-7 M
– [OH-] = 1  10-7 M
• What is the equilibrium expression for:
Remember, liquids are not included in equilibrium
expressions
H2O(l) + H2O(l) H3O+(aq) + OH-(aq)
Self-ionization of Water
Keq = [H3O+][OH-]

• The ion product constant for water as the
symbol Kw
• In pure water, [H3O+] = [OH-] = 1.0  10-7
M, what is the value for Kw?
1.0  10-14 (unitless)
• Kw depends on Temperature only. It has constant
value in all aqueous solutions
• The values of [H3O+] and [OH-] varies based on
the solution acidity and basicity
Ion Product Constant for Water
Kw = [H3O+][OH-]

pH: A Measurement Scale for Acids & Bases
• The common range for [H3O+] or [OH-] values is 1 to
1 x 10-14.
• In most solutions, [H3O+] is fraction (Ex: 1 x 10-7 =
0.0000001), therefore we express acidity using pH scale.
• pH scale – a scale that indicates the acidity or basicity of
a solution using this equation:-
• Common ranges from 0 (very acidic) to 14 (very basic)
• The pH scale is rather similar to the temperature scale
assigning relative values of hot and cold
pH = -log[H3O+]

Molarity and pH
Each 10 fold change
in concentration
changes the pH
by one unit

• Use these observations to develop a concept of
pH
– Acidic solution, [H3O+]  and [OH-]  (pH < 7)
– Basic solution, [OH-]  and [H3O+]  (pH > 7)
– Neutral solution, [H3O+] = [OH-]. (pH = 7)
• In all of these cases:
1.0  10-14 = [H3O+][OH-]
A Definition of pH
(All above numbers are based on Kw = 1.0 x 10-14
and Kw changes only with temperature)

Measuring pH
• Approximated using indicator / pH paper that develops a color
related to the solution pH
• Measured using pH meter whose sensor measures an
electrical property of the solution that is proportional to pH

pH Scale
pH = -log [H+] same as -log [H3O+]
Thus, for a solution in which
[H+]= 1 x 10 -7
pH = - log(10 -7) = 7

pH and pOH Scales
• A convenient way to express the acidity and basicity
of a solution is the pH and pOH scales.
• The pH of an aqueous solution is defined as:
 
pH = -log H O
3
+
• In general, a lower case p before a symbol is read as the
‘negative logarithm of’ the symbol.
• Thus we can write the following notations.
quantities
other
for
forth
so
and
+
Ag
-log
=
pAg
-
OH
-log
=
pOH













• How do we calculate the pH of a solution when either
the H3O+ or OH- ion concentration is known?
• How do we calculate the H3O+ or OH- ion
concentration when the pH is known?
• Use the expressions:
Calculating pH
[H3O+][OH-] = 1.0 x 10-14
pH = -log[H3O+]
14.00
pOH
pH 


Calculating pH from Acid Molarity
What is the pH of a 1.0  10-4 M HCl solution?
– HCl is a strong acid and dissociates in water
– 1 mol HCl produces 1 mol [H3O+] in solution
– Therefore, 1.0  10-4 M HCl solution has
[H3O+] = 1.0  10-4M
pH = -log [H3O+]
pH = -log [1.0  10-4]
pH = -[-4.00] = 4.00

Calculating [H3O+] from pH
What is the [H3O+] of a solution with pH = 6.00?
• An alternative mathematical form of this equation is:
[H3O+] = 10-pH
pH = -log[H3O+]
[H3O+] = 10-6.00
[H3O+] = 1.0 x 10-6 M

Calculating the pH of a Base
What is the pH of a 1.0  10-3 M KOH solution?
• KOH is a strong base, so it dissociates
completely
• 1 mol KOH produces 1 mol OH- ion in solution
• Therefore, 1.0 x 10-3 M KOH has
[OH-] = 1.0 x 10-3 M
• To use the pH formula, we need to know the
[H3O+] instead:
pH = -log[H3O+]

Calculating the pH of a Base
What is the pH of a 1.0  10-3 M KOH solution?
• Therefore the Kw equation is needed:
1.0 x 10-14 = [H3O+][OH-]
1.0 x 10-14 = [H3O+][1.0  10-3 ]
1.0 x 10-11 = [H3O+]
• Divide both sides by 1.0 x 10-3 :
• Now use the pH equation:
pH = -log (1.0 x 10-11)
pH = 11.00
pH = -log[H3O+]

Calculating [OH-] from pH
What is the [OH-] of a solution with pH = 5.00?
• First find [H3O+]
[H3O+] = 10-pH
[H3O+] = 10-5.00
[H3O+] = 1.0  10-5
• Next, use the Kw equation to find [OH-]
1.0 x 10-14 = [H3O+][OH-]
1.0 x 10-14 = 1.0  10-5 [OH-]
1.0 x 10-9 = [OH-]

5. Acid Ionization and the Acidity Constant Ka
The strength of an acid is represented by its
ionization constant (acidity constant), Ka
Ka= product of concentrations of ionized species
concentration of intact acid
HA + H2O A H3O
+
Ka =
A H3O
HA

The Acidity Constant, Ka
The Ka implies the concentrations of the acid and the ions
Ka > 1 Ionized products greater than intact acid.
Ka < 1 Ionized products less than intact acid.
Ka >> 1 Ionization goes to completion (strong acid).
(e.g., > 103)
Ka << 1 Ionization occurs partially.
(e.g., < 10–3)

The Acidity Constant, Ka
pKa = – log (Ka)
Since the Ka values for various acids have
such a wide range, a more manageable way to
discuss this measure of acidity is to use

Compare pKa and Ka Values
47
pKa
14
12
10
8
6
4
2
0
strong acids weak acids
Ka
10-14
10-10
10-6
10-2
The smaller the value of the pKa
the stronger the acid.
-2
102
47

Ionization of Weak Monoprotic Acids
 

  -
3
3
2
3 COO
CH
O
H
O
H
COOH
CH
• Let’s look at the dissolution of acetic acid, a weak
acid, in water as an example.
• The equation for the ionization of acetic acid is:
  
 
acid
acetic
for
10
8
.
1
COOH
CH
COO
CH
O
H
K 5
3
3
3
a
-
-





Weak Monoprotic Acids
Example 1: Write the equation for the ionization of the
weak acid HCN and the expression for its ionization
constant.
  
 
HCN
CN
H
K
CN
H
HCN
-
a
-



 


Weak Monoprotic Acids
• The ionization constant values for several acids are given
below.
– Which acid is the strongest? (Values not to be memorized)
Acid Formula Ka value
Acetic CH3COOH 1.8 x 10-5
Nitrous HNO2 4.5 x 10-4
Hydrofluoric HF 7.2 x 10-4
Hypochlorous HClO 3.5 x 10-8
Hydrocyanic HCN 4.0 x 10-10

Weak Bases
B(aq) + H2O(l) BH+
(aq) + OH–
(aq)
Kb is the basicity constant
Kb = [BH+][OH–]/[B]
base Conjugate
acid
Conjugate
base
acid
Kb always refers to the reaction of a base with water to
form the conjugate acid and the hydroxide ion.

52 52
Weak Monoprotic Bases
(Values not to be memorized)

NH3 is a base Kb = 1.8 x 10-5
NH3 + H2O NH4
+ + OH-
base1 acid1 acid2 base2
NH4
+ is an acid Ka = 5.6 x 10-10
NH4
+ NH3 + H+
acid2 base1
Ammonia as a Weak Base
Ka Kb = Kw = (5.6 x 10-10) (1.8 x 10-5) = 10-14
pKa + pKb = pKw = (9.25) + (4.75) = 14 53

54
Summary: Acid & Base Strength

6. Reactions Between Acids and Bases
Neutralization Reaction
• Neutralization reaction – the reaction of an acid
with a base to produce a salt and water
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Acid Base Salt Water
• Break apart into ions:
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) 
Na+(aq) + Cl-(aq) + H2O(l)
• Net ionic equation
– Show only the changed components
– Omit any ions appearing the same on both sides of
equation (Spectator Ions)
H+(aq) + OH-(aq)  H2O(l)

• The net ionic neutralization reaction is more
accurately written:
H3O+(aq) + OH-(aq)  2H2O(l)
• This equation applies to any strong acid / strong
base neutralization reaction
• An analytical technique to determine the
concentration of an acid or base is titration
• Titration involves the addition of measured
amount of a standard solution to neutralize the
second, unknown solution
• Standard solution – solution of known
concentration
Neutralization Reaction

Buret – long glass
tube calibrated in mL
which contains the
standard solution
Flask contains a
solution of unknown
concentration plus
indicator
Indicator – a
substance which
changes color as
pH changes
• Standard solution
is slowly added
until the color
changes
• The equivalence
point is when the
moles of H3O+
and OH- are equal
Acid – Base Titration

Titration Experiment
• Place a known volume of acid whose concentration
is not known into a flask
• Add an indicator
• Known concentration of base is placed in a buret
• Drip base into the flask until the indicator changes
color

• Indicator changes color
– equivalence point is reached
– mol OH- = mol H3O+ present in the unknown acid
• Volume dispensed from buret is determined
• Calculate acid concentration from the following
data:
– Volume of HCl: 25.00 mL
– Volume of NaOH added: 35.00 mL
– Concentration of NaOH: 0.1000 M
– Balanced reaction shows that 1 mol HCl reacts with
1 mol NaOH (a 1:1 ratio)

• At the equivalent point (endpoint):
moles OH- = moles H+
moles NaOH = moles of HCl
MNaOH x VNaOH = MHCl x VHCl
MHCl is unknown and can be calculated as follow:-
MHCl = (MNaOH x VNaOH) / VHCl

• The previous example have the acid and base at
a 1:1 combining ratio
– Not all acid-base pairs do this
• Polyprotic substance – donates or accepts more
than one proton per formula unit
– HCl is monoprotic, producing one H+ ion for each
unit of HCl
– Sulfuric acid is diprotic, each unit of H2SO4
produces 2 H+ ions
H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2 H2O(l)
Polyprotic Substances

Titration of polyprotic acids
H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2 H2O(l)
Example:- If 20.0 mL of H2SO4 is titrated with 36.5 mL of
0.10 M NaOH, what is the concentration of H2SO4.
Solution: According to the equation above, at the equivalent
point:
moles of NaOH = 2 moles of H2SO4
MNaOH.VNaOH = 2 MH2SO4.VH2SO4

7. Buffer Solutions
• Buffer solution - solution which resists large
changes in pH when either acids or bases are added
• These solutions are frequently prepared in
laboratories to maintain optimum conditions for
chemical reactions
• Blood is a complex natural buffer solution
maintaining a pH of ~7.4 using mainly carbonic
acid (H2CO3) and bicarbonate (HCO3
-) ions

• Buffers act to establish an equilibrium between a
conjugate acid/base pair
• Buffers consist of either
– a weak acid and its salt (conjugate base)
– a weak base and its salt (conjugate acid)
• An example is acetic acid, CH3COOH, and its
conjugate base CH3COO-
• An equilibrium is established in solution
between the acid and the salt anion
Buffer Solutions

Buffer Solutions
Buffer: A combination of substances that act together
to prevent a drastic change in pH; Example: Weak acid
and its conjugate base

• If OH– is added to a buffer solution, the pH increases
only slightly; the acid component of the buffer
neutralizes the added OH–.
• If H+ is added to a buffer solution, the pH decreases
only slightly; the base component of the buffer
neutralizes the added H+.
Buffer Solutions

• Buffering process is an equilibrium reaction
described by an equilibrium-constant
expression
– In acids, this constant is Ka
• If you want to know the pH of the buffer,
solve for [H3O+], then calculate pH
Determining Buffer Solution pH
Ka =
[H3O+][CH3COO-]
[CH3COOH]

Henderson-Hasselbalch Equation
• Equilibrium-constant expression and the pH
expression can be combined
– Henderson-Hasselbalch Equation is this combined
expression
• For the acetic acid/sodium acetate buffer system:
• Taking the – log of both sides:
Ka =
[H3O+][CH3COO-]
[CH3COOH]
- log Ka = - log [H3O+] - log
[CH3COO-]
[CH3COOH]

Henderson-Hasselbalch Equation
• The – log Ka is actually the pKa, much like
– log [H3O+] is actually the pH:
• Solving for pH:
• The generalized expression is:
pKa = pH - log
[CH3COO-]
[CH3COOH]
pH = pKa  log
[CH3COO-]
[CH3COOH]
pH = pKa  log
[conjugate base]
[weak acid]

Calculating the pH of a Buffer Solution
Calculate the pH of a buffer solution in which
– Both the acetic acid (acid) and sodium acetate (salt)
concentrations are 2.0  10-2 M
– The Ka for acetic acid is 1.75 x 10-5
pH = pKa  log [base]
[acid]
pH = pKa = -log(1.75 x 10-5)
pH = 4.76
Example:
Solution:

Example:
A buffered solution contains 0.25 M NH3
(Kb = 1.8 x 10-5 ) and 0.4 M NH4 Cl .
a. Calculate the pH of this solution.
Solution:
NH3 + H2 O ↔ NH4
+ + OH-
Ka = Kw / Kb = 10-14 / 1.8 x 10-5 = 5.6 x 10-10
pH = pKa + log (base/acid)
= 9.25 + log ( 0.25/0.40)
= 9.25 - 0.20 = 9.05

Buffer Capacity
• Buffer Capacity – a measure of the ability
of a solution to resist large changes in pH
when a strong acid or strong base is added
Higher Buffer
Capacity
Lower Buffer
Capacity

- Optimum (most effective) buffering will occur when
[HA] = [A-]. It is under this condition that their ratio is
most resistant to H+ or OH- added.
Buffer Capacity
- pKa of weak acid selected for buffer should be as
close as possible to desired pH (i.e. [HA] / [A-] = 1).
- pH of a buffered solution is determined by the ratio
[A-]/[HA]. The capacity of a buffered solution is
determined by the magnitudes of [HA] and [A-].

Example:
Which of the organic acids in the following table would be the most
appropriate for preparing a pH 4.15 buffer solution?
Solution:
Looking at the Ka and pKa values in the table, the ascorbic
acid (pKa = 4.10) will produce a buffer solution closest to
the desired pH of 4.15.

Example:
Which of the two solutions has higher buffer capacity?
(1) Solution A : 5.00 M HC2H3O2 and 5.00 M NaC2H3O2
(2) Solution B :0.05 M HC2H3O2 and 0.05 M NaC2H3O2
For acetic acid, Ka = 1.8 x 10-5
Solution A, with larger quantities of buffering components,
has a much higher buffering capacity than solution B.

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