Acids and Bases
2003 AP
#1 A-E
Write the Equilibrium,
Kb, for the reaction
represented
Writing an equilibrium expression
for 1a

Using the law of mass action given the chemical
equilibrium equation

Concentration Products over Reactants raised to their
stoichiometric coefficients excluding pure liquids and
solids!
Writing an equilibrium expression
for 1a

In this case, equilibrium expression consists of the
products of the concentrations of Conjugate acid of
Aniline and Hydroxide Ion over the concentration of
Aniline.
A Sample of aniline is
dissolved in water to
produce 25mL of a
0.10M. The pH of the
solution is 8.82. Calculate
the Equilibrium Constant
Calculate the Kb for the reaction

Use the equilibrium expression
above to determine the Kb.

The concentration of aniline is
0.1M
The Hydroxide and Conjugate acid
concentration can both be
determined after calculation of the
pOH (14-8.82).
Take the pOH and raise it to the
10^(-pOH)


Calculate the Kb for the reaction

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This will yield the
concentration for OHand Aniline Conjugate
because they are produce
in the same proportion
1:1.
Multiply these
concentrations and
divide them by the initial
concentration of Aniline.
Calculate Kb


Assume the initial
concentration of [OH-]
is negligible.
The reason x is not
subtracted from the
initial concentration of
aniline is because x is so
small that it is considered
negligible.
The solution prepared in
part b is titrated with a
0.10M. Calculate the pH
of the solution when
5.0mL of the acid has
been added.
Calculate pH after 5mL of HCl is
added

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
Set up a net chemical equation
for the reaction of Aniline and
H+ (Strong Acid)
Since the initial concentrations
of OH- and conjugate acid are
small they are not taken into
account.
Multiply the Molarity of
Aniline by the volume in liters
(same with the HCl)
From here you have the moles
of both Aniline and HCl
Calculate pH after 5mL of HCl is
added



The H+ will combine with
Aniline to form a conjugate
acid and goes to completion.
Subtract the number of
moles of H+ from the moles
of Aniline. This gives a new
number of moles of Aniline.
Since all of the H+ moles are
consumed, it is equal to the
number of moles of the
Conjugate acid.
Calculate pH after 5mL of HCl is
added




To determine the pH, we will
use a variation of the
Henderson-Hassalbauch.
Below
From here we take the pKb
calculated above and the
mole ratio of the acid over
the base.
The pOH can be determined
To get the pOH we subtract
the pOH from 14 to get the
pH.
Calculate the pH at the
equivalence point.
pH at the equivalence point




We already know the moles
of the Aniline.
The moles of Aniline must
equal the moles of HCl for
the equivalence point to be
reached. (Ratio 1:1)
From here we have the only
Aniline Conjugate Acid
moles.
We must determine the
concentration of conjugate
acid and the Ka of the
conjugate
pH at the equivalence point

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

25mL initially of aniline
To determine the volume of HCl take the moles of HCl and
divide it by the concentration which yields the volume required.
Convert all volumes to liters and add the initial volume of aniline
with the volume of HCl added.
Take the moles of conjugate acid and divide it by this new
volume
pH at the equivalence point

To determine the Ka dived the Kb into (1.0 x 10^-14).
pH at the equivalence point



Write the Chemical reaction
for the behavior of Aniline
conjugate in water and make
an equilibrium expression.
x is not subtracted from the
initial concentration because
it is considered negligible.
Multiply the concentration of
Conjugate aniline
concentration by the Ka.
Then take the square root of
the product.
pH at the equivalence point



Then take –log (x) of the
answer
This will give you the pH
at the equivalence point
pH = 2.97
Which of the following
indicators listed is most
suitable for this titration.
Selecting an Indicator



Based on the calculations above the pH at the
end point is 2.97
Erythrosine is optimal because based on the
color change in an acidic pH
It is a weak based titration with a strong acid.
2002
Calculate the value [H+]
in an HOBr solution that
has a pH of 4.95
Calculate [H+]

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
Given the pH is 4.95
Use the exponent 10 raised the –pH
This will yield the [H+]
Write the equilibrium
constant expression for
the ionization of HOBr
in an HOBr solution with
a [H+]= 1.8 x 10^(-5)
Equilibrium constant expression

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Given the chemical equilibrium
reaction and the concentration
of [H+]
[H+] = [OBr-]: This is true
because as HOBr dissociates
the products form in same
proportions
Exclude x because it is
negligible.
Use the Law of Mass action
Products over Reactants raised
to their stoichiometric
coefficients.
Use the equilibrium
constant.
 The product of the
concentration
divided by the Ka
yields the
concentration of
HOBr.

Calculate the volume of
0.115M Ba(OH)2 needed
to reach equivalence
when titrated into 65mL
sample of 0.146M of
HOBr
Calculate Volume Needed to Reach
Equivalent Point

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
Convert the Volume of HOBr to
liters
Multiply the volume in Liters by
the Molarity of HOBr in Solution
This will give you the moles of the
HOBr in solution
The titrant used is Ba(OH)2
(Strong Base) so the concentration
of OH- must be doubled therefore
you must multiply the
concentration of Ba(OH)2 to get
the concentration of OH-
Calculate Volume Needed to Reach
Equivalent Point


Take the Moles of HOBr
(HOBr = OH-) that was
calculated and divide it
by the molarity of OH([Ba(OH)2] x 2).
This will give you the
volume in liters
necessary to added to
reach eq point
Indicate whether the pH
at equivalence point is
less than 7, 7, or greater
than seven. Explain
What will the pH at the eq point be?


You can go through the calculations to
determine the pH specifically at the eq point.
The basic rule of thumb is that if you are
titrating a weak acid with a strong base then the
pH at the end point will be greater than 7.
Calculate the number of
the moles of NaOBr(s)
that would have to be
added to 125mL of
0.160M HOBr to produce
a buffer solution with a
[H+] of 5.00 x 10^(-9)
Number of Moles needed to produce
certain Concentration


The HendersonHasselbauch equation
can be used for this
scenario (See equation
Below)
Given the [H+] take the
–log([H+]) and
determine the pH for the
equation
Number of Moles needed to produce
certain Concentration


You have already been given the Ka so plug that
in the equation as well
Separate the ratio of concentrations into log(B)
– log (A) (Refer to Log Rules)
Number of Moles needed to produce
certain Concentration


Solve for the Log (Base)
and then eliminate the
log function by raising
the solution using the
base of 10.
This will yield your
concentration of Base
Number of Moles needed to produce
certain Concentration


Multiply the concentration by the volume of
solution in liters
This will yield the number of NaOBr moles
HOBr is a weaker acid
than HBrO3. Account
for this fact.
Strong Acid/Weak Acid

The number of oxygen atoms will effect the strength
of the acid. HOBr has a single oxygen atom while
HBrO3 contains three.

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Charge of X

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
Oxygen is a very electro negative atom and will draw
electrons away from the H-Br bond thus weakening the bond
making it easier to dissociate when in dissolved in water. The
more oxygen atoms the weaker the H-Br bond.
The charge on the X which in this case is the Br atom has a
different charge in HOBr (+1) than in HBrO3 (+5)
Strength of O-H Bond
Strength of X-O Bond
Arrhenius Acid Base Concept
Arrhenius Acid/Base Concept

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

Acids produce Hydrogen ions (H+) within an aqueous
solution
Bases produce Hydroxide Ions (OH-) in solution
Definition is limited because it applies only to acids and bases
that can dissociate OH- and H+ ions
Examples:
-
NaOH will dissociate into Na+ and OHHCl will dissociate into H+ and Cl-
Bronsted-Lowry Model

The model definition of Acid/Base

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Bronsted Acid – A proton “donor”
Bronsted Base – A proton “acceptor”
The definition applies to many more molecules that may
exhibit Acid/Base qualities but do not directly produce OHor H+ ions.
Every Acid and Base has a conjugate Acid or Base
Water can act as an acid and a base

H3O+ (Hydronium ion) (acid) and OH- (Base)
Bronsted-Lowry Model

Example of Bronsted Base

NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
Base

Acid
Conjugate Acid
Conjugate Base
Example of Bronsted Acid

HC2H3O2(aq) + H2O(l)  H3O+(aq) + C2H3O2-(aq)
Weak Acid
Base
Conjugate Acid
Conjugate Base
Lewis Acid/Base Definition

Lewis Acid/Base Definition
Lewis Acid – Electron Pair Acceptor
 Lewis Base – Electron Pair Donor
 Encompasses an even wider variety of molecules
(Bronsted and Arrhenius) even ones that do not
donate protons or produce OH- ions.
 Must be aware of the Lewis structure of a particular
molecule to determine whether it is a Lewis Acid or
Base.

Lewis Acid/Base Definition

Examples

BF3(g) + NH3(g)  F3BNH3(g)
BF3 is the Lewis Acid because it has no free unpaired
electrons with only has 6 electrons around the central
atom (Boron will require one more pair of electrons to
complete the valence shell)
 NH3 is the Lewis Base because the molecule has a
completed octet valence shell with free unpaired electrons
on the central atom. The BF3 will accept these unpaired
electrons and form a covalent bond.

Lewis Acid/Base Definition

Example

Ni2+(aq) + 6NH3(aq)  [Ni(NH3)6]2+(aq)
Ni2+ is the Lewis Acid because it is a cation which will
attract negatively charged electrons to toward itself.
 The NH3 is the Lewis Base because it provides the free
unpaired electrons for the Ni2+

Lewis Acid/Base Definition

Example
Even earlier definitions are encompassed in the
Lewis Acid Model
 H+ + H2O  H3O+
