ACID-BASE
EQUILIBRIUM
ERT 207 ANALYTICAL CHEMISTRY
SEMESTER 1, ACADEMIC SESSION 2015/16
Overview
2
ACID-BASE THEORIES
 ACID DISSOCIATION CONSTANT
 pH SCALE
 METHODS OF MEASURING pH
 POLYPROTIC ACIDS
 WEAK BASES
 BASE DISSOCIATION CONSTANT
 RELATIONSHIP BETWEEN Kw, Ka AND Kb

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Overview
3
BEHAVIOR OF SALTS IN WATER
 SALT SOLUTIONS
 ACIDS-BASE REACTIONS
 BUFFER SOLUTIONS
 HENDERSON-HASSELBALCH EQUATION
 PREPARING A BUFFER

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ACID-BASE THEORIES
4


Arrhenius 1903
Nobel Prize
Arrhenius (or Classical) Acid-Base Definition
 An acid is a substance that contains hydrogen and
dissociates in water to yield a hydronium ion :
H3O+
 A base is a substance that contains the hydroxyl
group and dissociates in water to yield : OH –
Neutralization is the reaction of an H+ (H3O+) ion
from the acid and the OH - ion from the base to
form water, H2O.
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ACID-BASE THEORIES
5
H3O+ =
H+(aq)
=
proton in water
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ACID-BASE THEORIES
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ACID-BASE THEORIES
7


The neutralization reaction is exothermic and
releases approximately 56 kJ per mole of acid
and base.
Brønsted-Lowry Acid-Base Definition
An acid is a proton donor, any species that
donates an H+ ion.
An acid must contain H in its formula; HNO3 and
H2PO4- are two examples, all Arrhenius acids are
Brønsted-Lowry acids.
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ACID-BASE THEORIES
8


A base is a proton acceptor, any species that accepts
an H+ ion.
 A base must contain a lone pair of electrons to bind
the H+ ion; a few examples are NH3, CO32-, F -, as
well as OH -.
 Brønsted-Lowry bases are not Arrhenius bases, but
all Arrhenius bases contain the Brønsted-Lowry base
OH-.
Therefore in the Brønsted-Lowry perspective, one
species donates a proton and another species
accepts it: an acid-base reaction is a proton
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transfer process.
ACID-BASE THEORIES
9
Figure 1: Acid-Base Theories
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ACID-BASE THEORIES
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ACID DISSOCIATION CONSTANT
ACID DISSOCIATION CONSTANT


Strong Acids: 100% dissociation
+
 good H donor
 equilibrium lies far to right (HNO3)
 generates weak base (NO3 )
Weak Acids: <100% dissociation
 not-as-good H+ donor
 equilibrium lies far to left (CH3COOH)
 generates strong base (CH3COO-)
ACID DISSOCIATION CONSTANT
ACID DISSOCIATION CONSTANT
ACID
DISSOCIATION
CONSTANT
Strength
vs. Ka
ACID DISSOCIATION CONSTANT
17
[H3O+] and [OH-]:
[H3O+]
 [OH-]

1x100 to 1x10-14 in water
1x10-14 to 1x100 in water
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ACID DISSOCIATION CONSTANT
18

Finding [H3O+] and [OH-]:
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pH SCALE
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
pH:
 pH is defined as the negative base-10 logarithm
of the hydronium ion concentration.
pH = –log [H3O+]
The “p” in pH tells us to take the negative log of
the quantity (in this case, hydronium ions).
 Some similar examples are :
pOH = –log [OH-]
pKw = –log Kw
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
pH SCALE
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pH SCALE
21

pH and pOH:
+
 As [H3O ] rises, [OH ] falls

As pH falls, pOH rises
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pH
SCALE
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pH
SCALE
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 Some pH
values:
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METHODS OF MEASURING pH
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pH indicator
pH meter
 It measures the voltage
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in the solution.
METHODS OF MEASURING pH
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Figure 1: pH indicators
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METHODS OF MEASURING pH
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
Relationship between Ka and pKa:
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METHODS OF MEASURING pH
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EXAMPLE 1: Calculating Ka from pH
 The pH of a 0.10 M solution of formic acid,
HCOOH, at 25°C is 2.38.
 Calculate Ka for formic acid at this temperature.

We know that
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METHODS OF MEASURING pH
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EXAMPLE 2: Calculating pH from Ka
 Calculate the pH of a 0.30 M solution of acetic
acid, C2H3O2H, at 25°C.

Ka for acetic acid at 25°C is 1.8  10-5.
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POLYPROTIC ACIDS
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Polyprotic acids have more than one acidic proton.
 If the difference between the Ka for the first
dissociation and subsequent Ka values is 103 or
more, the pH generally depends only on the first
dissociation.

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POLYPROTIC ACIDS
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WEAK BASES
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Strength of Bases:
 Strong:
 100% dissociation
 OH- supplied to solution
 NaOH(s)  Na+(aq) + OH-(aq)

Weak:
 <100% dissociation
 OH by reaction with water
 CH3NH2(aq) + H2O(l)
CH3NH2(aq) + OH-(aq)

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WEAK BASES
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Sustainable
 Sustainability.

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BASE DISSOCIATION CONSTANT
33

Bases react with water to produce hydroxide ion.

Sustainability.
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BASE DISSOCIATION CONSTANT
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
Kb can be used to find [OH–] and, through it,
pH.
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BASE DISSOCIATION CONSTANT
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BASE DISSOCIATION CONSTANT
36
EXAMPLE 3: Calculating pH of Basic Solutions
 What is the pH of a 0.15 M solution of NH3?

+] [OH−]
[NH
4
Kb =
= 1.8  10-5
[NH3]
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RELATIONSHIP BETWEEN Kw, Ka AND Kb
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BEHAVIOR OF SALTS IN WATER
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BEHAVIOR OF SALTS IN WATER
39

Properties of salt solutions:
 The spectator ions in acid-base reactions form
salts. Salts completely ionize in their aqueous
solutions.
 A salt NH4A, the ionizes
NH4A ↔ NH4+ + A-

The ions of salts interact with water (called
hydration),
A- + H2O ↔ HA + OHbblee@unimap
BEHAVIOR OF SALTS IN WATER
40

If the anions are stronger base than H2O, the
solution is basic.
NH4+ + H2O = NH3 + H3O+


If the cations are stronger acid than H2O, the
solution is acidic.
These competitive reactions make the solution
acidic or basic depending on the strength of the
acids and bases.
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SALT SOLUTIONS

41

Neutral Salt Solutions:
 Salts consisting of the Anion of a Strong Acid and
the Cation of a Strong Base yield a neutral
solutions because the ions do not react with
water.

HNO3 ( I )  H 2OI   NO3 ( aq )  H 3O ( aq )
Nitrate (NO3-) is a weaker base than water (H2O) ;
reaction goes to completion as the NO3- becomes
fully hydrated; does not react with water
NaNO3 ( s ) 
 Na aq   OH aq 
H 2O


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SALT SOLUTIONS
42


Na+ & NO3- do not react with water, leaving just
the “autoionization” of water, i.e., a neutral
solution.


2 H 2O  H 3O  OH
Salts that produce Acidic Solutions:
 A salt consisting of the anion of a strong acid
and the cation of a weak base yields an acidic
solution.
 The cation acts as a weak acid
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SALT SOLUTIONS
43
The anion does not react with water.
 In a solution of NH4Cl, the NH4+ ion that forms
from the weak base, NH3, is a weak acid.
 The Chloride ion, the anion from a strong acid
does not react with water

NH 4Cl( s )  H 2O 
 NH aq   Cl
H 2O

4

NH 4 aq   H 2O  NH 3 aq   H 3O  ( Acidic )
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SALT SOLUTIONS
44

Salts that produce Basic Solutions:
 A salt consisting of the anion of a weak aid and
the cation of a strong base yields a basic
solution.
 The anion acts as a weak base.
 The cation does not react with water.
 The anion of the weak acid accepts a proton
from water to yield OH- ion, producing a “Basic”
solution.
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SALT SOLUTIONS
45

Salts of Weakly Acidic Cations and Weakly Basic
Anions:
 Overall acidity of solution depends on relative
acid strength (Ka) or base strength (Kb) of the
separated ions.
 Eg. NH4CN - Acidic or Basic?
 Write equations for any reactions that occur
between the separated ions and water


NH 4 ( aq )  H 2OI   NH 3 ( aq )  H 3O
CN  ( aq )  H 2OI   HCN aq   OH 
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SALT SOLUTIONS

Compare Ka of NH4+ & Kb of CN-
46

Magnitude of Kb (Kb > Ka) = (1.6 x 10-5 / 5.7
x 10-10 = 3 x 104),
Kb of CN- >> Ka of NH4+
(Solution is Basic)
 Acceptance of proton from H2O by CN- proceeds
much further than the donation of a proton to
H2O by NH4+.
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SALTS OF WEAKLY ACIDIC CATIONS
AND WEAKLY BASIC ANIONS
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ACIDS-BASE REACTIONS
There are four classifications or types of reactions:
i. strong acid with strong base,
ii. strong acid with weak base,
iii. weak acid with strong base, and
iv. weak acid with weak base.
NOTE:
 For all four reaction types the limiting reactant
problem is carried out first.
 Once this is accomplished, one must determine
which reactants and products remain and write
an appropriate equilibrium equation for the
remaining mixture.
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
48
ACIDS-BASE REACTIONS
(i) STRONG ACID WITH STRONG BASE:
+
 The net reaction is: H
+ OH  H 2O
49
K net

1
14
=
= 1.0 10
Kw
The product, water, is neutral.
(ii) STRONG ACID WITH WEAK BASE:
+
+
+ H 2O
 The net reaction is: H 3O + B  HB
K bB
1
K net =
=
, with 1 < K net < 1.0 1014
K aHB+
Kw
+
 The product is HB and the solution is acidic.
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ACIDS-BASE REACTIONS
(iii) WEAK ACID WITH STRONG BASE:
 The net reaction is: HA + OH
 H 2O + A
K aHA
1
K net =
=
, with 1 < K net < 1.0 1014
KW
K bA The product is A and the solution is basic.
50
(iv) WEAK ACID WITH WEAK BASE:
 The net reaction is:
K aHA
1
14
K net =
=
, with 1 < K net < 1.0 10
KW
K bANotice that Knet may even be less than one.
This will occur when Ka HB+ > Ka HA.
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ACIDS-BASE REACTIONS
Example 4:
 You titrate 100 mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH to the
equivalence point (mol HBz = mol NaOH).
 What is the pH of (a) the final solution
(b) half way point?
51
Note: HBz and NaOH are used up!
HBz + NaOH ---> Na+ + Bz- + H2O
C6H5CO2H
= HBz
Ka = 6.3 × 10-5
Benzoate
ion = BzKb = 1.6 bblee@unimap
× 10-10
BUFFER SOLUTIONS
52
HCl is added to pure
water.
HCl is added to a
solution of a weak
acid H2PO4- and its
conjugate base
HPO42-.
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BUFFER SOLUTIONS

53
The function of a buffer is to resist changes
in the pH of a solution.
 Buffers are just a special case of the
common ion effect.
 Buffer Composition
Weak Acid
+ Conj. Base
HC2H3O2
+
C2H3O2H2PO4+
HPO42Weak Base + Conj. Acid
NH3
+
NH4+
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Acetic acid (HOAc) & a salt
BUFFER SOLUTIONS
of the acetate ion (OAc)
54
Consider HOAc/OAc- to see how buffers work.
 ACID USES UP ADDED OH .
 We know that
OAc + H2O
HOAc + OH
has Kb = 5.6 x 10-10
 Therefore, the reverse reaction of the WEAK
ACID with added OH- has
Kreverse = 1/ Kb = 1.8 x 109
 Kreverse is VERY LARGE, so HOAc completely uses
up the OH !!!!

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BUFFER SOLUTIONS
55
Consider HOAc/OAc- to see how buffers work.
+
 CONJUGATE BASE USES UP ADDED H
+
HOAc + H2O
OAc + H3O
has Ka = 1.8 x 10-5.
 Therefore, the reverse reaction of the WEAK
+
BASE with added H has

Kreverse = 1/ Ka = 5.6 x 104

Kreverse is VERY LARGE, so OAc- completely uses
+
up the H !
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BUFFER SOLUTIONS
56
Example 5
 What is the pH of a buffer that has [HOAc] =
0.700 M and [OAc-] = 0.600 M?
HOAc + H2O
OAc- + H3O+
Ka = 1.8 x 10-5
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BUFFER SOLUTIONS
+
 Notice that the expression for calculating the H
concentration of the buffer is
57
+
[H3O ] =

Orig. conc. of HOAc
This leads to a general equation for finding the H+
or OH- concentration of a buffer.

[Acid]
[H3O ] =
• Ka
[Conj. base]

Orig. conc. of OAc
• Ka

[Base]
[OH ] =
• Kb
[Conj. acid]
Notice that the H+ or OH- concentrations depend
on K and the ratio of acid and base
concentrations.
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HENDERSON-HASSELBALCH EQUATION
[H3O ] =
58

[Acid]
+
[Conj. base]
Take the negative log of both sides of this
equation:
[Acid]
pH pK - log
=
or

• Ka
a
[Conj. base]
[Conj. base]
pH = pKa + log
[Acid]
This is called the Henderson-Hasselbalch
equation.
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HENDERSON-HASSELBALCH EQUATION
[Conj. base]
pH = pK a + log
[Acid]
59
This shows that the pH is determined largely by
the pKa of the acid and then adjusted by the ratio
of acid and conjugate base.
 Sustainability.

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HENDERSON-HASSELBALCH EQUATION
60
EXAMPLE 6

What is the new pH when 1.00 mL of 1.00
M HCl is added to:
a) 1.00 L of pure water (before HCl, pH =
7.00)
b) 1.00 L of buffer that has [HOAc] = 0.700
M and [OAc-] = 0.600 M (pH = 4.68)
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PREPARING A BUFFER
You want to buffer a solution at pH = 4.30.
+
-pH = 5.0 x 10-5 M
 This means [H3O ] = 10
 It is best to choose an acid such that [H3O+] is
about equal to Ka (or pH pKa).
+
 You get the exact [H3O ] by adjusting the ratio of
acid to conjugate base.
 Buffer is prepared from:
 HCO3 weak acid
 CO32conjugate base

61
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PREPARING A BUFFER
62

You want to buffer a solution at pH = 4.30 or
[H3O+] = 5.0 x 10-5 M
POSSIBLE ACIDS
Ka
HSO4- / SO421.2 x 10-2
HOAc / OAc1.8 x 10-5
HCN / CN4.0 x 10-10
 Best choice is acetic acid / acetate.
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PREPARING A BUFFER
63

You want to buffer a solution at pH = 4.30 or
[H3O+] = 5.0 x 10-5 M

[H 3O ] = 5.0 x 10


-5
=
[HOAc]
-
(1.8 x 10
-5
)
[OAc ]
Solve for [HOAc]/[OAc-] ratio = 2.78/1
Therefore, if you use 0.100 mol of NaOAc and
0.278 mol of HOAc, you will have pH = 4.30.
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PREPARING A BUFFER
64
The concentration of the acid and conjugate base
are not important.
 It is the RATIO OF THE NUMBER OF MOLES of
each.
 This simplifying approximation will be correct for
all buffers with 3<pH<11, since the [H]+ will be
small compared to the acid and conjugate base.

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PREPARING A BUFFER
65

Preparing Buffers
1) Solid/Solid: mix two solids.
2) Solid/Solution: mix one solid and one solution.
3) Solution/Solution: mix two solutions.
4) Neutralization: Mix weak acid with strong base
or weak base with strong acid.
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PREPARING A BUFFER
66
Example 7:
 Preparing buffer: solid/solid
 Calculate the pH of a solution made by mixing
1.5 moles of phthalic acid and 1.2 moles of
sodium hydrogen phthalate in 500. mL of solution.
 It is given that Ka= 3.0 x 10-4
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PREPARING A BUFFER
Example 8:
 Preparing buffer: Solid/solution
 How many moles of sodium acetate must be
added to 500. mL of 0.25 M acetic acid to
produce a solution with a pH of 5.50?
67
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EXAMPLE 1
 To
calculate Ka, we need all equilibrium
concentrations.
 We can find [H3O+], which is the same
as [HCOO−], from the pH.
pH = –log [H3O+]
– 2.38 = log [H3O+]
10-2.38 = 10log [H3O+] = [H3O+]
4.2  10-3 = [H3O+] = [HCOO–]
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EXAMPLE 1
 In
table form:
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EXAMPLE 2
 The
equilibrium constant expression is:
 Use
the ICE (Initial Change and
Equilibrium) table:
70
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EXAMPLE 2
 Simplify:
x is relatively same,
71
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EXAMPLE 2
(1.8  10-5) (0.30) = x2
5.4  10-6 = x2
2.3  10-3 = x
pH = –log [H3O+]
pH = – log (2.3  10−3)
pH = 2.64
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EXAMPLE 3
 Tabulate
the data.
 Simplify:
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EXAMPLE 3
2
(x)
1.8  10-5 =
(0.15)
(1.8  10-5) (0.15) = x2
2.7  10-6 = x2
1.6  10-3 = x
 Therefore,
[OH–] = 1.6  10-3 M
pOH = –log (1.6  10-3)
pOH = 2.80
pH = 14.00 – 2.80
pH = 11.20
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EXAMPLE 4 (a)
 The
product of the titration of benzoic
acid, the benzoate ion, Bz-, is the
conjugate base of a weak acid.
 The final solution is basic.
+
+
Kb = 1.6 x 10-10
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EXAMPLE 4 (a)
 This
is a two-step problem:
1st:stoichiometry of acid-base reaction
2nd:equilibrium calculation
1. Calculate moles of NaOH required.
(0.100L HBz)(0.025M) = 0.0025mol HBz
This requires 0.0025 mol NaOH
2. Calculate volume of NaOH required.
0.0025 mol (1 L / 0.100 mol) = 0.025 L
25 mL of NaOH required
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EXAMPLE 4 (a)
3.Moles of Bz- produced = moles HBz
= 0.0025 mol Bz4.Calculate concentration of Bz-.
 There are 0.0025 mol of Bz- in a TOTAL
SOLUTION VOLUME of 125 mL [Bz-]
= 0.0025 mol / 0.125 L = 0.020 M
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EXAMPLE 4 (a)
-10
K b = 1.6 x 10
=
2
x
0.020 - x
 Solving:
x = [OH-] = 1.8 x 10-6,
pOH = 5.75,
pH = 8.25
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EXAMPLE 4 (b)
[H3O+] = { [HBz] / [Bz-] } Ka
 At
the half-way point,
[HBz] = [Bz-], so
[H3O+] = Ka = 6.3 x 10-5
pH = 4.20
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EXAMPLE 5
 Assuming that x << 0.700 and 0.600,
we have
 [H3O+] = 2.1 x 10-5 and pH = 4.68
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EXAMPLE 6
(a) Calculate [HCl] after adding 1.00 mL
of HCl to 1.00 L of water
M1 • V1 = M2 • V2
M2 = 1.00 x 10-3 M
pH = 3.00
(b) Step 1 — do the stoichiometry
H3O+ (from HCl) + OAc- (from buffer)
---> HOAc (from buffer)
 The
reaction occurs completely
because K is very large.
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EXAMPLE 6
 Step
2—Equilibrium.
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EXAMPLE 6
 Because [H3O+] = 2.1 x 10-5 M
BEFORE adding HCl, we again neglect x
relative to 0.701 and 0.599.
[HOAc]
0.701

-5
[H3O ] =
• Ka =
• (1.8 x 10 )
0.599
[OAc- ]
 [H3O+]
= 2.1 x 10-5 M
pH = 4.68

------>
The pH has not changed significantly
upon adding HCl to the buffer!
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EXAMPLE 7
Conjugates
do not
react!!
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EXAMPLE 8
 Let
X = moles NaC2H3O2,
Conjugates
do not
react!!
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