Transcript Chemistry 1011

Chapter 17
Additional Aspects of Acid-Base Equilibria
Dr. Peter Warburton
[email protected]
http://www.chem.mun.ca/zcourses/1051.php
The common-ion effect
Say we make two acid solutions:
0.100 M HCl (a strong acid) and
0.100 M CH3COOH (a weak acid).

HCl  H2O 
 H3O  Cl

A 0.100 M HCl solution by itself would
have a pH  1.0 ([H3O+] = 0.100 M) since
the reaction goes to completion.
2
The common-ion effect
CH 3COOH  H 2O
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
CH3COOH (aq) +
0.100
-x
0.100 – x




H2O (l)
N/A
N/A
N/A


H3O  CH3COO

H3O+ (aq) + CH3COO- (aq)
0.00
0.00
+x
+x
+x
+x
A 0.100 M CH3COOH solution
(Ka = 1.8 x 10-5) by itself would have a
pH  2.8 since an equilibrium is established
where the equilibrium concentration of acetic
acid  0.100 M
x = [H3O+] = [CH3COO-] = 1.3 x 10-3 M
3
The common-ion effect
4
The common-ion effect
Say we put together a solution that is BOTH
0.100 M HCl (a strong acid) and
0.100 M CH3COOH (a weak acid).
The two reactions of
the acids with water take place
in the same container
at the same time!
5
The common-ion effect
HCl  H 2O 
 H3O   Cl 




CH 3COOH  H 2O 
H
O

CH
COO

3
3
Both reactions are a source of
H3O+ and so we could expect that
[H3O+] = 0.100 M + 1.3 x 10-3 M
[H3O+] = 0.1013 M
and pH  0.99
which appears to be true
6
The common-ion effect
HCl  H 2O 
 H3O   Cl 
CH 3COOH  H 2O





H3O  CH 3COO

However, both reactions share the
common ion H3O+, and so they cannot be
treated as independent reactions.
Something that affects one reaction must
also affect the other reaction.
Le Chatalier’s Principle!
7
The common-ion effect
CH 3COOH  H 2O






H3O  CH 3COO
Imagine we start with the 0.100 M
CH3COOH and then we add 0.100 M H3O+
by adding 0.100 M HCl.
Le Chatalier’ Principle tells us our
reaction will shift back towards
reactants!
8
The common-ion effect
CH 3COOH  H 2O
(all in mol/L)
[HCl]initial
[CH3COOH]initial
Conc. changes
Equil. conc.






H3O  CH3COO
CH3COOH (aq) + H2O (l)  H3O+ (aq) + CH3COO- (aq)
0.000
N/A
0.100
0.000
0.100
N/A
0.000
0.000
+x
N/A
-x
+x
0.100 + x
N/A
0.100 - x
+x
If we assume x is much smaller than 0.100 M we
will quickly find that x = Ka
x = [CH3COO-] = 1.8 x 10-5 M
The value of x has decreased because of the
added H3O+!
9
H3O+ as a common-ion
10
OH- as a common-ion
11
Salts as source of a basic common-ion
12
Salts as source of a basic common-ion
On the left is a solution of
0.100 M CH3COOH while
on the right is a solution
that is both 0.100 M
CH3COOH and 0.100 M
CH3COONa which is a
source of the basic
common ion CH3COO-.
The reaction has shifted
back towards reactants!
13
Salts as source of an acidic common-ion
14
Salts as source of an acidic common-ion
On the left is a solution of
0.100 M NH3 while on the
right is a solution that is
both 0.100 M NH3 and
0.100 M NH4Cl which is
a source of the acidic
common ion NH4+.
The reaction has shifted
back towards reactants!
15
Buffer solutions
Solutions that contain both a weak
acid and its conjugate base are
buffer solutions.
These solutions are resistant to
changes in pH.
16
Buffer solutions
The system has enough of the
original acid and conjugate base molecules
in the solution to react with
added acid or added base,
so the new equilibrium mixture will be
very close in composition to the original
equilibrium mixture.
17
Buffer solutions
A 0.10 molL-1 acetic acid – 0.10 molL-1
acetate mixture has a pH of 4.74 and is a buffer
solution!
CH3COOH (aq) + H2O (l)  H3O+ (aq) + CH3COO- (aq)
(all in mol/L)
Initial
Conc. changes
Equil. conc.
CH3COOH (aq) + H2O (l)  H3O+ (aq) + CH3COO- (aq)
0.10
N/A
0.0
0.10
-x
N/A
+x
+x
0.10 - x
N/A
+x
0.10 + x


[H3O ][CH3COO ]
K a  1.8x10 
[CH3COOH]
5
18
Buffer solutions
If we rearrange the Ka expression
K a [CH3COOH]
5 [CH3COOH]
[H3O ] 
 1.8x10

[CH3COO ]
[CH3COO ]

and make the assumption that x is much less
than 0.10, we see
If [CH3COOH] = [CH3COO-],
then [H3O+] = 1.8 x 10-5 M = Ka
and pH = pKa = 4.74
19
Buffer solutions
20
Buffer solutions
What happens if we add 0.01 mol of NaOH (strong
base) to 1.00 L of the acetic acid – acetate buffer
solution?
CH3COOH (aq) + OH- (aq) → H2O (l) + CH3COO- (aq)
This goes to completion and keeps occurring until we
run out of the limiting reagent OH(all in moles)
Initial
Change
Final (where x = 0.01
due to limiting OH-)
CH3COOH (aq) + OH- (aq) → H2O (l) + CH3COO- (aq)
0.10
0.01
N/A
0.10
-x
-x
N/A
+x
0.10 – x
0.01 – x =
N/A
0.10 + x
= 0.09
0.00
= 0.11
New [CH3COOH] = 0.09 M and new [CH3COO-] = 0.11 M
21
Buffer solutions
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
CH3COOH (aq) +
0.09
-x
0.09 – x
H2O (l)
N/A
N/A
N/A

H3O+ (aq) + CH3COO- (aq)
0.0
0.11
+x
+x
+x
0.11 + x
[H3O ][CH3COO ]
(x)(0.11 x)
5
K a  1.8x10 
so 1.8x10 
With the assumption
that x is much smaller
than
[CH3COOH]
(0.09
 x)
5
0.09
mol
(an COOH]
assumption wealways
need to
K a [CH
0.09

5
5
3
[H3O
]


1.8x10

1.5x10
M

check after
calculations
are done!),
[CH COO
]
0.11 we find
3
Note we’ve made the assumption that x << 0.09!
pH = - log [H3O+]
pH = - log 1.5 x 10-5
pH = 4.82
22
Buffer solutions
Adding 0.01 mol of OH- to 1.00 L
of water would have given us a
pH of 12.0!
There is no significant amount
of acid in water for the base to
react with.
23
Buffer solutions
What happens if we add 0.01 mol of HCl (strong acid)
to 1.00 L of the acetic acid – acetate buffer solution?
CH3COO- (aq) + H3O+ (aq) → H2O (l) + CH3COOH (aq)
This goes to completion and keeps occurring until we
run out of the limiting reagent H3O+
(all in moles)
Initial
Change
Final (where x = 0.01
due limiting H3O+)
CH3COO- (aq) + H3O+ (aq) → H2O (l) + CH3COOH (aq)
0.10
0.01
N/A
0.10
-x
-x
N/A
+x
0.10 – x
0.01 – x =
N/A
0.10 + x
= 0.09
0.00
= 0.11
New [CH3COOH] = 0.11 M and new [CH3COO-] = 0.09 M
24
Buffer solutions
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
CH3COOH (aq) +
0.11
-x
0.11 – x
H2O (l)
N/A
N/A
N/A

H3O+ (aq) + CH3COO- (aq)
0.0
0.09
+x
+x
+x
0.09 + x


[H
O
][CH
COO
]
(x)(0.09 x)
5
5
3
3
K a  1.8x10 
so 1.8x10 
With the assumption
x is much smaller than(0.11
0.09mol
[CH3that
COOH]
x)
(an assumption we always need to check after
K a [CH3COOH]

5 0.11
5
calculations
are
done!),
we
find
[H3O ] 

1.8x10

2.2x10

M
[CH3COO ]
0.09
Note we’ve made the assumption that x << 0.09!
pH = - log [H3O+]
pH = - log 2.2 x 10-5
pH = 4.66
25
Buffer solutions
Adding 0.01 mol of H3O+ to 1.00 L
of water would have given us a pH
of 2.0!
There is no significant amount
of base in water for the acid to
react with.
26
Adding acid or base to a buffer
27
Problem
Calculate the pH of a 0.100 L buffer solution
that is 0.25 mol/L in HF and 0.50 mol/L in NaF.
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
HF (aq) +
0.25
-x
0.25 – x
H2O (l)
N/A
N/A
N/A

H3O+ (aq)
0.0
+x
+x
+
F- (aq)
0.50
+x
0.50 + x


[H
O
][F
]
(x)(0.50 x)
4
4
3
K a  3.5x10 
so 3.5x10 
[HF]
(0.25 x)
With the assumption that x is much smaller
than 0.25 mol (an assumption we always need
to check after calculations are done!), we find
28
Problem
K a [HF]
 4 0.25
4
[H3O ] 
 3.5x10
 1.75 x10 M

[F ]
0.50

pH = - log [H3O+]
pH = - log 1.75 x 10-4
pH = 3.76
29
Problem
a) What is the change in pH on addition of
0.002 mol of HNO3?
(all in moles)
Initial
Change
Final (where x = 0.002
due to limiting H3O+)
F- (aq)
0.050
-x
0.050 – x
= 0.048
+ H3O+ (aq) →
0.002
-x
0.002 – x
= 0.00
H2O (l) +
N/A
N/A
N/A
HF (aq)
0.025
+x
0.025 – x
= 0.027
New [HF] = 0.27 M and new [F-] = 0.48 M
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
HF (aq)
0.27
-x
0.27 – x
+
H2O (l)
N/A
N/A
N/A

H3O+ (aq) +
0.00
+x
+x
F- (aq)
0.48
+x
0.48 + x


[H
O
][F
]
(x)(0.48 x)
4
4
3
K a  3.5x10 
so 3.5x10 
[HF]
(0.27 x)
30
Problem
K a [HF]
 4 0.27
4
[H3O ] 

3.5x10

1.9
x10
M
7

[F ]
0.48

Notice we’ve made the assumption that x
<< 0.27. We should check this!
pH = - log [H3O+]
pH = - log 1.97 x 10-4
pH = 3.71
31
Problem
b) What is the change in pH on addition of
0.004 mol of KOH?
(all in moles)
Initial
Change
Final (where x = 0.004
due to limiting OH-)
HF (aq)
0.10
-x
0.025 – x
= 0.021
+ OH- (aq) → H2O (l) +
0.004
N/A
-x
N/A
0.004 – x
N/A
= 0.00
F- (aq)
0.10
-x
0.050 – x
= 0.054
New [HF] = 0.21 M and new [F-] = 0.54 M
(all in mol/L)
Initial conc.
Conc. change
Equil. conc.
HF (aq)
0.21
-x
0.21 – x
+
H2O (l)
N/A
N/A
N/A

H3O+ (aq) +
0.00
+x
+x
F- (aq)
0.54
+x
0.54 + x


[H
O
][F
]
(x)(0.54 x)
4
4
3
K a  3.5x10 
so 3.5x10 
[HF]
(0.21 x)
32
Problem
K a [HF]
 4 0.21
4
[H3O ] 

3.5x10

1.3
x10
M
6

[F ]
0.54

Notice we’ve made the assumption that x
<< 0.21. We should check this!
pH = - log [H3O+]
pH = - log 1.36 x 10-4
pH = 3.87
33
Predicting whether a solution is a buffer
Any solution that becomes a mixture of a
conjugate acid-base pair will be a buffer.
1) Weak acid-base conjugate pairs like
CH3COOH and CH3COO- or NH4+ and NH3.
2) Weak acid reacting with small amounts of
strong base like CH3COOH and NaOH.
3) Weak base reacting with small amounts of
strong acid like NH3 and HCl.
34
Problem
Describe how a mixture of a strong acid
such as HCl and a salt of a weak acid such
as CH3COONa can be a buffer solution.
35
Problem
What is the pH of a buffer solution
prepared by dissolving 23.1 g of NaCHO2
(molar mass is 68.01 gmol-1) in a sufficient
volume of 0.432 M HCHO2 to make 500.0
mL of the buffer?
Ka of formic acid is 1.8 x 10-4
Answer: pH = 3.94
36
The Henderson-Hasselbalch equation
We’ve seen that, for solutions with both
members of a conjugate acid-base pair,
that
[acid]
[H3O ]  K a
[base]

 [acid] 
[acid]
  logKa  log
 log[H3O ]  log K a
[base]
 [base]

pH = pKa + log [base] / [acid]
This is called the HendersonHasselbalch Equation.
37
The Henderson-Hasselbalch Equation
If we have a buffer solution of a conjugate
acid-base pair, then the pH of the solution
will be close to the pKa of the acid.
This pKa value is modified by the logarithm
of ratio of the concentrations of the base
and acid in the solution to give the actual
pH.
38
The Henderson-Hasselbalch equation
pH = pKa + log [base] / [acid]
ALWAYS remember when you use
the H-H eqn that there is the
assumption that the equilibrium
concentrations of acid and base
are relatively unchanged from the
initial concentrations.
That is we have assumed x is very
small compared to the initial
concentrations!
39
The Henderson-Hasselbalch equation
pH = pKa + log [base] / [acid]
Generally this assumption is valid
as long as we know
0.10 < [base] / [acid] < 10
AND
[base] / Ka > 100 AND [acid] / Ka > 100
40
Alternate Henderson-Hasselbalch equation
We can always look at a buffer solution as a
base combined with its conjugate acid
B (aq) + H2O (l)  OH- (aq) + BH+ (aq)


[BH ][OH ]
[B]
[base]

Kb 
so [OH ]  K b
 Kb

[B]
[BH ]
[acid]
 [base] 
[base]
  log K b  log
 log[OH ]  log K b
[acid]
 [acid] 

pOH = pKb + log [acid] / [base]
41
Alternate Henderson-Hasselbalch equation
If we have a buffer solution of a conjugate
acid-base pair, then the pOH of the
solution will be close to the pKb of the
base.
This pKb value is modified by the logarithm
of ratio of the concentrations of the acid
and base in the solution to give the actual
pH.
42
Problem
Use the Henderson-Hasselbalch
equation to calculate the pH of a buffer
solution prepared by mixing equal
volumes of 0.20 mol/L NaHCO3 and
0.10 mol/L Na2CO3.
Ka of HCO3- = 4.7 x 10-11
(see Table 16.4)
We should also check the validity of
using H-H at the end to be sure!
43
Problem answer
If we mix equal volumes, the total volume is
TWICE the volume for the original acid or base
solutions.
Since the number of moles of acid or base
DON’T CHANGE on mixing,
the concentrations will be
half the given values.
pH = pKa + log [base] / [acid]
pH = (-log 4.7 x 10-11) + log (0.05) / (0.10)
pH = 10.33 – 0.30
pH = 10.03
44
Preparing buffer solutions
If we want to create a buffer
solution of a specific pH,
the H-H equation tells us we
need to pick a conjugate
acid-base pair with a pKa for
the acid close to the pH we
want, and then we adjust
the amounts of the
conjugate acid and base.
45
How do we adjust the concentrations?
46
Problem
How many grams of (NH4)2SO4 (molar
mass is 132.141gmol-1) must be dissolved
in 0.500 L of 0.35 M NH3 to produce a
solution with pH = 9.00? Assume that the
solution volume remains at 0.500 L.
Kb for ammonia is 1.8 x 10-5.
Answer: 21 g
47
Buffer capacity
Buffer capacity is the measure of the
ability of a buffer to absorb acid or base
without significant change in pH.
Larger volumes of buffer solutions have
a higher buffer capacity, and buffer
solutions of higher initial concentrations
of the conjugate acid-base pair have a
larger buffer capacity.
48
Buffer range
We’ve seen that as long as
0.10 < [base] / [acid] < 10
then the assumption that the
Henderson-Hasselbalch equation
is based upon
(x << [base] and x << [acid])
is likely to be valid.
49
Buffer range
pH = pKa + log [base] / [acid]
pH = pKa + log 10
pH = pKa + 1.0
OR
pH = pKa + log [base] / [acid]
pH = pKa + log 0.10
pH = pKa - 1.0
50
Buffer range
In general buffer solutions have a
useful range of pH that is pKa  1.0
For instance an acetic acid - acetate buffer
has a useful pH range of about
3.7 to 5.7
since pKa is 4.7 (Ka = 1.8 x 10-5)
51
Application of buffers
Many biological processes can only
occur at very specific pH values
(usually between pH = 6 and pH = 8).
The reactions often take place in
buffered environments (e.g. human
blood is buffered to pH = 7.4 – see
text pg. 734).
52
Acid-base indicators
We often measure the pH of a solution with
a chemical acid-base indicator. Such
indicators are weak acids in their own
right (symbolized HIn) and indicate pH
because the acid form has a different
colour than the conjugate base form (In-)
HIn (aq) + H2O (l)  H3O+ (aq) + In- (aq)
colour A
colour B
53
Diprotic
indicator!
54
Acid-base indicators
HIn (aq) + H2O (l)  H3O+ (aq) + In- (aq)
colour A
colour B
If we increase the [H3O+] we shift this reaction
towards reactants. The colour will be that of HIn.
If we decrease the [H3O+] we shift this reaction
towards products. The colour will be that of In-.
A [H3O+] in between these two extremes will give
a colour that is a mixture of the two colours
because both HIn and In- are present in
significant amounts!
55
Acid-base indicators
HIn (aq) + H2O (l)  H3O+ (aq) + In- (aq)
colour A
colour B
More specifically, the HIn and In- form a
buffer so the indicator works in a pH range
of about 1 around the pKHIn of HIn.
56
57
Since indicators work in a range
of about 2 pH units we often put
several indicators that have
different ranges in a single
solution to give a universal
indicator for pH range of about
1 - 12
58
Universal indicator
59
Applications of indicators
Indicators are useful when
we want a general idea of
the value of the pH
without using a pH meter.
Pool chlorination (with Cl2
or NaOCl) is done to avoid
algae growth. This works
best at pH = 7.4, so we
might need to add some
acid or base…
60
Neutralization reactions
A reaction of an acid and a base
often produces water and an
aqueous salt as products. Such
reactions are called neutralization
reactions, and can be categorized by
the strengths of the acid and base
involved.
61
Strong acid – strong base neutralization
The reaction of a strong acid such as HCl and a
strong base such as NaOH becomes a reaction of
H3O+ and OH-,
Therefore while the overall reaction is
HCl (aq) + NaOH (aq)  H2O (l) + NaCl (aq)
the actual net ionic equation is
H3O+ (aq) + OH- (aq)  2 H2O (l)
where the Na+ and Cl- ions are not involved (they are
neutral spectator ions which don’t react with
water!)
See slides 102, 103 and 106 of last chapter!
62
Strong acid – strong base neutralization
If we mix equal numbers of moles of HCl and
NaOH, we will create twice as much water,
while leaving an excess of NEITHER ion.
In this case the [H3O+] = [OH-] = 1.0 x 10-7 M
at 25 °C (reaction goes to completion).
H3O+ (aq) + OH- (aq)  2 H2O (l)
is the reverse of the autoionization of water
reaction so
Kn 
1
H O  OH 

3

1
1
14




1.0
x
10
at
25
C
14
K w 1.0x10
63
Strong acid – strong base neutralization
Since the equilibrium constant is very
large, we see the reaction goes to
completion, and at the end, the
equilibrium mixture will consist of water
and an aqueous salt of ions that are
neutral in character because they
don’t react with water.
The pH after the reaction will be 7
(a neutral solution).
64
Strong acid titration by strong base
Say we add the strong base to the strong
acid solution drop by drop. Each drop
of base (the titrant) will react with some
of the acid to completion until the added
base (the limiting reagent!) is all gone.
If we measure the pH after we add each
drop of base, we can plot a pH versus
total volume of added base graph.
This is called a titration curve.
65
Titration curve – strong acid titrated by strong base
66
End point versus equivalence point
How do we know when we’ve reached the
end of our titration? We often will use an
indicator to tell us when the titrated
solution reaches a specific pH. This is
the end point of the titration.
The end point DEPENDS
on the indicator we use!
67
End point versus equivalence point
The equivalence point is that point in the
titration where we have added equal numbers
of MOLES of acid and base. For strong acid –
strong base neutralizations
the equivalence point occurs
when the pH is 7
(a neutral solution)
68
Titration curve – strong acid titrated by strong base
Different
end points
depending
on
indicator
69
End point versus equivalence point
Bromothymol blue is a good
indicator for strong acid-strong
base titrations because the end point
is very close to the equivalence
point.
Methyl red and phenolphtalein are
pretty good choices too because
the titration curve is very steep in
their effective pH ranges.
70
Titration curve – strong base titrated by strong acid
Equivalence
point
The curve looks
exactly the same, just
flipped vertically!
71
Millimoles (mmol)
Since we often set up titrations where our
titrant concentration is less than 1 M
and total titrant volume used is less
than 50 mL this means we are adding
about 5.00 x 10-3 moles of acid or base in
our titration.
Because of this, we often do calculations
in millimoles
(1 mmol = 1 x 10-3 mol)
72
Millimoles (mmol)
M = mol / L
M = mmol / mL
73
Problem
For the titration of 25.00 mL of 0.150 M HCl with
0.250 M NaOH, calculate
a) the initial pH;
b) the pH when the neutralization is 50.0%
complete;
c) the pH when the neutralization is 100%
complete; and
d) the pH when 1.00 mL of NaOH is added
beyond the equivalence point.
74
Problem answers
a) pH = 0.824 (Vtotal = 25.00 mL)
b) pH = 1.238 (Vtotal = 32.50 mL)
c) pH = 7.00 (Vtotal = 40.00 mL)
d) pH = 11.79 (Vtotal = 41.00 mL)
75
Weak acid – strong base
Because a weak acid HA is largely undissociated in
water, while the strong base is completely
dissociated in water (becoming a source of OH-), the
neutralization reaction of a weak acid and strong
base becomes that of
HA (aq) + OH- (aq)  H2O (l) + A- (aq)
Acetic acid is a weak acid, so when it reacts with
NaOH, the net ionic equation is
CH3COOH (aq) + OH- (aq)  H2O (l) + CH3COO- (aq)
Now remember that
Knet = K1 x K2 x K3 x … x Kn
76
Weak Acid – Strong Base




5
CH 3COOH (aq)  H 2 O (l)
H
O
(aq)

CH
COO
(aq)
K

1.8
x
10
3
3
a



H 3O  (aq)  OH (aq)
2 H 2O (l)

1
 1.0 x 1014
Kw
__________
__________
__________
__________
__________
__________
_



CH 3COOH (aq)  OH (aq)
H
O
(l)

CH
COO
(aq) K n  K a x
2
3

1
Kw
 1.8 x 109
The equilibrium constant shows us the
reaction goes to completion, so for equal
numbers of moles of weak acid and strong
base, we expect only water and the aqueous
salt in the equilibrium mixture.
77
Weak acid – strong base



CH 3COOH (aq)  OH (aq)
H
O
(l)

CH
COO
(aq) K n  K a x
2
3

1
Kw
 1.8 x 109
However, the salt comprises of a neutral
cation (Na+ in this case) and a weakly
basic anion (CH3COO-), meaning the
equilibrium mixture will be BASIC and have
a pH greater than 7.
See slide 107 of last chapter!
In a weak acid titration by a strong base
this means the equivalence point is NOT
at pH = 7 but rather at pH > 7!
78
Titration curve – weak acid titrated by strong base
3. pH = pKa when moles
added OH- = ½ initial
moles weak acid
4. buffer breaks!
2. creation of a buffer
Titration of
CH3COOH with
NaOH.
1. higher initial pH for a weak acid
6. Added strong
base dominates
weak base and
determines pH
5. pH > 7
because of water
hydrolysis by
conjugate base
79
Problem
For the titration of 20.00 mL of 0.150 M HF with
0.250 M NaOH, calculate
a) the initial pH; Ka = 6.6 x 10-4
b) the pH when the neutralization is 25.0%
complete;
c) the pH when the neutralization is 50.0%
complete; and
d) the pH when the neutralization is 100%
complete.
80
Problem answers
a) pH = 2.00 (Vtotal = 20.00 mL)
b) pH = 2.70 (Vtotal = 23.00 mL)
c) pH = 3.18 (Vtotal = 26.00 mL)
d) pH = 8.08 (Vtotal = 32.00 mL)
81
Weak base – strong acid
Because a weak base B is largely undissociated in
water, while the strong acid is completely
dissociated in water (becoming a source of H3O+),
the neutralization reaction of a strong acid and weak
base becomes that of
H3O+ (aq) + B (aq)  H2O (l) + BH+ (aq)
Ammonia is a weak base, so when it reacts with HCl,
the net ionic equation is
H3O+ (aq) + NH3 (aq)  H2O (l) + NH4+ (aq)
Again
Knet = K1 x K2 x K3 x … x Kn
82
Weak base – strong acid



5
NH 3 (aq)  H 2 O (l) 
NH
(aq)

OH
(aq)
K

1.8
x
10
4
b




H 3 O (aq)  OH (aq)  2 H 2 O (l)


H 3 O  (aq)  NH 3 (aq) 
NH 4 (aq)  H 2 O (l)

1
 1.0 x 1014
Kw
Kn  Kb x
1
Kw
 1.8 x 109
Again the equilibrium constant shows us
the reaction goes nearly to completion, so
for equal number of reactant moles, we
expect only water and the aqueous salt in
the equilibrium mixture.
83
Weak base – strong acid





H3O (aq)  NH3 (aq)
NH4

1
(aq)  H 2O (l) K n  K b x
Kw
 1.8 x 109
However, the salt comprises of a neutral anion
(Cl- in this case) and a weakly acidic cation
(NH4+), meaning the equilibrium mixture will be
ACIDIC and have a pH less than 7.
In a weak base titration by a strong acid this
means the equivalence point is NOT at pH = 7
but rather at pH < 7!
84
Titration curve – weak base titrated by strong acid
pH = 14.00 - pKb when
moles added H3O+ = ½
initial moles weak base
85
Problem
For the titration of 50.00 mL of 0.106 M
NH3 with 0.225 M HCl, calculate
a) the initial pH; Kb = 1.8 x 10-5
b) the pH when the neutralization is 25.0%
complete;
c) the pH when the neutralization is 50.0%
complete; and
d) the pH when the neutralization is 100%
complete.
86
Problem answers
a) pH = 11.14 (Vtotal = 50.00 mL)
b) pH = 9.74 (Vtotal = 55.89 mL)
c) pH = 9.26 (Vtotal = 61.78 mL)
d) pH = 5.20 (Vtotal = 73.55 mL)
87