ACIDS AND BASES
REVISE
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•
•
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Brønsted-Lowry acids and bases
Amphoteric substances
Conjugate acid base pairs
Neutralisation
Neutral:
Acidic:
Basic:
pH = 7
pH < 7
pH > 7
pH = -log [H+]
pOH = -log [OH-]
pH + pOH = 14
at 25oC
([H+] = [OH-])
([H+] > [OH-])
([H+] < [OH-])
Kw = [OH-][H+]
Kw = 14.0
at 25oC
Kw = KaKb
STRONG ACIDS AND BASES
Strong acids and bases  react nearly “completely” to
produce H+ and OH equilibrium constants are large
e.g.: HCl
H+ + Cl-
[H ][Cl  ]
K
[HCl]
Complete dissociation:
large
small
Common strong acids:
HCl, HBr, HI, H2SO4, HNO3, HClO4
(Why is HF not a strong acid?)
Common strong bases:
LiOH, NaOH, KOH, RbOH, CsOH, R4NOH
Example:
Calculate the pH of 0.1M LiOH.
LiOH  Li+
Start:
Complete rxn:
+
OH-
Problem:
What is the pH of 1x10-8 M KOH?
As before:
pOH = -log (1x10-8) = 8
pH = 14 – 8 = 6
BUT
pH 6  acidic conditions and KOH is a strong base
IMPOSSIBLE!!!!!!
Since the concentration of KOH is so low (1x10-8 M), we
need to take the ionisation of water into account.
In pure water [OH-] = 1x10-7 M, which is greater than the
concentration of OH- from KOH.
We do this by systematic treatment of equilibrium.
Charge balance:
[K+] + [H+] = [OH-]
Mass balance:
[K+] = 1x10-8 M
Equilibria:
[H+][OH-] = Kw = 1x10-14 M
3 equations + 3 unknowns  solve simultaneously
Find:
pH = 7.02
Hint: You end up with a quadratic equation which you solve using the formula.
Also note that:
• Only pure water produces 1x10-7 M H+ and OH-.
• If there is say 1x10-4 M HBr in solution,
pH = 4 and [OH-] = 1x10-10 M
• But the only source of OH- is from the dissociation
of water.
 if water produces 1x10-10 M OH- 
it can only produce 1x10-10 M H+ due to the
dissociation of water.
 pH in this case is due mainly to the dissociation of
HBr and not the dissociation of water.
• It is thus important to look at the concentration
of acid and bases present.
Some guidelines regarding the concentrations of acids
and bases:
1) When conc > 1x10-6 M
 calculate pH as usual
2) When conc < 1x10-8 M
 pH = 7
(there is not enough acid or base to affect the
pH of water)
3) When conc  1x10-8 - 1x10-6 M
 Effect of water ionisation and added acid and
bases are comparable, thus:
use the systematic treatment of equilibrium
approach.
WEAK ACIDS AND BASES
Weak acids and bases  react only “partially” to
produce H+ and OH equilibrium constants are small
HA
H+ + A-
[H ][A  ]
Ka 
[HA]
Partial dissociation
small
large
Acid dissociation constant
Common weak acids:
• carboxylic acids
(e.g. acetic acid = CH3COOH)
• ammonium ions
(e.g. RNH3+, R2NH2+, R3NH+)
Common weak bases:
• carboxylate anions
(e.g. acetate = CH3COO-)
• amines
(e.g. RNH2, R2NH, R3N)
Base hydrolysis:
B + H2O
BH+ + OHWeak base
[BH ][OH ]
Kb 
[B]
 partial dissociation
 Kb small
base hydrolysis constant/
base “dissociation” constant
NOTE:
pKa = -log Ka
pKb = -log Kb
As K increases, its p-function decreases and vice versa.
Problem:
Find the pH of a solution of formic acid given that the
formal concentration is 2 M and Ka = 1.80x10-4.
HCOOH
H2O
H+ + HCOOH+ + OH-
Systematic treatment of equilibria:
Charge balance:
Mass balance:
[H+] = [HCOO-] + [OH-]
2 M = [HCOOH] + [HCOO-]
Equilibria:
[H ][HCOO ]
Ka 
[HCOOH
K w  [H ][OH  ]  1 1014
4 equations  4 unknowns
 difficult to solve
Make an assumption:
[H+] due to acid dissociation  [H+] due to water dissociation
Produces HCOO-
Produces OH-
[HCOO-] large
[OH-] small
 [HCOO-] >> [OH-]
 Charge balance:
[H+]  [HCOO-]
Charge balance:
Mass balance:
[H+]  [HCOO-]
2 M = [HCOOH] + [H+]
Equilibria:
[H ][COOH  ]
Ka 
[HCOOH]
K w  [H ][OH  ]  1 1014
Let [H+] = [HCOO-] = x
x2
Ka 
F x
x.x
x2
Ka 

 1.8  10 4
[HCOOH] 2  x
x 2  1.8  104 x  3.6  104  0
x  0.019
[H+] = [HCOO-] = 0.019 M
Or x = -0.019
 No negative conc’s
 pH = 1.7
OR since [HCOOH] > 1x10-6, we can calculate pH as usual
Weak acid
equilibrium conditions
HCOOH
Start:
Equilibrium:
2M
2-x
H+
0
x
+
HCOO0
x
[H ][COOH  ]
Ka 
[HCOOH]
x.x
x2
Ka 

 1.8  104
2- x 2  x
Solve as
before
FRACTION OF DISSOCIATION, 
Fraction of acid in the form A
[A  ]

[A ]  [HA]

[A  ]

[A  ]  F  [A  ]

[A  ]
 
F
For the above problem:
[HCOO-]
 =
F
 Acid is ________ dissociated at 2 M formal concentration
Weak electrolytes dissociate more as they are diluted.
WEAK BASE EQUILIBRIA
B + H2O
BH+ + OH-
Charge balance:
[BH+] = [OH-]
Mass balance:
F = [B] + [BH+]
Equilibria:
[BH ][OH  ]
Kb 
[B]
Let [BH+] = [OH-] = x
x. x
x2
Kb 

[B] F  x
FRACTION OF ASSOCIATION
[BH ]

F
CONJUGATE ACIDS AND BASES
Relationship between Ka and Kb for a conjugate acidbase pair:
Ka.Kb = Kw = 1x10-14
at 25oC
If Ka is very large (strong acid)
Then Kb must be very small (weak conjugate base)
And vice versa
Base so weak it is not
a base at all in water
If Ka is very small, say 1x10-6 (weak acid)
Then Kb must be small, 1x10-8 (weak conjugate base)
Greater acid strength, weaker conjugate base strength,
and vice versa.
Problem:
Calculate the pH of 0.1 M NH3, given that pKa = 9.244 for
ammonia.
NH3 + H2O
base
acid
Kb
NH4+ + OH-
Problem:
Calculate the pH of 0.1 M NH3, given that pKa = 9.244 for ammonia.
pH = 11.12
BUFFERS
Mixture of an acid and its conjugate base.
Buffer solution  resists change in pH when acids
or bases are added or when dilution occurs.
Mix:
A moles of weak acid + B moles of conjugate base
Find:
• moles of acid remains close to A, and
• moles of base remains close to B
 Very little reaction
HA
H+ + A-
Le Chatelier’s principle
HENDERSON-HASSELBALCH EQUATION
For acids:
[A  ]
pH  pK a  log
[HA]
When [A-] = [HA],
pH = pKa
For bases:
pH  pK a  log
[B]
[BH ]
pKa applies
to this acid
Kb
 BH+ + OHB + H2O 
base
acid
Ka
acid
base
?
Why does a buffer resist change in pH when small
amounts of strong acid or bases is added?
The acid or base is consumed
by A- or HA respectively
A buffer has a maximum capacity to resist change to
pH.
Buffer capacity, :
 Measure of how well solution resists change in pH
when strong acid/base is added.
dCb  dCa


dpH dpH
Larger   more resistance to pH change
A buffer is most effective in
resisting changes in pH
when:
pH = pKa
i.e.:
[HA] = [A-]
 Choose buffer whose pKa
is as close as possible to
the desired pH.
pKa  1 pH unit
Problem:
Calculate the pH of a solution containing 0.200 M NH3
and 0.300 M NH4Cl given that the acid dissociation
constant for NH4+ is 5.7x10-10.
pH = 9.07
POLYPROTIC ACIDS AND BASES
Can donate or accept more than one proton.
In general:
Diprotic acid:
Diprotic base:
H2L
HL- + H+
Ka1  K1
L2- + H2O
HL- + OH-
Kb1
HL-
L2- + H+
Ka2  K2
HL- + H2O
H2L + OH-
Kb2
Relationships between Ka’s and Kb’s:
Ka1. Kb2 = Kw
Ka2. Kb1 = Kw
Using pKa values and mass balance equations, the
fraction of each species can be determined at a
given pH.
ACID BASE TITRATIONS
We will construct graphs to see how pH changes as
titrant is added.
Start by:
• writing chemical reaction between titrant and analyte
• using the reaction to calculate the composition and
pH after each addition of titrant
TITRATION OF STRONG BASE WITH
STRONG ACID
Example:
Titrate 50.00 ml of 0.02000 M KOH with 0.1000 M HBr.
HBr + KOH
KBr + H2O
What is of interest to us in an acid-base titration:
H+ + OH-
H2O
Mix strong acid and strong base
 reaction goes to completion
H+ + OH-  H2O
Example:
Titrate 50.00 ml of 0.02000 M KOH with 0.1000 M HBr.
* Calculate volume of HBr needed to reach the
equivalence point, Veq:
C1V1
C2V2
=
n1
n2
There are 3 parts to the
titration curve:
1) Before reaching the
equivalence point
1
 excess OH- present
2) At the equivalence
point
2
 [H+] = [OH-]
3
3) After reaching the
equivalence point
 excess H+ present
1) Before reaching the equivalence point
HBr + KOH
 excess OH- present
Say 2.00 ml HBr has been added.
COH- =
nunreacted
Vtotal
KBr + H2O
Starting nOH= (0.02 M)(0.050 L)
= 1x10-3 mol
nH+ added
= (0.1 M)(0.002 L)
= 2x10-4 mol
COH- = 0.01538 M
Vtotal = 50 + 2 mL
= 52 mL
= 0.052 L
 nOH- unreacted
= 8x10-4 mol
Kw = [H+][OH-]
1x10-14 = [H+](0.01538 M)
[H+] = 6.500x10-13 M
 pH = 12.19
2) At the equivalence point
 nH+ = nOHpH is determined by dissociation of H2O:
H2O
H+ + OHx
x
Kw = [H+][OH-]
1x10-14 = x2
x = 1x10-7 M
 [H+] = 1x10-7 M
 pH = 7
pH = 7 at the equivalence point ONLY for
strong acid – strong base titrations!!
3) After reaching the equivalence point
HBr + KOH
 excess H+ present
KBr + H2O
Say 10.10 ml HBr has been added.
nexcess
CH+ =
Vtotal
CH+ = 1.664x10-4 M
pH = 3.78
Vtotal = 50 + 10.1 mL
= 60.1 mL
= 0.0601 L
Starting nOH= 1x10-3 mol
nH+ added
= (0.1 M)(0.0101 L)
= 1.010x10-3 mol
 nH+ excess
= 1x10-5 mol
Note:
A rapid change in pH near
the equivalence point
occurs.
Equivalence point where:
• slope is greatest
dpH
slope 
dVa
• second derivative is
zero (point of inflection)
d 2pH
0
2
dVa
Calculate titration curve by calculating pH values after a number of
additions of HBr.
TITRATION OF WEAK ACID WITH
STRONG BASE
Example:
Titrate 50.00 ml of 0.02000 M formic acid with 0.1000 M
NaOH.
HCO2H + NaOH
OR
HCO2H +
OH-
HA
pKa = 3.745
Ka = 1.80x10-4
Kb = 5.56x10-11
HCO2Na + H2O
-
HCO2 + H2O
K
1
 1.80  1010
Kb
A-
Equilibrium constant so large
 reaction “goes to completion”
after each addition of OHStrong and weak react completely
HCO2H + OH-
Example:
HCO2- + H2O
Titrate 50.00 ml of 0.02000 M formic acid with 0.1000 M NaOH.
* Calculate volume of NaOH needed to reach the
equivalence point, Veq:
C1V1
C2V2
=
n1
n2
But n1 = n2 = 1
 CNaOHVeq = CFAVFA
(0.1000 M)Veq = (0.02000 M)(50.00 ml)
Veq = 10.00 ml
There are 4 parts to the titration curve:
1) Before base is added
 HA and H2O present. HA
weak acid,  pH
determined by equilibrium:
HA
Ka
H+ + A-
2) From first addition of
NaOH to immediately
before equivalence point 
mixture of unreacted HA
and AHA + OH-
2
A- + H2O
BUFFER!!  use
Henderson-Hasselbalch
eqn for pH
1
3) At the equivalence point
 all HA converted to A-.
A- is a weak base whose
pH is determined by
reaction:
A-
+ H 2O
Kb
HA +
OH-
4) Beyond the equivalence
point
 excess OH- added to A-.
Good approx:
pH determined by strong
base (neglect small effect
from A-)
4
3
1) Before base is added
 HA and H2O present. HA = weak acid.
HA
F- x
H+ + Ax
x
Ka = 1.80x10-4
[H ][A  ]
Ka 
[HA]
2
x
1.80x10-4 =
0.02 - x
x2 + 1.80x10-4x – 3.60x10-6 = 0
x = 1.81x10-3
 [H+] = 1.81x10-3
 pH = 2.47
2) From first addition of NaOH to immediately before
equivalence point  mixture of unreacted HA and A. BUFFER!!
HCO2H + OH- HCO2- + H2O
Say 2.00 ml NaOH has been added.
Starting nHA
= (0.02 M)(0.05 L)
= 1x10-3 mol
[A  ]
pH  pKa  log
[HA]
nA
pH  pK a  log
nHA
HA + OH-
A- + H2O
nOH- added
= (0.1 M)(0.002 L)
= 2x10-4 mol
Start
End
pH = 3.14
Special condition:
When volume of
titrant = ½ Veq
pH = pKa
Since:
pH  pK a  log
nHA = nA-
nA
nHA
3) At the equivalence point
 all HA converted to A-. A- = weak base.
(nHA = nNaOH)
HA +
Start
End
1x10-3
-
OH-
A-
1x10-3
-
-
+ H2O
1x10-3
Solution contains just A a solution of weak base
Starting nHA
= 1x10-3 mol
 nOH= 1x10-3 mol
A- + H2O
F- x
FA- =
HA + OHx
x
nA1x10-3 mol
=
V
0.060 L
Kb = 5.56x10-11
Vtotal = 50 + 10 mL
= 60 mL = 0.060 L
= 0.0167 M
[HA][OH  ]
Kb 
[A - ]
2
x
5.56x10-11 =
0.0167 - x
[OH-] = 9.63x10-7 M
x2 + 5.56x10-11x – 9.27x10-13 = 0
pOH = 6.02
x = 9.63x10-7
pH = 7.98
pH is slightly basic (pH above 7)
for strong base-weak acid titrations
CALCULATED TITRATION CURVE
pH
2.47
2.79
3.14
3.38
3.57
3.75
3.92
4.11
4.35
4.7
5.02
5.44
7.98
10.52
10.92
11.21
11.51
11.68
11.8
11.89
14
12
10
pH
Vol NaOH
0
1
2
3
4
5
6
7
8
9
9.5
9.8
10
10.2
10.5
11
12
13
14
15
8
6
4
2
0
0
5
10
Vol NaOH / ml
15
Titration curve
depends on Ka of HA.
As HA becomes a
weaker acid the
inflection near the
equivalence point
decreases until the
equivalence point
becomes too shallow
to detect
 not practical to
titrate an acid or base
that is too weak.
Titration curve
depends on extent of
dilution of HA.
As HA becomes a more
dilute the inflection
near the equivalence
point decreases until
the equivalence point
becomes too shallow
to detect
 not practical to
titrate a very dilute
acid or base.
TITRATION OF WEAK BASE WITH
STRONG ACID
This is the reverse of the titration of weak base with
strong acid.
The titration reaction is:
B + H+
BH+
Recall:
Strong and weak react completely
There are 4 parts to the titration curve:
1) Before acid is added
 B and H2O present.
B weak base  pH determined by equilibrium:
B + H2O
F-x
Kb
BH+ + OHx
x
2) From first addition of acid to immediately before
equivalence point
 mixture of unreacted B and BH+
B + H+
BH+
BUFFER!!
 use Henderson-Hasselbalch equation for pH
[B]
pH  pKa  log
[BH ]
pKa applies
to this acid
3) At the equivalence point
 all B converted to BH+.
BH+ is a weak acid  determined pH by
reaction:
BH+
Ka
F-x
FBH+ =
B + H+
x
x
n
Vtotal
Take dilution into account
pH is slightly acidic (pH below 7) for
strong acid-weak base titrations
4) Beyond the equivalence point
 excess H+ added to BH+.
Good approx: pH determined by strong acid
(neglect small effect from BH+)
Example:
50.00 ml of 0.05 M NaCN is titrated with 0.1 M HCl. Ka
for NaCN = 6.20x1010
Draw the titration curve by calculating pH at various
volumes of HCl.
pH
10.95
10.16
9.81
9.58
9.39
9.21
9.03
8.84
8.61
8.26
8.02
7.52
5.34
3.18
2.71
2.49
2.2
2.04
1.93
1.85
1.78
TITRATION CURVE OF WEAK
BASE WITH STRONG ACID
12
10
8
pH
Vol HCl
0
2.5
5
7.5
10
12.5
15
17.5
20
22.5
23.5
24.5
25
25.5
26.5
27.5
30
32.5
35
37.5
40
6
4
2
0
0
10
20
30
Volume HCl/ml
40
TITRATIONS IN DIPROTIC SYSTEMS
Example - a base that is dibasic:
pKb1 = 4.00
pKb2 = 9.00
With corresponding reactions:
B + H+  BH+
BH+ + H+  BH22+
Two end points will be observed.
FINDING END POINTS WITH A pH
ELECTRODE
After each small addition of titrant the pH is recorded
and a titration curve is plotted.
2 ways of determining end points from this:
• using derivatives
• using a Gran plot
Setup
But there are
autotitrators!
Titrando from
Metrohm
USING DERIVATIVES
End point is taken where
the slope is greatest
dpH
dV
Or where the 2nd
derivative is zero
d 2pH
0
2
dV
USING A GRAN PLOT
A problem with using derivatives
 titration data is most to obtain near the end point
Example – titration of a weak acid, HA
HA
H+ + A-
Ka =
[H+]H+[A-] A[HA] HA
NOTE:
pH electrode responds to hydrogen ion ACTIVITY, not
concentration
Say we titrated Va ml of HA (formal conc = Fa) with Vb ml
of NaOH (formal conc = Fb) :
HA + OH- A- + H2O
[A-]
=
[HA] =
nOH(titrated)
VbFb
=
VTotal
Va + Vb
nHA(initial) – nOH(titrated)
=
VTotal
VaFa- VbFb
Va + Vb
Substitute into the equilibrium constant:
[H+]H+[A-] AKa =
[HA] HA
 VF 
[H ] H  b b  A 
Va  Vb 

Ka 
 VaFa  VbFb 

 HA
 Va  Vb 
[H ] H VbFb A 
Ka 
VaFa  VbFb  HA
[H ] H VbFb A 
Ka 
VaFa  VbFb  HA
Rearrange:
Vb [H ] H

 A  VaFa  VbFb 


 Ka
 HA 
Fb

VaFa
[H+]H+ = 10-pH
Vb 10 pH  K a
Fb
 A
Ve  Vb 
 HA
- Vb = Ve - Vb
Gran plot equation:
Vb 10 pH  K a
 A
Ve  Vb 
 HA
Gran plot  Graph of Vb10-pH vs Vb
If A- is constant, then:
HA
ASlope = -Ka 
HA
and
x-intercept = Ve
• Use data taken before end point to find end point
• Can determine Ka from slope
Use only linear
portion of graph
Extrapolate
graph to get Ve
FINDING END POINTS WITH
INDICATORS
Acid-base indicator  acid or base itself
Various protonated species have different colours
HIn
H+ + In-
Choose indicator
whose colour
change is as
close as possible
to the pH of the
end point
Indicators
transition range
overlaps the
steepest part of
the titration
curve
Indicator error: difference between the observed end
point (colour change) and the true equivalence point.
Systematic error
Random error
Visual uncertainty associated with distinguishing the
colour of the indicator reproducibly
Why do we only add a few drops of indicator?
Indicator is an acid/base itself  will react with
analyte/titrant
Few drops  neglible relative to amount of analyte