ACIDS AND BASES

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ACIDS AND BASES
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REVISE

• Brønsted-Lowry acids and bases
• Amphoteric substances
• Conjugate acid base pairs
• Neutralisation

Neutral pH 7 (H OH-) Acidic pH lt
7 (H gt OH-) Basic pH gt 7 (H lt OH-)
pH -log H pOH -log OH- pH pOH 14
at 25oC
Kw OH-H Kw 14.0 at 25oC Kw KaKb
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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-
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
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Example Calculate the pH of 0.1M LiOH.
LiOH ? Li OH-
Start
Complete rxn
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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!!!!!!
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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 HOH- 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.
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• 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.

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Some guidelines regarding the concentrations of
acids and bases

• When conc gt 1x10-6 M
• ? calculate pH as usual

• When conc lt 1x10-8 M
• ? pH 7
• (there is not enough acid or base to affect the
  pH of water)

• 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.

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WEAK ACIDS AND BASES
Weak acids and bases ? react only partially to
produce H and OH-
? equilibrium constants are small
HA H A-
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)

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Base hydrolysis B H2O BH OH-

• Weak base
• 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.
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Problem Find the pH of a solution of formic acid
given that the formal concentration is 2 M and Ka
1.80x10-4.
HCOOH H HCOO-
H2O H OH-
Systematic treatment of equilibria
Charge balance H HCOO- OH-
Mass balance 2 M HCOOH HCOO-
Equilibria
4 equations ? 4 unknowns
? difficult to solve
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?Make an assumption
H due to acid dissociation ? H due to water
dissociation
Produces HCOO-
Produces OH-
HCOO- large
OH- small
? HCOO- gtgt OH-
? Charge balance H ? HCOO-
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Charge balance H ? HCOO-
Mass balance 2 M HCOOH H
Equilibria
Let H HCOO- x
Or x -0.019 ? No negative concs
?H HCOO- 0.019 M
? pH 1.7
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OR since HCOOH gt 1x10-6, we can calculate pH as
usual
Weak acid ?equilibrium conditions
HCOOH H HCOO-
2M
0
0
Start
2-x
x
x
Equilibrium
Solve as before
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FRACTION OF DISSOCIATION, ?
Fraction of acid in the form A-
For the above problem
? Acid is ________ dissociated at 2 M formal
concentration
Weak electrolytes dissociate more as they are
diluted.
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WEAK BASE EQUILIBRIA
Charge balance BH OH-
Mass balance F B BH
Equilibria
Let BH OH- x
FRACTION OF ASSOCIATION
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CONJUGATE ACIDS AND BASES
Relationship between Ka and Kb for a conjugate
acid- base 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.
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Problem Calculate the pH of 0.1 M NH3, given
that pKa 9.244 for ammonia.
Kb
NH3 H2O NH4 OH-
acid
base
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Problem Calculate the pH of 0.1 M NH3, given
that pKa 9.244 for ammonia.
pH 11.12
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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 Chateliers principle
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HENDERSON-HASSELBALCH EQUATION
For acids
When A- HA, pH pKa
For bases
pKa applies to this acid
Kb
?
B H2O BH OH-
?
Ka
acid
base
acid
base
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Derivation
HA H A-
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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.
Larger ? ? more resistance to pH change
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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
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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
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POLYPROTIC ACIDS AND BASES
Can donate or accept more than one proton.
In general
Diprotic acid H2L HL- H Ka1 ? K1
HL- L2- H Ka2 ? K2
Diprotic base L2- H2O HL- OH- Kb1
HL- H2O H2L OH- Kb2
Relationships between Kas and Kbs
Ka1. Kb2 Kw
Ka2. Kb1 Kw
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Using pKa values and mass balance equations, the
fraction of each species can be determined at a
given pH.
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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

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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
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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
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There are 3 parts to the titration curve
1

• Before reaching the equivalence point
• ? excess OH- present

• At the equivalence point
• ? H OH-

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3

• After reaching the equivalence point
• ? excess H present

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• Before reaching the equivalence point
• ? excess OH- present

HBr KOH KBr H2O
Starting nOH- (0.02 M)(0.050 L) 1x10-3
mol
Say 2.00 ml HBr has been added.
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 HOH- 1x10-14 H(0.01538 M)
H 6.500x10-13 M
? pH 12.19
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• At the equivalence point
• ? nH nOH-

pH is determined by dissociation of H2O H2O
H OH-
x
x
Kw HOH- 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!!
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• After reaching the equivalence point
• ? excess H present

HBr KOH KBr H2O
Say 10.10 ml HBr has been added.
Starting nOH- 1x10-3 mol
nH added (0.1 M)(0.0101 L) 1.010x10-3 mol
CH 1.664x10-4 M
pH 3.78

• nH excess
• 1x10-5 mol

Vtotal 50 10.1 mL 60.1 mL
0.0601 L
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Note A rapid change in pH near the equivalence
point occurs.
Equivalence point where

• slope is greatest

• second derivative is
• zero (point of inflection)

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Calculate titration curve by calculating pH
values after a number of additions of HBr.
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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 HCO2Na H2O
OR HCO2H OH- HCO2- H2O
HA
A-
pKa 3.745
Equilibrium constant so large ? reaction goes
to completion after each addition of OH-
Ka 1.80x10-4 Kb 5.56x10-11
Strong and weak react completely
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Example Titrate 50.00 ml of 0.02000 M formic
acid with 0.1000 M NaOH.
HCO2H OH- HCO2- H2O
Calculate volume of NaOH needed to reach the
equivalence point, Veq
But n1 n2 1
? CNaOHVeq CFAVFA
(0.1000 M)Veq (0.02000 M)(50.00 ml)
Veq 10.00 ml
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There are 4 parts to the titration curve

• From first addition of NaOH to immediately before
  equivalence point ? mixture of unreacted HA and
  A-
• HA OH- A- H2O
• BUFFER!! ? use Henderson-Hasselbalch eqn for pH

2
1
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4
3
4) Beyond the equivalence point ? excess OH-
added to A-. Good approx pH
determined by strong base (neglect small effect
from A-)
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• Before base is added
• ? HA and H2O present. HA weak acid.

x2 1.80x10-4x 3.60x10-6 0
x 1.81x10-3
? H 1.81x10-3
? pH 2.47
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• 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
nOH- added (0.1 M)(0.002 L) 2x10-4 mol
pH 3.14
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But VA VHA VTot
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Special condition
When volume of titrant ½ Veq pH pKa
Since
nHA nA-
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• At the equivalence point
• ? all HA converted to A-. A- weak base.
  (nHA nNaOH)

Starting nHA 1x10-3 mol
HA OH- A- H2O
? nOH- 1x10-3 mol

• Solution contains just A- ? a solution of weak
  base

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Kb 5.56x10-11
A- H2O HA OH-
F- x
x
x
Vtotal 50 10 mL 60 mL 0.060 L
0.0167 M
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
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CALCULATED TITRATION CURVE
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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.
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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.
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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
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There are 4 parts to the titration curve
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• 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

pKa applies to this acid
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Take dilution into account
pH is slightly acidic (pH below 7) for strong
acid-weak base titrations
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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.
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TITRATION CURVE OF WEAK BASE WITH STRONG ACID
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TITRATIONS IN DIPROTIC SYSTEMS
Example - a base that is dibasic
Two end points will be observed.
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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

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Setup
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But there are autotitrators!
Titrando from Metrohm
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USING DERIVATIVES
End point is taken where the slope is greatest
Or where the 2nd derivative is zero
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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-
NOTE pH electrode responds to hydrogen ion
ACTIVITY, not concentration
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Say we titrated Va ml of HA (formal conc Fa)
with Vb ml of NaOH (formal conc Fb)
HA OH- A- H2O
Substitute into the equilibrium constant
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Rearrange
H?H 10-pH
Ve - Vb
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Gran plot equation
Gran plot ? Graph of Vb10-pH vs Vb
and
x-intercept Ve

• Use data taken before end point to find end
  point
• Can determine Ka from slope

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Use only linear portion of graph
Extrapolate graph to get Ve
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FINDING END POINTS WITH INDICATORS
Acid-base indicator ? acid or base itself
Various protonated species have different colours
HIn H In-
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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
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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