ACIDS AND BASES
REVISE
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•
•
•
Brø
Brønstednsted-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
STRONG ACIDS AND BASES
Strong acids and bases → react nearly “completely” to
produce H+ and OH-
([H+] = [OH-])
([H+] > [OH-])
([H+] < [OH-])
Kw = [OH-][H+]
Kw = 14.0
at 25oC
Kw = KaKb
Example:
Calculate the pH of 0.1M LiOH.
equilibrium constants are large
e.g.: HCl
H+ + Cl-
[H + ][Cl − ]
K=
[HCl]
Complete dissociation:
large
small
In fact, we assume the reaction goes to completion:
HCl → H+ + Cl-
LiOH → Li+
+
OH-
Start:
Complete rxn:
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
Problem:
What is the pH of 1x10-8 M KOH?
In pure water [OH-] = 1x10-7 M, which is greater than the
concentration of OH- from KOH.
As before:
pOH = -log (1x10-8) = 8
pH = 14 – 8 = 6
BUT
pH 6
Since the concentration of KOH is so low (1x10-8 M), we
need to take the ionisation of water into account.
acidic conditions and KOH is a strong base
IMPOSSIBLE!!!!!!
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.
1
Also note that:
• Only pure water produces 1x10-7 M H+ and OH-.
Some guidelines regarding the concentrations of acids
and bases:
• If there is say 1x10-4 M HBr in solution,
pH = 4 and [OH-] = 1x10-10 M
1) When conc > 1x10-6 M
→ calculate pH as usual
• But the only source of OH- is from the dissociation
of water.
2) When conc < 1x10-8 M
→ pH = 7
(there is not enough acid or base to affect the
pH 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.
WEAK ACIDS AND BASES
Base hydrolysis:
B + H2 O
Weak acids and bases → react only “partially” to
produce H+ and OH-
H+ + A-
[H+ ][A − ]
Ka =
[HA]
Partial dissociation
small
HA + H2O
H3O+ + A-
Acid dissociation constant
Common weak acids:
• carboxylic acids
(e.g. acetic acid = CH3COOH)
• ammonium ions
(e.g. RNH3+, R2NH2+, R3NH+)
Weak base
partial dissociation
Kb small
base hydrolysis constant/
base “association” constant
[H O + ][A − ]
Ka = 3
[HA]
large
BH+ + OH-
[BH+ ][OH− ]
Kb =
[B]
equilibrium constants are small
HA
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.
NOTE:
Common weak bases:
• carboxylate anions
(e.g. acetate = CH3COO-)
• amines
(e.g. RNH2, R2NH, R3N)
pKa = -log Ka
pKb = -log Kb
As K increases, its p-function decreases and vice versa.
FRACTION OF DISSOCIATION, α
Problem:
Find the pH of a solution of formic acid given that the
formal concentration is 2 M and Ka = 1.80x10-4.
Fraction of acid in the form A-
Since [HCOOH] > 1x10-6, we can calculate pH as usual
HCOOH
H+
+
α=
HCOO-
Start:
[A − ]
[A − ] + [HA]
=
∴α =
Equilibrium:
[A − ]
(
[A − ] + F − [A − ]
)
−
[A ]
F
For the above problem:
[HCOO-]
α =
F
Acid is ………..% dissociated at 2 M formal concentration
Weak electrolytes dissociate more as they are diluted.
pH = 1.7
2
WEAK BASE EQUILIBRIA
B + H2 O
Charge balance:
[BH+] = [OH-]
Mass balance:
F = [B] + [BH+]
Equilibria:
Let
[BH+]
CONJUGATE ACIDS AND BASES
BH+ + OH-
Kb =
=
[OH-]
Relationship between Ka and Kb for a conjugate acidbase pair:
Ka.Kb = Kw = 1x10-14
at 25oC
∴If Ka is very large (strong acid)
[BH+ ][OH− ]
[B]
Then Kb must be very small (weak conjugate base)
=x
And vice versa
x. x
x2
Kb =
=
[B] F − x
Base so weak it is not
a base at all in water
FRACTION OF ASSOCIATION
α=
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.
[BH+ ]
F
Problem:
Calculate the pH of 0.1 M NH3, given that pKa = 9.244 for
NH4+.
BUFFERS
Mixture of a weak 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
Le Chatelier’s principle
H+ + A-
pH = 11.12
HENDERSONHENDERSON-HASSELBALCH EQUATION
Derivation:
HA
H+ + A-
For acids:
pH = pK a + log
[A − ]
[HA]
When [A-] = [HA],
pH = pKa
pH = pK a + log
[B]
[BH ]
pKa applies
to this acid
Kb
“Ka”
acid
[H+ ][A− ]
[HA]
= − log [H+ ] − log
+
→ BH+ + OHB + H2 O ←
acid
[H+ ][A− ]
[HA]
- logKa = − log
For bases:
base
Ka =
base
pKa = pH − log
[A− ]
[HA]
[A− ]
[HA]
pH = pKa + log
[A− ]
[HA]
3
?
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
HA
H+ + A-
A buffer has a maximum capacity to resist change to
pH.
Buffer capacity, β:
→ Measure of how well solution resists change in pH
when strong acid/base is added.
β=
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.
dCb − dCa
=
dpH dpH
Larger β
pKa ± 1 pH unit
more resistance to pH change
POLYPROTIC ACIDS AND BASES
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.
Can donate or accept more than one proton.
In general:
Diprotic acid:
Diprotic base:
H2 L
HL- + H+
Ka1 ≡ K1
L2- + H2O
HL-
L2-
Ka2 ≡ K2
HL-
+
H+
HL- + OH-
Kb1
OH-
Kb2
+ H2 O H2 L +
Relationships between Ka’s and Kb’s:
Ka1. Kb2 = Kw
Ka2. Kb1 = Kw
pH = 9.07
Using pKa values and mass balance equations, the
fraction of each species can be determined at a
given pH.
ACIDACID-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
4
TITRATION OF STRONG BASE WITH
STRONG ACID
Example:
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:
Titrate 50.00 ml of 0.02000 M KOH with 0.1000 M HBr.
HBr + KOH
C1 V 1
C2 V 2
=
n1
n2
KBr + H2O
What is of interest to us in an acid-base titration:
H+ + OH-
H2 O
Mix strong acid and strong base
reaction goes to completion
H+ + OH- → H2O
1) Before reaching the equivalence point
HBr + KOH
→ excess OH- present
There are 3 parts to the
titration curve:
Say 2.00 ml HBr has been added.
1
1) Before reaching the
equivalence point
KBr + H2O
→ excess OH- present
2) At the equivalence
point
2
→ [H+] = [OH-]
3
3) After reaching the
equivalence point
→ excess H+ present
pH = 12.19
3) After reaching the equivalence point
HBr + KOH
→ excess H+ present
2) At the equivalence point
→ nH+ = nOHpH is determined by dissociation of H2O:
H2 O
KBr + H2O
Say 10.10 ml HBr has been added.
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!!
pH = 3.78
5
Calculate titration curve by calculating pH values after a number of
additions of HBr.
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
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-
HCO2Na + H2O
HCO2- + H2O
HA
A-
pKa = 3.745
Ka = 1.80x10-4
Kb = 5.56x10-11
1
K=
= 1.80 × 1010
Kb
Equilibrium constant so large
reaction “goes to completion”
after each addition of OH-
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:
C1 V 1
C2 V 2
=
n1
n2
But n1 = n2 = 1
∴ CNaOHVeq = CFAVFA
(0.1000 M)Veq = (0.02000 M)(50.00 ml)
Veq = 10.00 ml
Strong and weak react completely
There are 4 parts to the titration curve:
3) At the equivalence point
1) Before base is added
→ HA and H2O present. HA
weak acid, ∴ pH
determined by equilibrium:
HA
Ka
→ all HA converted to A-.
A- is a weak base whose
pH is determined by
reaction:
H+ + A-
A- + H2O
2) From first addition of
NaOH to immediately
before equivalence point
→ mixture of unreacted HA
and AHA +
OH-
A-
HA + OH-
3
4) Beyond the equivalence
point
→ excess OH- added to A-.
2
+ H2O
BUFFER!! ∴ use
Henderson-Hasselbalch
eqn for pH
Kb
4
1
Good approx:
pH determined by strong
base (neglect small effect
from A-)
6
1) Before base is added
2) From first addition of NaOH to immediately before
equivalence point → mixture of unreacted HA and ABUFFER!!
HCO2H + OH- HCO2- + H2O
→ HA and H2O present. HA = weak acid.
HA
F- x
H+ + Ax
x
Ka = 1.80x10-4
Say 2.00 ml NaOH has been added.
[H+ ][A − ]
Ka =
[HA]
pH = pK a + log
[A − ]
[HA]
pH = pK a + log
nA
nHA
HA + OH-
A- + H2O
Start
End
pH = 2.47
pH = pK a + log
pH = 3.14
[A − ]
[HA]
Special condition:
nA
[A − ]
VA
=
[HA] nHA
VHA
When volume of
titrant = ½ Veq
pH = pKa
But VA = VHA = VTot
nA
[A − ]
VTot
=
n
[HA]
HA
VTot
Since:
pH = pK a + log
−
[A ] n A
=
[HA] nHA
pH = pK a + log
nHA = nA-
nA
nHA
3) At the equivalence point
→ all HA converted to
(nHA = nNaOH)
HA + OHStart
End
1x10-3
-
1x10-3
-
nA
nHA
A-.
A-
A- + H2O
1x10-3
∴Solution contains just Aa solution of weak base
A- + H2O
F- x
= weak base.
FA- =
HA + OHx
x
nA-
=
V
Starting nHA
= 1x10-3 mol
∴ nOH-
1x10-3 mol
0.060 L
Kb = 5.56x10-11
Vtotal = 50 + 10 mL
= 60 mL = 0.060 L
= 0.0167 M
Kb =
= 1x10-3 mol
5.56x10-11 =
[HA][OH− ]
[A - ]
x2
0.0167 - x
x2 + 5.56x10-11x – 9.27x10-13 = 0
x=
9.63x10-7
[OH-] = 9.63x10-7 M
pOH = 6.02
pH = 7.98
pH is slightly basic at equivalence point
for strong base-weak acid titrations
7
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
Titration curve
depends on Ka of HA.
14
12
As HA becomes a
weaker acid the
inflection near the
equivalence point
decreases until the
equivalence point
becomes too shallow
to detect
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
not practical to
titrate an acid or base
that is too weak.
0
0
5
10
15
Vol NaOH / ml
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.
There are 4 parts to the titration curve:
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
2) From first addition of acid to immediately before
equivalence point
→ mixture of unreacted B and BH+
1) Before acid is added
→ B and H2O present.
B weak base ∴ pH determined by equilibrium:
B + H2 O
F-x
Kb
BH+ + OHx
x
B + H+
BH+
BUFFER!!
∴ use Henderson-Hasselbalch equation for pH
pH = pK a + log
[B]
[BH+ ]
pKa applies
to this acid
8
3) At the equivalence point
4) Beyond the equivalence point
→ all B converted to BH+.
→ excess H+ added to BH+.
BH+ is a weak acid ∴ determined pH by
reaction:
Good approx: pH determined by strong acid
(neglect small effect from BH+)
BH+
F′′-x
F′′BH+ =
Ka
B + H+
x
x
n
Vtotal
Example:
Take dilution into account
50.00 ml of 0.05 M NaCN is titrated with 0.1 M HCl. Ka
for NaCN = 6.20x1010
pH is slightly acidic (pH below 7) for
strong acid-weak base titrations
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
TITRATIONS IN DIPROTIC SYSTEMS
Example - a base that is dibasic:
12
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
Draw the titration curve by calculating pH at various
volumes of HCl.
10
pKb1 = 4.00
8
pKb2 = 9.00
With corresponding reactions:
6
4
B + H+ → BH+
2
BH+ + H+ → BH22+
0
0
10
20
30
40
Two end points will be observed.
Volume HCl/ml
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
9
Setup
But there are
autotitrators!
Titrando from
Metrohm
USING DERIVATIVES
USING A GRAN PLOT
End point is taken where
the slope is greatest
A problem with using derivatives
→ titration data is the most difficult to obtain near the
end point
dpH
dV
Example – titration of a weak acid, HA
Or where the 2nd
derivative is zero
HA
d 2pH
=0
dV 2
pH electrode responds to hydrogen ion ACTIVITY, not
concentration
BEFORE EQUIVALENCE:
[HA] =
nOH(titrated)
VTotal
=
VbFb
VaFa- VbFb
Va + Vb
Substitute into the equilibrium constant:
[H+]γγH+[A-] γAKa =
[HA] γHA
Ka =
VbFb
γ A−
Va + Vb
VaFa − VbFb
γ HA
Va + Vb
Ka =
[H+ ]γ H+ VbFbγ A −
(VaFa − VbFb )γ HA
Rearrange:
Va + Vb
nHA(initial) – nOH(titrated)
=
VTotal
[H+ ]γ H+
[H+]γγH+[A-] γA[HA] γHA
Ka =
RECALL:
Say we titrated Va ml of HA (formal conc = Fa) with Vb ml
of NaOH (formal conc = Fb) :
HA + OH- A- + H2O
[A-] =
H+ + A-
[H + ]γ H + VbFb γ A −
Ka =
(VaFa − VbFb )γ HA
Vb [H+ ]γ H+ = K a
γ A − VaFa − VbFb
γ HA
Fb
Va F a
[H+]γγH+ = 10-pH
Vb 10 −pH = K a
Fb
- Vb = Ve - Vb
γ A−
(Ve − Vb )
γ HA
10
We are titrating HA with NaOH
Gran plot equation:
Vb 10 −pH = K a
Gran plot
γ A−
(Ve − Vb )
γ HA
Use only linear
portion of graph
Graph of Vb10-pH vs Vb
If γA- is constant, then:
γHA
Slope = -Ka
γAγHA
and
x-intercept = Ve
• Use data taken before end point to find end point
Extrapolate
graph to get Ve
• Can determine Ka from slope
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
negligible relative to amount of analyte
11