Brønsted-Lowry Acids and Bases
Acid - Base Equilibria
After Arrhenius
After Lewis
After Brønsted-Lowry
1923
The essential feature of an
acid-base reaction is the
transfer of a proton (H+)
from one species to
another.
Brønsted
Oxtoby Chapter 8, Sections 8.1-8.4
Lowry
Acids and bases occur as conjugated pairs
A Brønsted Acid: A Proton Donor
Acid - H+ = Base
Acid1
A Brønsted Base: A Proton Acceptor
HSO4-(aq) + OH-(aq)
acid
H2O(l)
SO42-(aq)
+
HF(aq) + H2O(l)
Acid2
Base1
H3 O+
+ F-(aq)
When a species donates a proton, it becomes its conjugate base;
When a species gains a proton, it becomes its conjugate acid.
Conjugate acids and bases are in equilibrium in solution.
base
PO43-(aq) + H2O(l)
base
Base2
Base + H+ = Acid
HPO42-(aq)
+
OH-(aq)
H2O(l) + H2O(l)
acid
H3O+
+ OH-(aq)
Hydronium ion
Water and the pH Scale- a Reminder
OH-(aq) + H3O+(aq)
2 H2O (l)
For which the equilibrium constant, Kw is
Kw =
[H3O+][OH-]
[H2O]2
However, since water is a pure liquid we
replace [H2O] by 1. (Oxtoby 7.6)
Hence, Kw = [H3O+][OH-]
T (°C)
0
25
40
60
which depends on temperature:
Kw
0.114 x 10-14
1.01 x 10-14
2.92 x 10-14
9.61 x 10-14
And the pH of a solution is given by
pH = -log10[H3O+]
And the pOH of a solution is given by
pOH = -log10[OH-]
And the pKw of a solution is given by
pKW = -log10[KW] = pOH + pH
1
The Strength of Brønsted Acids
The Strength of Brønsted Bases
The strength of a Brønsted acid in aqueous solution is expressed by
its acidity constant, Ka:
H3 O+
HF(aq) + H2O(l)
+ F-(aq)
[H3O+][F-]
Ka =
The strength of a Brønsted base in aqueous solution is expressed
by basicity constant, Kb:
pKa = -log10[Ka]
[HF]
NH4+
NH3(aq) + H2O(l)
Kb =
[NH4+][OH-]
+ OH-(aq)
Kb(NH3) = 1.8x10-5 (at 298 K)
[NH3]
If Ka<<1: proton retention by the acid is favored
If Kb<<1: the base is a weak proton acceptor
If Ka>>1: proton donation by the acid is favored
Relationship between Ka and Kb
NH4+
NH3(aq) + H2O(l)
NH4
+
+ H2O(aq)
+ OH-(aq)
NH3(aq) + H3
(aq)
Kb =
Ka =
[NH4+][OH-]
[NH3]
[NH4+][OH-]
[NH3][H3O+]
[NH3]
[NH4+]
[NH4+]
= [OH-][H3O+] = Kw
Polyprotic Acids
Conj base
Kb
Hydrochloric
HCl
107
Cl-
10-21
Nitric
HNO3
20
NO3-
5 x 10-16
Acetic
CH3COOH 1.8 x 10-5
Sulfuric (1)
H2SO4
Sulfuric (2)
-
HSO4
102
pKa + pKb = pKw
The strength of base is inversely proportional to the strength of
its conjugated acid:
the weaker the acid, the stronger its conjugated base.
Some Acidity constants for common acids
Ka
KaKb = Kw
[NH3][H3O+]
OH-(aq) + H3O+(aq)
2 H2O (l)
KaKb =
O+
In other words …
CH3COOHSO4
-
10-3
H2PO4
H3O+(aq) +
H2PO4-(aq) pKa= 2.1
H2PO4-(aq) + H2O(l)
H3O+(aq) +
HPO42-(aq) pKa= 7.2
HPO42-(aq) + H2O(l)
H3O+(aq) +
10-16
SO42-
1.2 x
10-2
5.5 x 10-8
H3PO4(aq) + H2O(l)
8.3 x 10-13
-
1.3 x
PO43-(aq) pKa= 12.7
10-12
Phosphoric (1)
H3PO4
7.5 x
Phosphoric (2)
H2PO4-
6.2 x 10-8
HPO42-
1.6 x 10-7
Phosphoric (3)
HPO42-
2.2 x 10-13
PO43-
0.045
Phosphoric acid is a mixture of four phosphate species
The strength of the acid usually gets weaker with each loss of a proton
2
Indicators
Indicators
Compounds that change color noticeably over a rather short pH range
Methylorange
Cyanidine: Why roses are red and violets are blue.
pH=3.2
pH=4.4
Acidic Solution, red (not absorbed)
Basic Solution, blue (not absorbed)
http://www.chemguide.co.uk/physical/acidbaseeqia/indicators.html
Indicators in Equilibrium: how pH affects colour
Some Indicator Ranges
HIn(aq) + H2O(l)
red
[H3O+]
=
Ka
H3 O+
[H3O+][In-]
+ In-(aq) Ka =
[HIn]
10-6
[HIn]
=
[In-]
= 10
10-7
blue for litmus
For litmus indicator with pKa=7 in a solution of pH=6
10-8
= 0.1
=
10-7
For litmus indicator with pKa=7 in a solution of pH=8
Equilibria involving weak acids
A weak acid has a small Ka (< Ka(H3O+) = 1) and a large
pKa (> pKa(H3O+) = 0).
In aqueous solution:
HA(aq) + H2O(l)
H3
O+
+
A-(aq)
But also:
H2O (aq) + H2O(l)
0.75 mol of acetic acid is contained in 1 dm3 of an aqueous solution.
Acetic acid has a Ka of 1. 8 µ 10-5 at 25oC. Calculate the pH of the
solution, and the fraction of acid that has dissociated.
initial
0.75
equbm
0.75-x
Ka =
H3O+
+ OH-(aq)
If you know the amount of acid originally added to your solution,
then you can calculate the pH of the solution.
H3 O+
HA(aq) + H2O(l)
[H3O+][A-]
[HA]
x = 0.0037 = [H3O+];
+ A-(aq)
x
x2
Ka =
[0.75-x]
x
= 1. 8 µ 10-5
pH = -log10 [H3O+] = 2.4
Fraction dissociated: x µ 1 dm3 /0.75 = 0.0049
(Can neglect
effect of
autoionization)
3
Hydrolysis
Equilibria involving weak bases
A weak base has a small Kb (< 1) and a large pKb (> 0).
In general: the reaction of a substance with water.
Here: the reaction of an ion with water leading to change of the pH of
the water away from being neutral.
In aqueous solution:
NH4+
NH3(aq) + H2O(l)
X-(aq) + H2O(l)
+ OH-(aq)
OH-
PO43-(aq) + H2O(l)
But also:
H3O+
H2O (aq) + H2O(l)
+ HX(aq)
+ HPO42-(aq)
OH-
+ OH-(aq)
All treatment is now analogous to the situation in weak acids.
Y+(aq) + H2O(l)
YOH(aq) + H3O+
Cu[(H2O)4]2+(aq) + H2O(l)
Cu[(H2O)3(OH)]+(aq) + H3O+(aq)
Buffer Solutions
Buffers are solutions that maintain the pH constant despite small
amounts of acid or base added to system.
Typical Buffers:
Buffer Solutions
and Acid-Base Titrations
Conjugated acid / conjugated base mixtures
Weak acids / conjugated weak base
Example
CH3COOH / CH3COOCH3COO-
CH3COOH + H2O
Ka =
Chapter 8, Section 8.5
[H3O+][A-]
[H3O+] =
[HA]
+
H3O+
Ka [HA]
[A-]
Calculations with buffers
Calculations with buffers
1.00 mole of acetic acid and 0.50 mole of sodium acetate are added to
water and the resulting solution diluted to 1 L with water at 25 °C.
Calculate the pH of the final solution.
1.00 mole of acetic acid and 0.50 mole of sodium acetate are added to
water and the resulting solution diluted to 1 L with water at 25 °C.
Calculate the pH of the final solution.
Strictly, draw up reaction table:
CH3COOH + H2O Ý CH3COO– + H3O+
Initially
1.00
0.500
At equbm
1.00 – x
0.500 + x
but x is small, so to a good approximation
1.00
0.500
x
⎡⎣CH3COO ⎤⎦ ⎡⎣H3O ⎤⎦
⎡⎣CH3COO ⎤⎦
⎡H O ⎤⎦
eq
eq
initial ⎣ 3
eq
=
[CH3COOH]eq
[CH3COOH]initial
−
Ka =
0
x
+
⎛ [CH COOH]
3
initial
pH = pK a − log ⎜
⎜ ⎡CH3COO− ⎤
⎦ initial
⎝⎣
−
⎞
⎟
⎟
⎠
+
Henderson-Hasselbach equation
[H3O+] =
Ka [HA]
[A-]
⎡⎣H3O + ⎤⎦ =
K a [HA ]
⎡⎣ A - ⎤⎦
1.8 × 105 × 1.00 M
=
0.50 M
= 3.6 × 10 −5
pH = 4.44
4
Titration of a strong acid with a strong base
Acid - Base Titrations
Strong acid:
HA +
Strong base:
BOH
H3 O+
+
H2O
H3O+
+
A-
B+
+
OH-
OH-
K = 1/KW = 1014
2 H2O
reaction goes virtually
to completion
pH at point a
Titration of a strong acid with a strong base
pH
100 ml of 0.1M HCl
(strong acid)
Titrated with
pH
0.1M NaOH
(strong base)
12
11
10
9
8
7
6
5
4
3
2
1
0
a
pH = -log10[H3O+] = -log10(0.1) = 1
If we have 100 ml of HCl solution to start with,
we know that we have to neutralise 0.01 mol H3O+
b
nO(H3O+ ) = 0.01 mol
100
Volume 0.1M NaOH added to acid
pH
pH at point b
pH at point c (equivalence point)
30 ml of NaOH added
H3O+ + OH–
So
=
c
2 H2 O
b
V
(K = 1/KW = 1014)
n(H3O+) = nO(H3O+ ) – nadded(OH-)
= (0.0100 – 0.0030) mol = 0.0070 mol
[H3O+]
pH
100 ml of NaOH added
Amount of hydoxide added:
nadded(OH–)
= volume µ concentration
= 0.030 dm3 µ 0.10 M = 0.0030 mol
But
Pure HCl - Titration not commenced
c
50
Strongly acidic
a
Strongly basic
d
0.007 mol = 0.054M
(0.1 + 0.03)l
pH = -log10[H3O+] = 1.27
Amount of hydoxide added:
nadded(OH–) = volume µ concentration
= 0.100 dm3 µ 0.10 M = 0.0100 mol
= n0(H3O+)
But
[H3O+]
H3O+ + OH– Ý 2 H2O
= [OH–] = 10–7 M
V
K = 1/KW = 1014
pH = 7
5
pH
pH at point d (end point)
Titration of a Weak Acid with a Strong Base
d
100.05 ml of NaOH added
100 ml of 0.1M CH3COOH
(strong acid)
12
Amount of hydoxide added:
nadded(OH–) = volume µ concentration
= 0.10005 dm3 µ 0.10000 M = 0.010005 mol
H3O+ + OH– Ý 2 H2O
X/s
OH– after
[OH–] =
rxn
n(OH–)
V
K = 1/KW = 1014
(OH–)
Titrated with
0.1M NaOH
(strong base)
pH
(H+)
= nadded
– n0
= (0.10005 – 0.10000) mol = 5 µ 10–6 mol
weakly acidic
5µ
mol
= 2.5µ10–5M
(0.100 + 0.10005) dm3
pH = pKw – pOH = 9.40
10–6
11
10
9
8
7
6
5
4
3
2
1
0
Strongly basic
d
b
c
a
50
Strongly acidic
100
Volume 0.1M NaOH
pH at point b
pH at point a
30 ml of NaOH added
b
a
CH3COOH(aq) + OH-(aq) Ý CH3COO-(aq) +
H2O(l)
Pure CH3COOH - Titration not commenced
K=1/Kb=1/(5.6 x 10-10)
After reaction:
⎡H3O+ ⎤⎦ ⎡⎣ A − ⎤⎦
x2
Ka = ⎣
=
= 1.8 × 10 −5
HA
0.1000
−x
[ ]
(
)
(
n CH3COO− = nadded OH−
)
= 0.00300 mol
0.00300 mol
−
⎣⎡CH3COO ⎦⎤ = ( 0.100 + 0.030 ) dm3
x = ⎡⎣H3O+ ⎤⎦ = 0.00134 M
pH = 2.9
= 2.31× 10 −2 M
This solution is a Buffer - we know how to deal with
them…
(
n ( CH3COOH) = n0 ( CH3COOH) − n CH3COO −
)
= 0.00700 mol
[CH3COOH] =
0.00700 mol
( 0.100 + 0.030 ) dm3
= 5.38 × 10 −2 M
pH at point c (half equivalence point)
c
50 ml of NaOH added
CH3COOH(aq) + H2O(l)
Hence,
Ka =
CH3COO-(aq) +
H3O+(aq)
n(CH3COO-) = n(CH3COOH)
= 0.050 M µ 0.100 mol = 0.00500 mol
[H3O+][CH3COO-]
[CH3COOH]
[CH3COOH] = 1.8 × 10 −5 × 0.0538 M M=4.2 × 10−5 M
⎡⎣H3O+ ⎤⎦ = K a
0.0231M
⎡⎣CH3COO− ⎤⎦
pH = -log10 [H3O+] = 4.4
[CH3COO-] =
Ka =
0.0050 mol
= [CH3COOH] =
0.15 l
[H3O+][CH3COO-]
[CH3COOH]
0.033 M
Ka = [H3O+]
pKa = pH = 4.7
6
pH at point d (equivalence point)
d
100 ml of NaOH added
CH3COO- + H2O
Ý
0.05-x
CH3COOH
+
OH-
x
[OH-][CH
[CH3
e
100.05 ml of NaOH added
0.01moles of NaCH3COO in 200 ml. [NaCH3COO]= 0.05M
Kb =
pH at point e (end point)
3COOH]
COO-]
=
x2
=
0.05-x
x = 5.27 µ 10-6 M pOH = 5.3 , pH = 8.7
x
Kw
Ka
=
Excess OH– after rxn:
n(OH-) = 5 µ 10-5 dm3 µ 0.1 M = 5 µ 10-6 mol of OH[OH-] =
5 µ 10-6 mol
0.205 dm3
= 2.4 µ 10-5 M
10-14
1.8 µ 10-5
Not neutral
CH3COO(aq) + H2O(l)
CH3COOH(aq) +
OH-(aq)
pOH = -log10(2.4 µ 10-5) = 4.6 pH = pKW - pOH = 9.4
7