acids and bases acid – a compound that produces hydrogen ions in a water solution hcl (g) → h +...
TRANSCRIPT
ACIDS AND BASES
ACID – A compound that produces hydrogen ions in a water solution
HCl (g) → H+(aq) + Cl-(aq)
BASE – A compound that produces hydroxide ions in a water solution
NaOH (s) → Na+(aq) + OH-(aq)
1884 SVANTE ARRHENIUS
Proposed the first definitions of acids and bases
1E-1 (of 25)
FORMATION OF ACIDS AND BASES
Bases are produced from the reaction of metal oxides with water
Na2O (s) + H2O (l) → 2NaOH (aq)
Acids are produced from the reaction of nonmetal oxides with water
CO2 (g) + H2O (l) → H2CO3 (aq)
1E-2 (of 25)
Na2O MgO Al2O3 SiO2 P2O5 SO3 Cl2O7
NaOH Mg(OH)2 Al(OH)3 Si(OH)4 PO(OH)3 SO2(OH)2 ClO3(OH)
H4SiO4 H3PO4 H2SO4 HClO4
AMPHOTERIC – A substance that can act as an acid or a base
base base base or
acid
acid acid acid acid
strongest base
strongest acid
Products of elemental oxides with water:
1E-3 (of 25)
base
Products of elemental oxides with water:
1+ 2+ 3+ 4+ 5+ 6+ 7+
The higher the element’s oxidation number, the more acidic its hydroxide
Cr(OH)2
Cr(OH)3
Cr(OH)6
amphoteric
acid
1E-4 (of 25)
Na2O MgO Al2O3 SiO2 P2O5 SO3 Cl2O7
NaOH Mg(OH)2 Al(OH)3 Si(OH)4 PO(OH)3 SO2(OH)2 ClO3(OH)
H4SiO4 H3PO4 H2SO4 HClO4
base base base or
acid
acid acid acid acid
ACID – A hydrogen ion (or proton) donor
BASE – A hydrogen ion (or proton) acceptor
1923
Expanded the definitions of acids and bases
JOHANNES BRØNSTED
1E-6 (of 25)
HCl + H2O → Cl- + H3O+
acid base
HYDRONIUM ION – H3O+, formed when a hydrogen ion attaches to a water
+-
1E-8 (of 25)
conjugate base
of HCl
conjugate acid
of H2O
Acids turn into bases, and bases turn into acids
HCl + H2O Cl- + H3O+
acid base
←
NH3 + H2O → NH4+ + OH-
base acid conjugate acid
of NH3
conjugate base of H2O
1E-12 (of 25)
conjugate base
of NH4+
conjugate acid
of NO2-
NH4+ + NO2
- → NH3 + HNO2
acid base
←
The favored reaction direction turns strong acids and bases into weak acids and bases
HNO2 is a stronger acid than NH4+
NH3 is a stronger base than NO2-
1E-13 (of 25)
AUTOIONIZATION OF WATER
Water ionizes itself to a small extent
2H2O (l) → H3O+(aq) + OH-(aq)
-+
Makes solutions acidic Makes solutions basic
Equal amounts of H3O+ and OH- make a solution NEUTRAL
pure water is neutral
1E-15 (of 25)
The ionization of water is an endothermic process
2H2O (l) ⇆ H3O+ (aq) + OH-
(aq)
Write the equilibrium constant expression for the autoionization of water
Keq = [H3O+][OH-]
ION-PRODUCT CONSTANT FOR WATER (Kw) – The equilibrium constant for the ionization of water
20 25 30
0.68 x 10-14
1.00 x 10-14
1.47 x 10-14
Temp. (°C) Kw
energy +
1E-16 (of 25)
Find [H3O+] and [OH-] in pure water at 25°C.
x x
2H2O (l) ⇆ H3O+ (aq) + OH- (aq)
Initial M’sChange in M’sEquilibrium M’s
0 0
+ x + x
Kw = [H3O+][OH-]
1.00 x 10-14 M2 = x2
1.00 x 10-7 M = x = [H3O+] = [OH-]
1E-17 (of 25)
THE pH SCALE
pH – The negative logarithm of [H3O+] of a solution
The common logarithm of a number is:the exponent to which 10 must be raised to equal the number
100 0.01 0.001
100 = 102 0.01 = 10-2 0.001 = 10-3
0.002
0.002 = 10-2.699
log 100 = 2 log 0.01 = -2 log 0.001 = -3 log 0.002 = -2.699
pOH – The negative logarithm of [OH-] of a solution
1E-18 (of 25)
Calculate the pH of orange juice if its [H3O+] = 2.5 x 10-4 M.
= -log(2.5 x 10-4 M)
pH = -log[H3O+]
= -(0.40 + -4.000000…)
1E-19 (of 25)
= 3.60
For logarithmic numbers, only the digits after the decimal point are significant figures, not the digits before
= 3.6 incorrect number of significant figures
= -[log(2.5) + log(10-4)]
Calculate the pH and pOH of pure water at 25ºC.
= -log(1.00 x 10-7 M)
pH = -log[H3O+]
For pure water:
[H3O+] = 1.00 x 10-7 M
= 7.000
= -log(1.00 x 10-7 M)
pOH = -log[OH-]
[OH-] = 1.00 x 10-7 M
= 7.000
1E-20 (of 25)
In any water solution:
Kw = [H3O+][OH-]
Kw = [OH-] ________
[H3O+]
at 25°C
1.00 x 10-14 = [OH-] _____________
[H3O+]
1E-21 (of 25)
For orange juice:
[H3O+] = 2.5 x 10-
4 M
1.00 x 10-14 M2 = [OH-] _________________
2.5 x 10-4 M
= 4.0 x 10-11 M
pH 7 = Neutral< 7 is Acidic > 7 is Basic
7 8 9 10 11 12 13 14650 3 421
[H3O+] = 10-7 M[OH-] = 10-7 M
[H3O+] = 10-3 M[OH-] = 10-11 M
[H3O+] = 10-12 M[OH-] = 10-2 M
-1 15
[H3O+] = 101 M[OH-] = 10-15 M
1E-22 (of 25)
In any water solution:
Kw = [H3O+][OH-]
log Kw = log ([H3O+][OH-])
-log Kw = -log ([H3O+][OH-])
-log Kw = - log[H3O+] - log[OH-]
pKw = pH + pOH
at 25°C, -log(1.00 x 10-14) = 14.000
∴ 14.000 = pH + pOH
1E-23 (of 25)
Calculate the pH and pOH of soda if its [H3O+] = 1.6 x 10-3 M.
= -log(1.6 x 10-3 M)
pH = -log[H3O+]
[H3O+] = 1.6 x 10-3 M
= 2.80
= 14.000 - 2.80
pOH = pKw - pH
pH + pOH = pKw
= 11.20
1E-24 (of 25)
Calculate the [H3O+] of blood, which has a pH of 7.4.
= 0.0000000398 M
pH = -log[H3O+]
-pH = log[H3O+]
antilog (-pH) = [H3O+]
antilog (-7.4) = [H3O+]
For logarithmic numbers, only the digits after the decimal point are significant figures, not the digits before
= 3.98 x 10-8 M = 4 x 10-8 M
1E-25 (of 25)
IONIZATION OF ACIDS
HF (aq) + H2O(l) ⇆ H3O+(aq) + F-(aq)
Write the Keq expression for the ionization of hydrofluoric acid
= [H3O+][F-]
____________
[HF]ACID IONIZATION CONSTANT (Ka) – The equilibrium constant for an acid ionizing
Keq
1F-1 (of 19)
IONIZATION OF ACIDS
HF (aq) + H2O(l) ⇆ H3O+(aq) + F-(aq)
Write the Keq expression for the ionization of hydrofluoric acid
= [H3O+][F-]
____________
[HF]ACID IONIZATION CONSTANT (Ka) – The equilibrium constant for an acid ionizing
Ka
1F-2 (of 19)
Acids are
a) strong if every acid molecule gives up a hydrogen ionHCl, HBr, HI, any acid with 2 or more O’s than H’sKa = large
b) weak if less than every acid molecule gives up a hydrogen ionall other acidsKa = small
1F-3 (of 19)
()
IONIZATION OF BASES
NH3 (aq) + H2O(l) ⇆ NH4+(aq) + OH-(aq)
Write the Keq expression for the ionization of ammonia
= [NH4+][OH-]
______________
[NH3]BASE DISSOCIATION CONSTANT (Kb) – The equilibrium constant for a base ionizing
Keq
1F-4 (of 19)
IONIZATION OF BASES
NH3 (aq) + H2O(l) ⇆ NH4+(aq) + OH-(aq)
Write the Keq expression for the ionization of ammonia
= [NH4+][OH-]
______________
[NH3]BASE DISSOCIATION CONSTANT (Kb) – The equilibrium constant for a base ionizing
Kb
1F-5 (of 19)
Bases are
a) strong if every molecule/ion accepts a hydrogen ionalkali metal hydroxides, dilute Ca(OH)2, Sr(OH)2, Ba(OH)2 solutions
Kb = large
b) weak if less than every molecule/ion accepts a hydrogen ionall other bases, including NH3
Kb = small
1F-6 (of 19)
Calculate the pH of 0.010 M nitric acid.
1F-7 (of 19)
CALCULATING THE pH OF STRONG ACID OR BASE SOLUTIONS
Assume complete ionization or dissociation
0.010 0.010
HNO3 (aq) + H2O (l) → H3O+ (aq) + NO3
- (aq)
Initial M’sChange in M’sFinal M’s
0.010 0 0
- x + x + x
0
= -log(0.010 M)
pH = -log[H3O+]
= 2.00
- 0.010 + 0.010 + 0.010
Calculate the pH of 0.0010 M sodium hydroxide.
1F-8 (of 19)
0.0010 0.0010
NaOH (aq) → Na+ (aq) + OH- (aq)
Initial M’sChange in M’sFinal M’s
0.0010 0 0
- x + x + x
0
= -log(0.0010 M)
pOH = -log[OH-]
= 3.00 = 14.000 - 3.00
pH = pKw - pOH
pH + pOH = pKw
= 11.00
- 0.0010 + 0.0010 + 0.0010
the Ka for HNO2 can be looked up
Calculate the pH of 0.010 M nitrous acid.
1F-9 (of 19)
CALCULATING THE pH OF WEAK ACID OR BASE SOLUTIONS
Assume INCOMPLETE ionization or dissociation, reaching a state of equilibrium
x x
HNO2 (aq) + H2O (l) ⇆ H3O+ (aq) + NO2
- (aq)
Initial M’sChange in M’sEquilibrium M’s
0.010 0 0- x + x + x
0.010 - x
: 4.0 x 10-4
Ka = [H3O+][NO2
-] ________________
[HNO2]
Ka = x2 _____________
(0.010 – x)
4.0 x 10-4 = x2 ______________
(0.010 – x)
If the Ka is < 10-2 the acid does not ionize much, so you may assume that x is very small, and ignore it when it is subtracted from the initial molarity
4.0 x 10-4 = x2 _______
0.010
2.00 x 10-3 = x
For x values that are >1% of the original acid molarity, they should be put back in place of x in the denominator of the Ka expression, and solved again
1F-10 (of 19)
4.0 x 10-4 = x2 _________________________
(0.010 – 2.00 x 10-3)
1.79 x 10-3 = x
Repeat until the answer (to 2 sig fig’s) is the same twice in a row
4.0 x 10-4 = x2 _________________________
(0.010 – 1.79 x 10-3)
1.81 x 10-3 = x
= -log(1.81 x 10-3 M)
pH= -log[H3O+] = 2.74
= [H3O+]
1F-11 (of 19)
Calculate the pH of 0.20 M acetic acid if its Ka = 1.8 x 10-5.
x x
HC2H3O2 (aq) + H2O (l) ⇆ H3O+ (aq) + C2H3O2
- (aq)
Initial M’sChange in M’sEquilibrium M’s
0.20 0 0- x + x + x
0.20 - x
Ka = [H3O+][C2H3O2-]
____________________
[HC2H3O2]
1.8 x 10-5 = x2 ____________
(0.20 – x)
1.8 x 10-5 = x2 ______
0.20
1.90 x 10-3 = x
1F-12 (of 19)
1.8 x 10-5 = x2 ________________________
(0.20 – 1.90 x 10-3)
1.89 x 10-3 = x
= -log(1.89 x 10-3 M)
pH = -log[H3O+] = 2.72
Calculate the pH of 0.20 M acetic acid if its Ka = 1.8 x 10-5.
= [H3O+]
Calculate the percent ionization of acetic acid.
1.89 x 10-3 M _________________
0.20 M
x 100 = 0.95%
1F-13 (of 19)
Calculate the pH of a 0.10 M ammonia solution if its Kb = 1.8 x 10-5.
x x
NH3 (aq) + H2O (l) ⇆ NH4+
(aq) + OH- (aq)
Initial M’sChange in M’sEquilibrium M’s
0.10 0 0- x + x + x
0.10 - x
Kb = [NH4+][OH-]
______________
[NH3]
1.8 x 10-5 = x2 ____________
(0.10 – x)
1.8 x 10-5 = x2 ______
0.10
1.34 x 10-3 = x
1F-14 (of 19)
1.8 x 10-5 = x2 ________________________
(0.10 – 1.34 x 10-3)
1.33 x 10-3 = x
= -log(1.33 x 10-3 M)
pOH= -log[OH-] = 2.88
Calculate the pH of a 0.10 M ammonia solution if its Kb = 1.8 x 10-5.
= [OH-]
= 14.000 - 2.88
pH = pKw - pOH
pH + pOH = pKw
= 11.12
1F-15 (of 19)
A 0.500 M solution of the weak acid HA has a pH 2.010. Calculate its Ka.
x x
HA (aq) + H2O (l) ⇆ H3O+ (aq) + A-
(aq)
Initial M’sChange in M’sEquilibrium M’s
0.500 0 0- x + x + x
0.500 - x
Ka = [H3O+][A-] _____________
[HA]
Ka = x2 ______________
(0.500 – x)
x = [H3O+]
= antilog (-2.010)
= antilog (-pH)
= 0.009772 M
1F-16 (of 19)
CALCULATING THE Ka OF A WEAK ACID FROM pH
A 0.500 M solution of the weak acid HA has a pH 2.010. Calculate its Ka.
Ka = (0.009772 M)2
_____________________________
(0.500 M - 0.009772 M)
= 1.95 x 10-4
What is the pKa of HA?
= -log(1.947 x 10-4)
pKa = -log Ka
= 3.710
The smaller the Ka (or the bigger the pKa) the weaker the acid
1F-17 (of 19)
CALCULATING THE Ka OF A WEAK ACID FROM pH
A 0.500 M solution of the weak acid HX is 3.15% ionized. Calculate its Ka.
x x
HX (aq) + H2O (l) ⇆ H3O+ (aq) + X- (aq)
Initial M’sChange in M’sEquilibrium M’s
0.500 0 0- x + x + x
0.500 - x
Ka = [H3O+][X-] _____________
[HX]
Ka = x2 ______________
(0.500 – x)
3.15 = x (100) _______
0.500
0.01575 = x
1F-18 (of 19)
CALCULATING THE Ka OF A WEAK ACID FROM PERCENT IONIZATION
A 0.500 M solution of the weak acid HX is 3.15% ionized. Calculate its Ka.
Ka = (0.01575 M)2
_____________________________
(0.500 M - 0.01575 M)
= 5.12 x 10-4
What is the pKa of HX?
= -log(5.123 x 10-4)
pKa = -log Ka
= 3.291
1F-19 (of 19)
CALCULATING THE Ka OF A WEAK ACID FROM PERCENT IONIZATION
1G-1 (of 13)
POLYPROTIC ACIDS
Polyprotic acids have Ka values for the ionization of each H+
H2CO3 (aq) + H2O (l) ⇆ H3O+ (aq) + HCO3- (aq)
Ka1 = [H3O+][HCO3-]
__________________
[H2CO3]
Ka1 = 4.3 x 10-7
HCO3- (aq) + H2O (l) ⇆ H3O+ (aq) + CO3
2- (aq)
Ka2 = [H3O+][CO32-]
_________________
[HCO3-]
Ka2 = 5.6 x 10-11
Successive H+’s are harder to remove
∴ H2CO3 is a stronger acid than HCO3-
H2CO3 (aq) + H2O (l) ⇆ H3O+ (aq) + HCO3- (aq) Ka1 = 4.3 x 10-7
HCO3- (aq) + H2O (l) ⇆ H3O+ (aq) + CO3
2- (aq) Ka2 = 5.6 x 10-11
H2CO3 (aq) + 2H2O (l) ⇆ 2H3O+ (aq) + CO32- (aq) Ka1,2 = ?
Ka1,2 = (4.3 x 10-7)(5.6 x 10-11)
= 2.4 x 10-17
1G-2 (of 13)
[H3O+][HCO3-]
__________________
[H2CO3]
x[H3O+][CO3
2-]_________________
[HCO3-]
[H3O+]2[CO32-]
__________________
[H2CO3]
=
Ka1Ka2 = Ka1,2
Find the concentrations of each species in a 0.10 M H2CO3 solution.
x x
H2CO3 (aq) + H2O (l) ⇆ H3O+ (aq) + HCO3- (aq)
Initial M’sChange in M’sEquilibrium M’s
0.10 0 0- x + x + x
0.10 - x
Ka1 = [H3O+][HCO3-]
_________________
[H2CO3]
4.3 x 10-7 = x2 ____________
(0.10 – x)
x = 2.07 x 10-4
[H2CO3] = 0.10 M – 0.000207 M = 0.10 M
[H3O+] = = 0.00021 M
[HCO3-] = = 0.00021 M
1G-3 (of 13)
Find the concentrations of each species in a 0.10 M H2CO3 solution.
0.000207 + y y
HCO3- (aq) + H2O (l) ⇆ H3O+ (aq) + CO3
2- (aq)
Initial M’sChange in M’sEquilibrium M’s
0.000207 0.000207 0- y + y + y
0.000207 - y
Ka2 = [H3O+][CO32-]
_________________
[HCO3-]
5.6 x 10-11 = (0.000207 + y) y __________________
(0.000207 – y)
y = 5.6 x 10-11
[H2CO3] = = 0.10 M
[H3O+] = 0.000207 + 5.6 x 10-11 = 0.00021 M
[HCO3-] = 0.000207 – 5.6 x 10-11 = 0.00021 M
[CO32-] = = 5.6 x 10-11 M
1G-4 (of 13)
pH OF SALT SOLUTIONS
Acid
HNO3
HNO2
Strong acids have non conjugate basesWeak acids have weak conjugate bases
Strong
Weak
Conjugate Base
NO3-
NO2-
Non
Weak
Base
NH3
NaOH
NaOH(H2O)5
Strong bases have non conjugate acidsWeak bases have weak conjugate acids
Weak
Strong
Conjugate Acid
NH4+
Na+
Na(H2O)6+
Weak
Non
1G-5 (of 13)
from LiOH
which is a strong base
∴ Li+ is a non acid
from H3PO4
which is a weak acid
∴ PO43- is a weak base
Li3PO4
∴ pH > 7
1G-6 (of 13)
Li+ PO43-
PO43- (aq) + H2O (l) ⇆ HPO4
- (aq) + OH- (aq)
HYDROLYSIS – The reaction of a dissolved ion with water
from NH3
which is a weak base
∴ NH4+ is a weak acid
from HCl
which is a strong acid
∴ Cl- is a non base
NH4Cl
∴ pH < 7
1G-7 (of 13)
NH4+ Cl-
NH4+ (aq) + H2O (l) ⇆ NH3 (aq) + H3O+ (aq)
from KOH
which is a strong base
∴ K+ is a non acid
from HNO3
Which is a strong acid
∴ NO3- is a non base
KNO3
∴ pH = 7
1G-8 (of 13)
K+ NO3-
Find the pH of a 0.10 M potassium acetate solution
x x
C2H3O2- (aq) + H2O (l) ↔ HC2H3O2 (aq) + OH- (aq)
Initial M’sChange in M’sEquilibrium M’s
0.10 0 0- x + x + x
0.10 - x
Kb = [HC2H3O2][OH-] ___________________
[C2H3O2-]
5.6 x 10-10 = x2 ____________
(0.10 – x)
x = 7.48 x 10-6
Potassium ion is a non acid ; acetate ion is a weak base
Need the Kb for acetate : 5.6 x 10-10
1G-9 (of 13)
CALCULATING THE pH OF A SALT SOLUTION
7.48 x 10-6 = x
= -log(7.48 x 10-6 M)
pOH = -log[OH-] = 5.13
= [OH-]
= 14.000 – 5.13
pH = pKw - pOH
pH + pOH = pKw
= 8.87
Find the pH of a 0.10 M potassium acetate solution
1G-10 (of 13)
CALCULATING THE pH OF A SALT SOLUTION
HC2H3O2 (aq) + H2O (l) ⇆ H3O+ (aq) + C2H3O2- (aq) Ka for acetic acid
C2H3O2- (aq) + H2O (l) ⇆ HC2H3O2 (aq) + OH- (aq) Kb for acetate
2H2O (l) ⇆ H3O+ (aq) + OH- (aq) Kw
KaKb = Kw
1G-11 (of 13)
Find the pH of a 0.25 M ammonium chloride solution if the Kb for ammonia is 1.8 x 10-5.
Ammonium ion is a weak acid; chloride ion is a non base
Need the Ka for ammonium
Have the Kb for ammonia : 1.8 x 10-5
Ka = Kw
____
Kb
KaKb = Kw
= 1.00 x 10-14
_______________
1.8 x 10-5
= 5.56 x 10-10
1G-12 (of 13)
Find the pH of a 0.25 M ammonium chloride solution if the Kb for ammonia is 1.8 x 10-5.
x x
NH4+
(aq) + H2O (l) ⇆ H3O+ (aq) + NH3 (aq)
Initial M’sChange in M’sEquilibrium M’s
0.25 0 0- x + x + x
0.25 - x
Ka = [H3O+][NH3] ______________
[NH4+]
5.56 x 10-10 = x2 ____________
(0.25 – x)
= -log(1.18 x 10-5 M)
pH = -log[H3O+] = 4.93
x = 1.18 x 10-5 M
1G-13 (of 13)