Chemistry 130
Acid and Base equilibria
Dr. John F. C. Turner
409 Buehler Hall
[email protected]
Chemistry 130
Acids and bases
The Brønsted­Lowry definition of an acid states that any material that produces the hydronium ion is an acid. A Brønsted­Lowry acid is a proton donor: +
­
HA  H2 Ol  H3 Oaq  A aq
+
+
H 3 Oaq or Haq 
The hydronium ion, has the same structure as ammonia. It is pyramidal and has one lone pair on O. Chemistry 130
Acids and bases
The Brønsted­Lowry definition of a base states that any material that accepts a proton is a base. A Brønsted­Lowry base is a proton acceptor: ­
­
+
Baq  H2 Ol  HOaq  BHaq
In aqueous solution, a base forms the hydroxide ion: Chemistry 130
Conjugate acids and bases
For any acid­base reaction, the original acid and base are complemented by the conjugate acid and conjugate base:
­
+
NH3aq  H 2 Ol  HOaq  NH4 aq
On the RHS, On the LHS, Water is the proton donor
Hydroxide ion is the proton acceptor
Ammonia is the proton acceptor
Ammonium ion is the proton donor
Water and hydroxide ion are conjugate acid and base
Ammonia and ammonium ion are also conjugate acid and base Chemistry 130
Conjugate acids and bases
The anion of every acid is the conjugate base of that acid
The cation of every base is the conjugate acid of that base
Acid­base conjugates exist due to the dynamic equilibrium that exists in solution.
­
+
­
+
NH3aq  H 2 Ol  HOaq  NH4 aq
NH3aq  H 2 Ol  HOaq  NH4 aq
Chemistry 130
Acid and base constants
We use equilibrium constants to define the position of equilibrium and to reflect the dynamic nature of the system.
For an acid, we define
+
­
HClaq   H2 Ol  H3 Oaq  Claq
KA =
[Products]
=
[Reactants]
+
­
[H3 Oaq ][Claq ]
[HClaq ]
Water is not included as the change in concentration of water is negligible (water is ~55 M when pure) for a dilute solution.
Chemistry 130
Acid and base constants
We use equilibrium constants to define the position of equilibrium and to reflect the dynamic nature of the system.
For a base, we define
+
­
NaOHaq  H2 Ol  Naaq  OH aq
KB =
[Products]
=
[Reactants]
­
[Na+aq ][OHaq
]
[NaOHaq ]
Again, water is not included as the change in concentration of water is negligible (water is ~55 M when pure) for a dilute solution.
Chemistry 130
Acid and base constants
We also use the logarithm of the acid or base constant as an indicator of acid or base strength:
+
KA =
­
[H 3 Oaq ][Claq  ]
[HClaq ]
pK A = −lg K A
+
KB =
­
[Na aq ][OHaq ]
[NaOHaq ]
pK B = −lg K B
Note that the logarithm used here is to base 10, not the natural log
lg = log 10
ln = log e
Chemistry 130
Conjugate acid and base strengths
The dynamic nature of the equilibrium between the conjugate acid­base pair means that a strong acid will have a weak conjugate base
A strong base will have a weak conjugate acid.
In both these cases, 'strong' and 'weak' are defined qualitatively the position of the acid or base equilibrium. A strong acid forms almost exclusively the hydronium ion and the concentration of the undissociated acid is negligible; the conjugate base must be weak.
Chemistry 130
Conjugate acid and base strengths
There are a variety of strong acids – any material that has a larger pKA than the hydronium ion will form the hydronium ion in solution. Any material with an acid constant smaller than the hydronium ion will establish a measurable equilibrium between the acid and the dissociated hydronium ion and associated conjugate base – the anion.
Water therefore acts as a leveling solvent and restricts the degree of acidity possible in aqueous solution. Chemistry 130
Acid
pK A
HI
HBr
HCl
H2 SO4
−9
−8
−6
−3
+
H3 Oaq −1.7
HNO3 −1.3
Autoprotolysis of water
Water can act as both an acid and a base – it can form the hydronium ion as well as the hydroxide ion in solution.
Pure water also undergoes a 'self­equilibrium':
+
­
H 2 Ol  H2 Ol  H3 Oaq  OH aq
KW =
[Products]
=
[Reactants]
+
­
­
[H 3 O+aq ][OHaq
]
2
[H 2 Ol]
−14
K W = [H 3 Oaq ][OHaq ] = 1×10
pK W = 14
This is autoprotolysis or autionization.
Chemistry 130
pKW , pH and pOH
The autoprotolysis of water relates the hydronium ion concentration and the hydroxide ion concentration to the autoprotolysis constant of water:
+
­
H 2 Ol  H2 Ol  H3 Oaq  OH aq
­
K W = [H 3 O+aq ][OHaq
] = 1×10−14



 [H O ]−lg  [OH ]  = p H
+
­
+


­

pK w = −lg10 [H 3 Oaq ][OHaq ] = −lg 10 [H 3 Oaq ] −lg 10 [OHaq ] = 14
pK w = −lg10
3
+
aq
10
­
aq
 pOH = 14
pK w = p H  p OH = 14
This relationship dictates the concentrations of hydroxide and hydronium ion in all aqueous solutions
Chemistry 130
Simple pH and pOH calculations
Q: What is the pH of a solution of 0.0002 M solution of HI?
−1
Molarity of solution = 0.0002 mol L
The ionization equation of HI in water is
+
­
HIaq   H2 Ol  H3 Oaq  Iaq
The K A of HI is very large and we know that HIaq is a strong acid
+
Therefore, [H3 Oaq ] = 0.0002 mol L


+
lg 10 [H3 Oaq
] = −3.699


p H = −lg 10 [H3 O+aq ] = 3.699
Chemistry 130
−1
Simple pH and pOH calculations
Q: Calculate pOH, the pH and the concentration of hydroxide ion in a solution of 0.0002 M solution of HI?
We have already calculated the p H of the solution
+
p H = −lg 10 [H3 Oaq ] = 3.699


The relationship between p H, pOH and pK W is
pK w = p H  pOH = 14
3.699  pOH = 14
pOH = 14 − 3.699 = 10.301

­

pOH = −lg 10 [OHaq ] = 10.301
[OH­aq ] = 10−10.301 = 5 × 10−11 mol L−1
Chemistry 130
Simple pH and pOH calculations
Q: Calculate pOH, the pH and the concentration of hydroxide ion in a solution of 0.00075 M solution of KOH?
pOH calculation
­
−1
[OHaq ] = 0.00075 mol L



­

lg [OHaq  ] = −3.125
­
pOH = −lg 10 [OHaq
] = 3.125
As
pK w = p H  pOH = 14
p H  3.125 = 14
p H = 14 − 3.125 = 10.875
As
+
p H = −lg 10 [H3 Oaq ]

+
lg 10 [H3 Oaq


]  = −10.875
[H3 O+aq ] = 10−10.875 = 1.33 × 10−11 mol L−1
Chemistry 130
Weak acids and bases
These calculations are straightforward as HI is a strong acid and KOH is a strong base. For weak acids and bases, which do not fully ionize in solution, we have to use the rules for equilibria with which we are already familiar.
We also use
pK A = −lg K A
pK B = −lg K B
which we have defined for the generic acid reaction and base reaction in water. Chemistry 130
Weak acid/base calculations
Q: Calculate pOH, the pH and the concentration of hydroxide ion in a solution of 0.5 M solution of acetic acid?
+
­
−5
K A = 1.8 × 10
CH 3 CO2 Haq  H 2 Ol  H3 Oaq  CH3 CO2aq
Acetic acid
Acetate ion
CH3 CO2 Haq
Initial concentrations
Change
Equilibrium concentrations
0.5
−x
0.5− x
At equilibrium, the new concentrations are:
+
­
H 3 Oaq CH3 CO2aq
0
x
x
≈0
x
x
[CH 3 CO2 Haq ] = 0.5− x
­
[CH 3 CO2aq ] = x
+
[H3 Oaq ] = x
Chemistry 130
Weak acid/base calculations
−5
The acid constant for acetic acid is K A = 1.8 × 10
−14
whereas the autoprotolysis constant for water is K W = 1 × 10
The contribution of the self­ionization of water is of the order of 10­7 to the concentration of hydronium ion.
In general, if the acid constant for the weak acid is sufficiently large in comparison to the autoprotolysis constant for water, we can ignore the autoprotolysis of water with respect to the hydrogen ion concentration Chemistry 130
Weak acid/base calculations
Q: Calculate pOH, the pH and the concentration of hydroxide ion in a solution of 0.5 M solution of acetic acid?
+
­
CH 3 CO 2 Haq  H2 Ol  H 3 Oaq  CH3 CO2aq
Acetic acid
Acetate ion
−5
K A = 1.8 × 10
We now set up the equilibrium constant for aqueous acetic acid:
­
+
[CH 3 CO2aq ][H3 Oaq ]
KA =
[CH 3 CO2 Haq ]
and use the concentrations that we have already calculated:
[CH 3 CO2 Haq ] = 0.5− x
­
+
[CH
CO
][H
O
 x  x 
­
3
2 aq
3 aq ]
[CH 3 CO2aq ] = x
KA =
=
[CH 3 CO2 Haq ]
0.5− x 
+
[H 3 Oaq ] = x
Chemistry 130
Weak acid/base calculations
Q: Calculate pOH, the pH and the concentration of hydroxide ion in a solution of 0.5 M solution of acetic acid?
+
­
CH3 CO2 Haq  H 2 Ol  H3 Oaq  CH3 CO2aq
Acetic acid
Acetate ion
−5
K A = 1.8 × 10
We now solve the equilibrium equation that we have set up:
KA =
2
 x  x 
−5
= 1.8 × 10
0.5− x 
x
−5
= 1.8 × 10
0.5− x 
There are two ways of solving this equation
1. use the quadratic formula
Chemistry 130
2. make the assumption that x ≪0.5
Weak acid/base calculations
Q: Calculate pOH, the pH and the concentration of hydroxide ion in a solution of 0.5 M solution of acetic acid?
+
­
CH3 CO 2 Haq  H2 Ol  H3 Oaq  CH3 CO2aq 
Acetic acid
Acetate ion
−5
K A = 1.8 × 10
Using assumption 2, that x ≪0.5
x2
x2
−5
KA =
≈
= 1.8 × 10
0.5− x 
0.5
x 2 ≈ 0.5 × 1.8 × 10−5 ≈ 9 × 10−6
1
−6 2
x ≈  9 × 10  ≈ 0.003
+
−1
[H3 Oaq ] = 0.003 mol L
p OH = 14− p H = 11.48
Chemistry 130
+
p H = −lg [H3 Oaq ] = 2.52
­
−12
[OHaq ] = 3.33 × 10
−1
mol L
Weak acid/base calculations
Q: Calculate pOH, pH and the concentrations of both hydroxide ion and hydronium in a solution of 1.5 M solution of hydroxylamine?
­
+
−9
NH2 OHaq  H 2 Ol  OHaq  NH3 OHaq
K B = 9.1 × 10
Hydroxylamine
Hydroxylammonium
NH 2 OHaq
Initial concentrations
Change
Equilibrium concentrations
1.5
−x
1.5−x
At equilibrium, the new concentrations are:
­
+
OHaq NH 3 OHaq
0
x
x
≈0
x
x
[NH2 OHaq ] = 1.5− x
+
[NH3 OHaq ] = x
­
[OHaq ] = x
Chemistry 130
Weak acid/base calculations
Q: Calculate pOH, pH and the concentrations of both hydroxide ion and hydronium in a solution of 1.5 M solution of hydroxylamine?
­
+
−9
NH2 OHaq  H 2 Ol  OHaq  NH3 OHaq
K B = 9.1 × 10
Hydroxylamine
Hydroxylammonium
We now set up the base equilibrium constant for aqueous hydroxylamine:
+
­
[NH3 OHaq ][OHaq ]
KA =
[NH2 OHaq  ]
and use the concentrations that we have already calculated:
[NH2 OHaq ] = 1.5− x
+
­
2
[NH
OH
][OH
x
+
3
aq
aq ]
[NH3 OHaq ] = x
KB =
=
[NH2 OHaq  ]
1.5− x 
­
[OHaq ] = x
Chemistry 130
Weak acid/base calculations
Q: Calculate pOH, pH and the concentrations of both hydroxide ion and hydronium in a solution of 1.5 M solution of hydroxylamine?
­
+
−9
NH2 OHaq  H 2 Ol  OHaq  NH3 OHaq
K B = 9.1 × 10
Hydroxylamine
Hydroxylammonium
The equation that we need to solve is:
+
­
2
[NH 3 OHaq ][OHaq ]
x
−9
KB =
=
= 9.1 × 10
[NH2 OHaq ]
1.5−x 
Using the quadratic method,
x2
−9
= 9.1 × 10
1.5− x 
2
−9
−8
−9
so x = 1.5− x ⋅9.1 × 10 = 1.365 × 10 − 9.1 × 10 x
2
−9
−8
x  9.1 × 10 x − 1.365 × 10 = 0
Chemistry 130
Weak acid/base calculations
Q: Calculate pOH, pH and the concentrations of both hydroxide ion and hydronium in a solution of 1.5 M solution of hydroxylamine?
­
+
−9
NH2 OHaq  H2 Ol  OHaq   NH3 OHaq
K B = 9.1 × 10
Hydroxylamine
Hydroxylammonium
2

−b±
b
− 4ac 
2
For a quadratic of the form ax  bx  c = 0, x =
2a
−9
−8
In this case, a = 1 b = 9.1 × 10 c = −1.365 × 10 and so
−9.1 × 10 ± 9.1 × 10  − 4⋅1⋅−1.365 × 10  
x =
2
−9
−9 2
−8
1
−8 2
−9.1 × 10 ± 5.460000008 × 10
x =
2
−9
Chemistry 130

1
2
−4
−1
= 1.168 × 10 mol L
1
2
Weak acid/base calculations
Q: Calculate pOH, pH and the concentrations of both hydroxide ion and hydronium in a solution of 1.5 M solution of hydroxylamine?
­
+
−9
NH2 OHaq  H2 Ol  OHaq   NH3 OHaq
K B = 9.1 × 10
Hydroxylamine
Hydroxylammonium
The equation that we need to solve is:
+
­
2
[NH3 OHaq ][OHaq ]
x
−9
KB =
=
= 9.1 × 10
[NH2 OHaq ]
1.5− x 
Using the assumption that x ≪1.5
2
2
x
x
≈
= 9.1 × 10−9
1.5− x 
1.5
2
−9
−8
so x = 1.5⋅9.1 × 10 = 1.365 × 10
1
−8 2
x ≈ 1.365 × 10  ≈ 1.168 × 10−4 mol L−1
Chemistry 130
Weak acid/base calculations
Q: Calculate pOH, pH and the concentrations of both hydroxide ion and hydronium in a solution of 1.5 M solution of hydroxylamine?
­
+
−9
NH2 OHaq  H2 Ol  OHaq  NH3 OHaq
K B = 9.1 × 10
Hydroxylamine
Hydroxylammonium
In this case there is no difference between the methods to the third place of
­
−4
−1
decimal and so x = [OHaq ] = 1.168 × 10 mol L
p OH = −lg 10  [OHaq ]  = −lg10  1.168 × 10
­
As p OH  p H = 14
p H = 14 − p OH = 14 − 3.93 = 10.07
+
and so as p H = −lg10 [H3 Oaq ] 
−4
[H3 O+aq ] = 10−10.07 = 8.56 × 10−11 mol L−1
Chemistry 130

= 3.93
Weak acid/base calculations
Given that pH and pOH are related via the relationship
+
­
−14
K W = [H3 Oaq  ][OHaq ] = 1×10
pK w = pH  pOH = 14
then in aqueous solution, only one of these quantities is required for all to be calculable.
Similarly, if KA is known for an acid or KB is known for a base, then using the change on the establishment of equilibrium will give the hydronium ion concentration or the hydroxide concentration and so all is known. If we know the pH or pOH, we can also calculate KA or KB for the system, using similar methods.
Chemistry 130
Weak acid/base calculations
Q: Calculate pKB for dimethylamine given that a 0.164 M solution has a pH of 11.98.
­
+
Me2 NHaq  H2 Ol  OHaq  Me2 NH2aq
Dimethylamine
Dimethylammonium
−1
When [Me2 NHaq ] = 0.164 mol L , p H = 11.98
p H  p OH = 14
so 11.98  pOH = 14 so p OH = 14 − 11.98 = 2.02
­
Given that pOH = −lg 10 [OHaq
]
­
−2.02
[OHaq ] = 10
Chemistry 130


−3
−1
= 9.55 × 10 mol L
Weak acid/base calculations
Q: Calculate pKB for dimethylamine given that a 0.164 M solution has a pH of 11.98.
­
+
Me2 NHaq  H2 Ol   OHaq   Me2 NH2aq
Dimethylamine
Dimethylammonium
­
−3
−1
From the p H calculation, [OHaq ] = 9.55 × 10 mol L
­
+
Me2 NHaq 
OHaq
Me2 NH2aq 
Initial concentrations
0.164
≈0
0
−3
−3
−3
Change
−9.55 × 10
9.55 × 10
9.55 × 10
−3
−3
−3
Equilibrium concentrations 0.164−9.55 × 10
9.55 × 10
9.55 × 10
Chemistry 130
Weak acid/base calculations
Q: Calculate pKB for dimethylamine given that a 0.164 M solution has a pH of 11.98.
­
+
Me2 NHaq  H2 Ol  OHaq  Me2 NH2aq
Dimethylamine
Dimethylammonium
We now know all the concentrations requred to calculate K B
−3
[Me2 NHaq ] = 0.164−9.55 × 10
−1
−1
= 1.5445 × 10 mol L
[OH­aq ] = 9.55 × 10−3 mol L−1
+
−3
−1
[Me2 NH2aq ] = 9.55 × 10 mol L
+
KB =
­
[Me2 NH2aq ][OHaq ]
[Me2 NHaq ]
pK B = −lg 10  K B  = 3.23
Chemistry 130
−3
−3
9.55 × 10 9.55 × 10 
−4
=
=
5.9
×
10
1.5445 × 10−1 
Polyprotic acids
A polyprotic acid is one that can ionize more than once. Common examples include Sulfuric acid
H 2 SO 4
Phosphoric acid
H 3 PO4
Carbonic acid
H 2 CO3
Note that acids such as acetic acid are not polyprotic
+
­
CH 3 CO2 Haq  H2 Ol  H3 Oaq  CH 3 CO 2aq
­
+
2­
CH3 CO2aq  H2 Ol   H3 Oaq  CH2 CO2aq
Chemistry 130
Polyprotic acids
Each ionization of a polyprotic acid has an associated acid constant. For phosphoric acid H 3 PO 4
+
­
K A 1 = 7.1 × 10
+
2­
K A 2 = 6.3 × 10
H 3 PO4 aq  H2 Ol   H3 Oaq  H 2 PO4 aq
­
H 2 PO4 aq  H2 Ol   H3 Oaq  HPO4 aq
2­
+
3­
−3
−8
−13
K A 3 = 4.3 × 10
HPO4 aq  H2 Ol   H3 Oaq  PO4 aq
Note that each acid constant differs from the one before and, although the first ionization, in this case, is strong, the others are not.
−3
H 3 PO4 aq
K A 1 = 7.1 × 10
Strong acid
H 2 PO­4 aq
2­
HPO4 aq
Chemistry 130
K A 2 = 6.3 × 10−8
−13
K A 3 = 4.3 × 10
Weak acid
Weak acid
Polyprotic acids
If the first ionization of a polyprotic acid is described by a large acid constant, then equal concentrations of the hydronium ion and dihydrogenphosphate ons are produced.
The equilibrium for the second ionization is ­
+
2­
−8
K A 2 = 6.3 × 10
H 2 PO4 aq  H2 Ol   H3 Oaq  HPO4 aq
and the equilibrium constant is
2­
K A2 =
+
[HPO4 aq ][H3 Oaq  ]
­
[H2 PO4 aq ]
2­
2­
=
­
[HPO4aq ][H2 PO4 aq ]
­
[H2 PO4 aq  ]
= 6.3 × 10−8
−8
K A 2 = [HPO4 aq ] = 6.3 × 10
and so the acidity of the second ionization is independent of the initial concentration of phosphoric acid Chemistry 130
Chemistry 130
Acid and Base equilibria
Dr. John F. C. Turner
409 Buehler Hall
[email protected]
Chemistry 130
Salts of strong and weak acids
When a salt is dissolved, the equilibria for the conjugate acid and base are established.
Dissolving the salt introduces the conjugate acid or the conjugate base into the solution and the normal equilibria occur
Salts of strong acids and strong bases form neutral solutions
Salts of strong acids and weak bases form acidic solutions
Salts of weak acids and strong bases form basic solutions
The pH of solutions of salts of weak acids and weak bases depend on the acid constant of the acid and the base constant of the base Chemistry 130
Salts of strong and weak acids
The reaction that occurs when an anion associated with a weak acid is dissolved in water changes the pH of the solution. This happens because the anion is the conjugate base of the associated acid and the acid­base equilibrium for that acid is established.
For nitrous acid, the associated anion is nitrite, NO­2
­
­
NO­2aq  H2 Ol  OHaq
 HNO2aq 

+
KB =
[OHaq ][HNO2aq ]
­
NO2aq
­
[H3 Oaq ]
+
[OHaq ][HNO2aq ] [H3 Oaq ]
We can multiply K B by to give K B =
⋅
+
­
+
[H3 Oaq ]
NO2aq
[H3 Oaq ]
­
+
[OHaq ][HNO2aq ] [H3 Oaq ]
KB =
⋅
­
NO2aq
[H3 O+
]
aq
−4
For nitrous acid, K A,HNO = 7.2 × 10
2
Chemistry 130
Salts of strong and weak acids
For nitrous acid, the associated anion is nitrite, ­
NO­2aq   H2 Ol  OH­aq  HNO2 aq
KB =
+
[OHaq ][HNO2aq ]
­
NO2aq
­
[H3 Oaq ]
+
[OHaq ][HNO2aq ] [H 3 Oaq ]
We can multiply K B by to give K B =
⋅
+
­
+
[H3 Oaq ]
NO2aq 
[H 3 Oaq ]
­
[OHaq
][HNO2aq ] [H3 O+aq ]
KW
KB =
⋅
=
­
+
K A,HNO
NO2aq
[H3 Oaq ]
−4
2
For nitrous acid, K A,HNO = 7.2 × 10 and so 2
1 × 10−14
−11
KB =
=
1.39
×
10
−4
7.2 × 10
Chemistry 130
Salts of strong and weak acids
For nitrous acid, the associated anion is nitrite, ­
­
NO­2aq   H2 Ol  OHaq
 HNO2aq 

KB =
+
[OHaq ][HNO2aq ]
­
NO2aq
­
[H3 Oaq ]
+
[OHaq ][HNO2aq ] [H3 Oaq ]
We can multiply K B by to give K B =
⋅
+
­
+
[H3 Oaq ]
NO2aq 
[H3 Oaq ]
[OH­aq ][HNO2aq ] [H3 O+aq ]
KW
KB =
⋅
=
­
+
K A,HNO
NO2aq
[H3 Oaq ]
−4
2
For nitrous acid, K A,HNO = 7.2 × 10 and so 2
−14
1 × 10
−11
KB =
=
1.39
×
10
−4
7.2 × 10
Chemistry 130
Strength of conjugate acids and bases The acid and base strength of a conjugate acid­base pair, such as ­
nitrite­nitrous acid
NO2 − HNO2
acetic acid­acetate
CH 3 CO2 H − CH3 CO2
ammonia­ammonium
NH3 − NH4
­
+
are related by the relationship K A KB = K W
pK A  pK B = pK W = 14
Chemistry 130
Strength of conjugate acids and bases This situation occurs because the conjugate base of a weak acid and the conjugate acid of a weak base are both appreciably strong.
The acid or base strength of a conjugate acid or base of a strong base or acid is extremely weak and is negligible in most applications.
The appreciable strength of a conjugate acid or base and the presence of an equilibrium, because the base or acid is weak means that additions of acid or base to a solution that contains the acid­base conjugate pair will not effect the pH of the solution greatly.
These solutions are termed 'buffers'. Chemistry 130
Buffer solutions
A buffer solution is one that contains a conjugate acid­base pair and is used to provide a relatively constant pH in chemical reactions, biological and medical systems and in industrial settings.
Because of the presence of the conjugate acid­base, we can write
HA aq

H2 Ol
Weak acid
­
KA =
A ­aq

H3 O+aq
Weak acid
+
[A aq  ][H3 Oaq  ]
[HA aq ]

and so K A
[HA aq ]
­
[ A aq ]
+
= [H3 Oaq
]

from which we can calculate the pH if we know the concentrations of the acid and conjugate base and the acid constant for the acid.
Chemistry 130
Buffer solutions
Given that
HAaq
Weak acid
[H3 O
+
aq
H2 Ol

] = KA


+
H3 Oaq
[HA aq ]
­
[ A aq ]
+
aq
p H = −lg 10 [H3 O


] = −lg10 K A
p H = −lg 10  K A   lg10
Chemistry 130
­
A aq
Weak acid


­
[A aq
]

[HA aq ]

[HAaq ]
[A
­
aq
]

= −lg10  K A  − lg 10

[HAaq  ]
­
[ A aq ]

Buffer solutions
For a buffer, the pH is given by
p H = p K A  lg 10

which is the Henderson­Hasselbalch equation.
Chemistry 130
­
[ A aq ]
[HA aq ]

Chemistry 130