Models for acids and bases
(a) Arrhenius model
Acids produce H+ ions in solution
Bases produce OH- ions in solution
(b) Bronsted-Lowry model
An acid is a proton donor
A base is a proton acceptor
In the Bronsted-Lowry model, we have the concept of conjugate acid-base pairs.
An acid has an extra H+ than its conjugate base – these two form a pair
A base has one less H+ than its conjugate acid – these two form a pair
Example:
Base – Conjugate acid pair
HA (aq) + H2O (l) H3O+ (aq) + A- (aq)
Acid - Conjugate base pair
Here, HA is an acid (one extra H+) and forms a pair with A -, its conjugate base.
and H2O is a base (one less H+) and forms a pair with H 3O+, its conjugate acid.
(c)Lewis model
A Lewis acid is an electron-pair acceptor
A Lewis base is an electron-pair donor
Concept of pH
ACIDS
Since the original Arrhenius model defined an acid as a hydrogen ion donor (donates H+, which
is essentially a proton) in solution, a scale that measures this concentration is used to define
acids. This is a logarithmic scale, designated pH. There is a decrease of 1 unit for every 10-fold
increase in [H+] concentration.
pH = - log [H+]
,To derive [H+], the H+ concentration from pH the formula is
[H+] = 10 – pH
A strong acid dissociates completely in water solutions to produce equimolar amounts of H+
and conjugate base:
Example: HCl H+ + Cl –
(or)
HCl + H2O H3O + + Cl –
Ka represents the equilibrium dissociation constant of an acid, which reflects the ability of an
acid to dissociate into H+ ions or into [H3O]+ ions in the presence of water. Ka of a strong acid is
large (> 1) and lies far to the right (mostly ionized). The stronger the acid, the higher its K a
A weak acid dissociates only partially into H+ and conjugate base and has a smaller Ka
Ka, the equilibrium dissociation constant of a weak acid, is small ( < 1) and lies far to the left
(mostly unionized). The weaker the acid, the smaller its Ka
Most organic acids contain a carboxyl group (COOH) which contributes the H+ ion and are weak
acids.
Example: Acetic acid (CH3COOH also written as HC2H3O2)
BASES
The concentration of the base (a OH- ion producer or a proton acceptor) in solution is defined
by pOH (The base extracts a hydrogen ion from water, leaving the OH - ion)
pOH = - log [OH -]
To derive [OH-], the OH- concentration from pOH, the formula is
[OH-] = 10 – pOH
A strong base has OH- ions in its formula (examples NaOH, KOH) and the OH - ions are fully
dissociated in solution.
Example:
NaOH Na+ + OH -
A weak base (B) does not contain OH - ions, but reacts with water to extract a hydrogen ion and
releases OH- ions. This reaction is base hydrolysis.
, Example:
B (aq) + H2O BH+ + OH -
Formulas to remember:
MOLARITY [M] = moles of solute / Liter of solution (mol/L) gives the
concentration of acids and bases in solution
For acids, [H+] > [OH-] and pH < 7, higher the [H+], lower the pH
For bases, [H+] < [OH-] and pH > 7, lower the [H+], higher the pH
For neutral solutions, [H+] = [OH-] and pH = 7
pH + pOH = 14
[H+][OH-] = 10 -14 = Kw, the equilibrium constant for the auto-ionization of water.
Ka x Kb = Kw = 10 -14, where ‘a’ and ‘b’ are acid and conjugate-base pairs.
Also,
The stronger the acid, weaker is its conjugate base.
The stronger the base, weaker is its conjugate acid.
Kb represents the equilibrium dissociation constant of a base, which reflects the ability of a base
to accept a proton from water to form OH- ions in solution. Strong bases have OH- ions in their
formula. Kb of a strong base is large (> 1) and lies far to the right (mostly ionized). The stronger
the base, the higher its Kb
A weak base dissociates only partially and reacts with water to produce OH- ions and has a
smaller Kb.
Substances like water, ammonia are amphoteric and can act as acids or bases.
Example problems:
(1) What is the pOH of a solution whose [H+] is 2.75 x 10 -4 M?
pH = - log [H+] = - log (2.75 x 10-4 M) = 3.56
pOH = 14 – 3.56 = 10.44
(2) What is the [H+] of a solution whose pOH = 2.86?
pH = 14 – pOH = 14 - 2.86 = 11.14
[H+] = 10 -11.14 = 7.24 x 10 -12 M
(3) What is the pH of a 0.0001M solution of NaOH?
pOH = - log[OH- ] = - log [0.0001M] = 4
(a) Arrhenius model
Acids produce H+ ions in solution
Bases produce OH- ions in solution
(b) Bronsted-Lowry model
An acid is a proton donor
A base is a proton acceptor
In the Bronsted-Lowry model, we have the concept of conjugate acid-base pairs.
An acid has an extra H+ than its conjugate base – these two form a pair
A base has one less H+ than its conjugate acid – these two form a pair
Example:
Base – Conjugate acid pair
HA (aq) + H2O (l) H3O+ (aq) + A- (aq)
Acid - Conjugate base pair
Here, HA is an acid (one extra H+) and forms a pair with A -, its conjugate base.
and H2O is a base (one less H+) and forms a pair with H 3O+, its conjugate acid.
(c)Lewis model
A Lewis acid is an electron-pair acceptor
A Lewis base is an electron-pair donor
Concept of pH
ACIDS
Since the original Arrhenius model defined an acid as a hydrogen ion donor (donates H+, which
is essentially a proton) in solution, a scale that measures this concentration is used to define
acids. This is a logarithmic scale, designated pH. There is a decrease of 1 unit for every 10-fold
increase in [H+] concentration.
pH = - log [H+]
,To derive [H+], the H+ concentration from pH the formula is
[H+] = 10 – pH
A strong acid dissociates completely in water solutions to produce equimolar amounts of H+
and conjugate base:
Example: HCl H+ + Cl –
(or)
HCl + H2O H3O + + Cl –
Ka represents the equilibrium dissociation constant of an acid, which reflects the ability of an
acid to dissociate into H+ ions or into [H3O]+ ions in the presence of water. Ka of a strong acid is
large (> 1) and lies far to the right (mostly ionized). The stronger the acid, the higher its K a
A weak acid dissociates only partially into H+ and conjugate base and has a smaller Ka
Ka, the equilibrium dissociation constant of a weak acid, is small ( < 1) and lies far to the left
(mostly unionized). The weaker the acid, the smaller its Ka
Most organic acids contain a carboxyl group (COOH) which contributes the H+ ion and are weak
acids.
Example: Acetic acid (CH3COOH also written as HC2H3O2)
BASES
The concentration of the base (a OH- ion producer or a proton acceptor) in solution is defined
by pOH (The base extracts a hydrogen ion from water, leaving the OH - ion)
pOH = - log [OH -]
To derive [OH-], the OH- concentration from pOH, the formula is
[OH-] = 10 – pOH
A strong base has OH- ions in its formula (examples NaOH, KOH) and the OH - ions are fully
dissociated in solution.
Example:
NaOH Na+ + OH -
A weak base (B) does not contain OH - ions, but reacts with water to extract a hydrogen ion and
releases OH- ions. This reaction is base hydrolysis.
, Example:
B (aq) + H2O BH+ + OH -
Formulas to remember:
MOLARITY [M] = moles of solute / Liter of solution (mol/L) gives the
concentration of acids and bases in solution
For acids, [H+] > [OH-] and pH < 7, higher the [H+], lower the pH
For bases, [H+] < [OH-] and pH > 7, lower the [H+], higher the pH
For neutral solutions, [H+] = [OH-] and pH = 7
pH + pOH = 14
[H+][OH-] = 10 -14 = Kw, the equilibrium constant for the auto-ionization of water.
Ka x Kb = Kw = 10 -14, where ‘a’ and ‘b’ are acid and conjugate-base pairs.
Also,
The stronger the acid, weaker is its conjugate base.
The stronger the base, weaker is its conjugate acid.
Kb represents the equilibrium dissociation constant of a base, which reflects the ability of a base
to accept a proton from water to form OH- ions in solution. Strong bases have OH- ions in their
formula. Kb of a strong base is large (> 1) and lies far to the right (mostly ionized). The stronger
the base, the higher its Kb
A weak base dissociates only partially and reacts with water to produce OH- ions and has a
smaller Kb.
Substances like water, ammonia are amphoteric and can act as acids or bases.
Example problems:
(1) What is the pOH of a solution whose [H+] is 2.75 x 10 -4 M?
pH = - log [H+] = - log (2.75 x 10-4 M) = 3.56
pOH = 14 – 3.56 = 10.44
(2) What is the [H+] of a solution whose pOH = 2.86?
pH = 14 – pOH = 14 - 2.86 = 11.14
[H+] = 10 -11.14 = 7.24 x 10 -12 M
(3) What is the pH of a 0.0001M solution of NaOH?
pOH = - log[OH- ] = - log [0.0001M] = 4
