Acid Base Equilibria
Acid Ionization, also known as acid dissociation, is the process in where an acid
reacts with water to produce a hydrogen ion and the conjugate base ion.
H+(aq) + C2H3O2-(aq)
HC2H 3O2(aq)
For a weak acid the concentration of ions in solution is determined by the acid
ionization constant which is the equilibrium constant for the ionization of a weak
acid.
H3O+(aq) + A-(aq)
HA(aq) + H2O(l)
Ka =
[H3O+] [A-]
[HA]
There are two methods to determine the ionization constant for a weak acid are:
1. Determination of Ka from the degree of ionization - is the fraction of molecules
that react with water to give ions.
2. Determination of Ka from the solution pH - finding the pH of a solution will give
you the hydronium concentration.
Example: Nicotinic acid is a monoprotic acid with the formula, HC6H4NO2. A solution that is
0.012M has a pH of 3.39 at 25ºC. What is the acid ionization constant, Ka, for this acid at 25ºC?
What is the degree of ionization of nicotinic acid in this solution?
Calculations with Ka
Give the Ka for a weak acid
you can determine the
concentrations of HA, H+ and
A-. Using the method that we
used to determine the
equilibrium constant, Kc, we
can apply that theory in a
more simplified form to
determining the different
molarities of our weak acid.
Example: What are the concentrations of nicotinic acid, hydrogen ion, and nicotinate ion in the
solution of a 0.10 M nicotinic acid? What is the pH of the solution? What is the degree of
ionization?
To determine whether you can use the simplified assumption check your answer. It
can be shown that this simplifying assumption gives an error of less than 5% if the
concentration of the acid, ca, divided by Ka equal 100 or more.
Example: an acid concentration of 10-2M and the Ka of 10-5
Ca / Ka = 10-2 / 10-5 = 10-3
so the assumption is valid.
If the simplifying assumption is not valid, you can solve the equilibrium equation
with the quadratic formula.
What is the pH at 25ºC of the solution obtained by dissolving a 5.00 grain tablet of aspirin in
0.500L of water? The tablet contains 5.00 grains, or 0.325 g of acetylsalicylic acid, HC9H7O4.
The acid is monoprotic and the Ka = 3.3 x 10-4 at 25ºC.
Polyprotic Acids are acids that have more than one ionizable hydrogen
Example: Carbonic acid
H2CO3(aq)
H+(aq) + HCO-3(aq)
Ka1 = 4.3 x 10-7
HCO3-(aq)
H+(aq) + CO-23(aq)
Ka2 = 4.8 x 10-11
The Ka1 is much larger than the Ka2 which indicates that the carbonic acid loses the
first proton more easily than the second one.
Ascorbic acid is a diprotic acid, H2C6H6O6. what is the pH of a 0.10 M solution? What is the
concentration of ascorbate ion? The acid ionization constants are Ka1 = 7.9 x 10-5 and
Ka2 = 1.6 x 10-12.
Base Ionization Equilibria involves weak bases and are treated similarly to those
of weak acids.
NH4+(aq) + OH-(aq)
NH3(aq) + H2O(l)
[NH4+][ OH-]
Kb =
[NH3]
Morphine, C17H19NO3, is a naturally occurring base, or alkaloid. What is the pH of a 0.0075 M
solution of morphine at 25C? The base ionization constant, Kb, is 1.6 x 10-6.
A salt can be regarded as an ionic compound from a neutralization reaction in
aqueous solution. Often it is acidic or basic because it reacts with the water. The
hydrolysis of an ion is the reaction of an ion with water to produce the conjugate
acid and hydroxide ion or the conjugate base and the hydrogen ion.
Generally : The anions of a weak acid are basic and the cations of a weak base are
acidic.
To predict the acidity or basicity of a salt solution you need to examine the acidity
or basicity of the ions composing the salt.
1. A salt of a strong base and a strong acid gives a neutral aqueous solution
2. A salt of a strong base and a weak acid gives a basic solution.
3. A salt of a strong acid and a weak base gives an acidic solution.
4. A salt of a weak acid and a weak base is dependent on the relative acid-base
strengths of the two ions.
Calculating the Ka from Kb or Kb from Ka
Ka Kb = Kw
Determine the Kb for CN- and the Ka for NH4+
Common Ion Effect is the shift in an ionic equilibrium caused by the addition of a
solute that provides an ion that takes part in the equilibrium.
CH3COOH(aq) + H2O(l)
CH3COO-(aq) + H3O+(aq)
Buffers
A buffer is a solution characterized by the ability to resist changes in pH when
limited amounts of acid and base are added to it.
Buffers contain either a weak acid and its conjugate base or a weak base and its
conjugate acid.
A buffer solution resists change in pH through the ability to combine with both
H3O+ ions and OH- ions.
Two important characteristics of a buffer are :
pH
buffer capacity
Given instructions on how to make up a buffer we can calculate the pH of the final
solution
Instructions for making a buffer say to mix 60 mL of 0.100 M NH3 with 40 mL of 0.100 M
NH4Cl. What is the pH of the buffer?
Henderson-Hasselbalch Equation is an equation relating the pH of a buffer for
different concentrations of conjugate acid and base.
[௕௔௦௘]
pH = pKa + log [௔௖௜ௗ]
You add 1.5 mL of a 1.0M HCl to each of the following solutions. Which one will
show the least change of pH?
a. 15 mL of 0.1 M NaOH
b. 15 mL of 0.1 M CH3COOH
c. 30 mL of 0.1 M NaOH and 30 mL of 0.1 M CH3COOH
d. 30 mL of 0.1 M NaOH and 60 mL of 0.1 M CH3COOH
Acid Base Titration Curves is a plot of the pH of a solution of acid (or Base)
against a volume of added base (or acid).
Curve for the titration of a strong acid by a strong base
Equivalence point - the point in a titration when a stoichiometric amount of
reactant has been added.
Calculating the pH of a solution of a strong acid and a strong base.
Calculate the pH of a solution in which 15 mL of a 0.10 M NaOH has been added
to 25 mL of a 0.10 M HCl.
Curve for the titration of a weak acid by a strong base
Calculate the pH of the solution at the equivalence point when 25 mL of 0.10 M
nicotinic acid is titrated by 0.10 M NaOH. Ka for nicotinic acid is 1.4 x 10-5
Curve for the titration of a weak base by a strong acid
Calculate the pH of the NH3 solution at the following points during the titration:
a) Prior to the addition of any HCl
b) After the addition of 12 mL of 0.100 M HCl
c) At the equivalence point
d) After the addition of 31 mL of 0.100 HCl