Dissociation of waterACID-BASE EQUILIBRIA
In water the concentrations of H+ and OH-
are
finite but small and equal
Can solve for their concentrations:
[H+] = [OH-] = 1 x 10-7
To avoid using small powers of ten, i.e. 10-7, to describe
the concentrations of [H+],
a new quantity is defined called the pH = -log[H+]
In water the pH is equal to 7. This is a neutral solution.
If the pH is less than 7 then the concentration of H+ in
water is more than that of pure
water and the solution is acidic. For a 0.10M HCl solution the
pH is -log(0.1) = 1
One can also describe the concentration of OH- in water in
terms of pOH.
Addition of 0.01M NaOH to water: pOH = -log(0.01) = -log(10-2)
= 2
Equilibrium constants that describe the ionization of weak acids, Ka,
are relatively
small numbers, say Ka = 2 x 10-5 for acetic acid.
This number can be defined in
terms of pKa and for acetic acid pKa = -log(2x10-5)
= 4.7
For water pKw = -logKw = -log(10-14) = 14.
Also pH + pOH = pKw = 14.
In an acidic solution with a pH of 3, the pOH is 14 - 3 = 11.
The pH of some common solutions vary from 1 to 13.
ACID
FORMULA IONIZATION CONSTANT
Acetic
CH3COOH
1.8 x 10-5
Nitrous
HNO2
4.5 x 10-4
Hydrofluoric HF
7.2 x 10-4
Hypochlorous HClO
3.5 x 10-8
Hydrocyanic HCN
4.0 x 10-10
Can measure the pH of a solution using a pH meter. Can also use
indicators to
roughly establish the pH of a solution.
Indicator
Color in acidic range pH range
Color in basic range
Methyl violet
yellow
0-2
purple
Methyl orange pink
3.1-4.4
yellow
Litmus
red
4.7-8.2
blue
Phenol-phthalein colorless
8.3-10.0
red
Bronsted-Lowry Acids and Bases
A Bronsted acid is a proton donor
HCl + H2O ---> H3O+ + Cl-
HCl is a strong acid
A Bronsted base is a proton acceptor NH3 + H2O = NH4+ + OH- NH3 is a weak base
For the first acid-base reaction, HCl + H2O ---> H3O+
+ Cl- , HCl is the acid and H2O is the
Bronsted base because H2O accepted the proton (H+)
from HCl. H3O+ is called the conjugate acid
of the base, H2O, and Cl- is called the conjugate
base of the acid HCl.
(slide 3)
For the second acid-base reaction, NH3 + H2O
= NH4+ + OH-, NH3 is the
base and H2O is the
Bronsted acid because H2O donated the proton (H+)
to NH3. NH4+ is called the conjugate acid
of the base, NH3, and OH- is called the conjugate
base of the acid H2O.
Relative strengths of acids and bases.
(slide 4, 5, 6)
The strong acids are HCl,
H2SO4, and HNO3
Some weak acids are HF and
HC2H3O2.
Notice that HSO4- is
what is formed after the strong acid, H2SO4, loses
one H+. HSO4-is considered to be
a weak acid because its
dissociation is not 100% and a Ka is assigned to HSO4-,
Ka2 = 1.2 x 10-2. The
subscript 2 refers to the
second dissociation of the hydrogen ion. H2SO4 is
a diprotic acid. Another diprotic
acid is H2CO3
however
both forms, H2CO3 and HCO3-
, are weak acids with Ka1 and Ka2 equal to
4.3 x 10-7 and
5.6 x 10-11 respectively. Other polyprotic acids are
(slide 7)
The concentration of H+ in water after adding a strong acid
is equal to the concentration of the strong acid => it is 100%
ionized. Therefore in a 0.05M HNO3 solution ==>
H+ = 0.05 and pH = -log(0.05) = 1.30
Whereas in a 0.05M HC2H3O2 solution, one
needs to calcluate the H+ using the Ka for acetic
acid.
To do this I will offer a procedure that can be used to solve any equilibrium
type problem.
1. DESCRIBE (write down all the equilibria
needed) HC2H3O2
== H+ + C2H3O2-
.
H2O == H+ + OH-
2. DEFINE (write down all the definitions
for the K's) Ka = [H+][C2H3O2-]/[HC2H3O2]
Kw = [H+][OH-]
3. ASSIGN (this is the most challenging
step and depends upon how the problem is worded) In the above problem
you are told the initial [HC2H3O2] is
0.05. This is called the analytical concentration - the amount you
added before equilibration took place. When equilibration takes place some
of the acetic acid was "lost"
to form equal amounts of H+ and C2H3O2-.
Since you are not told [H+], you assign it an unknown value
or x. This unknown value or x can be solved for using the definition
above. Now you assign equilibrium
values. [H+] = x
[C2H3O2 -] = x
[HC2H3O2] = 0.05 - x
4. PLUG IN (substitute values and unknowns
into the definition)
Ka = [H+][C2H3O2-]/[HC2H3O2]
1.2 x 10-5 = (x)(x)/(0.05 - x)
5. SOLVE (solve for the unknowns)
In this case the unknown is x. It appears that you need to solve
a quadratic
equation. But in this case x will be very small compared to 0.05
so 0.05 - x is approximately equal
to 0.05.
x2/0.05 = 1.2 x 10-5and x = {(0.05)(1.2 x 10-5)1/2
= 7.75 x 10-4.
6. RELATE (now use this value of x to
determine the concentration of all the species)
[H+] = x = 7.75 x 10-4
[C2H3O2 -] = x = 7.75
x 10-4 [HC2H3O2]
= (0.05 - x) = 0.05
If you know [H+] you can determine pH, [OH-]
and pOH.
You can also determine the percent dissociation which is {7.75
x 10-4 /0.05}x100 = 1.55%
Identifying the three types of problems involving weak acids and weak bases
When given a problem involving a weak acid (weak base) it helps to identify this problem as one of three basic types of problems - weak acid only, weak acid + salt of weak acid (a buffer solution), and salt of weak acid only ( a hydrolysis problem).
If the problem is stated such that the species present
in relatively large amount is the weak acid only then this is a weak acid
type problem.
Examples - 1. What is the
pH of 0.10M formic acid solution? Ka = 1.8 x 10-4
2. What is the pH when 2.3 g of aspirin is dissolved in 250 mL of water?
MW = 180 Ka = 3 x 10-5.
If the problem is stated such that the species present
in relatively large amounts are the weak acid and the anion of the weak
acid then this is a buffer type problem.
Examples - 1. What is the
pH of a 0.10M formic acid and 0.20M sodium formate solution.
2. What is the pH of a solution made by adding 1.2 g of potassium formate
to 150 mL of 0.25M formic acid.
3. What is the pH of a solution made by adding 120 mL of 0.22M sodium hydroxide
solution to 230 mL of
0.20M formic acid?
4. How many grams of sodium acetate must be added to 150 mL of 0.25M
acetic acid to produce solution
with a pH of 5?
If the problem is stated such that the species present
in relatively large amounts is the anion of the weak acid only then this
problem is a "hydrolysis" type problem.
Examples - 1. What is the
pH of a 0.30M sodium acetate solution?
2. 200 mL of a 0.20M acetic acid solution has been neutralized with a 0.30M
sodium hydroxide solution.
This is the equivalence point of a weak acid-strong base titration. What
is the pH of this solution?