Transcript Analytical Chemistry Acid-Base Arrhenius Theory: H+ and OH This theory states that an acid is any substance that ionizes (partially or completely) in water.

Analytical Chemistry
Acid-Base
Arrhenius Theory: H+ and OH
This theory states that an acid is any substance that ionizes (partially or
completely) in water to give hydrogen ions (which associate the solvent to give
hydronium ions, H3O+):
HA + H2O

H3O+ + A-
A base ionizes in water to give –hydroxyl ions. Weak (partially ionized) bases
generally ionize as follows:
B + H2O
BH+ + OH-
Bronsted-Lowry Theory:
Taking and Giving Protons

1.
2.
3.
This theory states that an acid is any substance that can donate a proton,
and a base is any substance that can accept a proton. Thus, we can write a
“half-reaction”:
acid= H+ + base
The acid and base of a half-reaction are called conjugate pairs. Free
protons do not exist in solution.
There must be a proton acceptor (base) before a proton donor (acid) will
release its proton.
HA
H+ +
A(acid)
(proton)
(conjugate base)
B
+
H+
BH+
(base) (proton)
(conjugate acid)
There must be a combination of two half-reactions.
HA +
B
BH+ +
A(acid 1) (base 1) (acid 2) (base 2)
Note that (acid 2), (base 2) are weaker than (acid 1), (base 1).
Lewis Theory: Taking and Giving Electrons

In the Lewis theory, an acid is a substance that can accept an electron
pair, and a base is a substance that can donate an electron pair.
Acid- Base Equilibria in Water
1.
Acid-Base Ionization in Aqueous Solution:
H2O + HCl
H3+O + ClH2O + NH3
OH- + NH4+
Pure water ionizes slightly to give hydronium and hydroxide ions
2H2O
H3+O + OHFor simplifying purposes the equation above is written in this way:
H2O
H+ + OHConstant equilibrium is written:
keq = [H+] [OH-] = 1.82 x 10-16
[H2O]
Acid- Base Equilibria in Water
2.
Since the amount of dissociated water is very little, it is assumable that its
concentration is stable and equal to the concentration of water (1g/ ml)
(1000g/1) /18 = 55.56 H2O
keq [H2O] = 1.82 x 10-16 x 55.56 = kw = [H+] [OH-]
kw = [H+] [OH-] = 10-14
where kw is the thermodynamic autoprotolysis of water or self ionization of
water. Accordingly, we find that the hydrogen ion concentration is equal to
the hydroxide ion concentration in water:
[H+] = [OH-] = kw = 10-14 = 10-7M
Acid- Base Equilibria in Water
3.
Water also has another characteristic, which is the ability to equalize different
strengths of strong acids such as, HCl and HNO3. That is due to its reaction
with these acids that produces H3+O (acid 2), which is weaker than the
original acid.
HCl + H2O
H3O+ + Cl(acid 1) (base 1)
(acid 2)
Because they all transform to hydronium ion acid, they are all equal in
strength. This transformation is known as the leveling effect.
The same situation applies to strong bases, where they all transform to the
base [OH-].
By that, all strong bases appear equally strong in water.
Acid- Base Equilibria in Water


When [H+] = [OH-], then a solution is said to be neutral. If [H+]
[OH-] then the
solution is acidic. And if [H+]
[OH-], the solution is alkaline. The hydrogen ion

hydroxyl ion concentrations in pure water at 25 C are each 10-7 M, and the pH
of water is 7. A pH of 7 is therefore neutral.
The scientist Sorensen invented what is known as pH, which is equal to
negative logarithm for [H] for the specific solution.
pH = -log [H+]
Because [H+] and [OH-] are related to each other by the kw equation, knowing
one of them can determine the other. So, it has been customary that pH is used
to determine the acid-base surroundings.
kw = [H+] [OH-] = 1 x 10-14
by taking the logarithm for both ends:
log kw = log [H+] + log [OH-]
when multiplying by (-):
-log kw = log [H+] + (-log [OH-]) = -log 1 x 10-14
since log is known as p, we find that:
pkw pH + pOH = 14
Acid- Base Equilibria in Water



This equation strongly bonds together pH, pOH. That is why the neutral
aqueous solution’s pH number is:
pH = log 10-7 = pOH
if the surrounding is an acid
pH 7
If the surrounding is a base
pH 7
pH scale
1. pH Calculation for Aqueous Solution
A.
Strong Acid and Base Solutions:
Strong acids and bases are ionized completely in water, for e.g.,
HCl
H+ + ClCa
O
O
Ca
O
Ca
before ionization
after ionization
pH = -log [H+] = log Ca
Ca = HCl concentration
Dissolving of NaOH:
NaOH
Na+ + OHCb
Cb
O
Cb
O
Cb
pOH = -log [OH-] = log Cb
Cb = NaOH concentration
before ionization
after ionization
Strong Acid and Base
Solutions:
1. pH Calculation for Aqueous Solution
B.
Weak Acids and Bases:
Weak acids and bases are ionized incompletely in water:
Example 1:
CH3 COOH
CH3COO- + H+
Ca
Ca- 
O

O

before ionization
after ionization
1. pH Calculation for Aqueous Solution
Example 2: Acid:
HA
H+ +
A-
Ka = [H+] [A-] = [H+] [A-] . . . . . . . . . . . 
[HA]
Ca- [H+]
Ka = weak acid contuent dissociation
[ ] = Molar concentration
Ca= acid initial concentration before dissociation
[HA]= remaining acid concentration
at equilibrium  = [A-] = [H+]
Ka =
[H+]
Ca – [H+]
If we assume that the acid is very weak and Ka very small, then we are neglecting [H+]
compared to Ca, we obtain:
Ka = [H+]2
Ca
[H+] = Ka Ca
pH = -log Ka Ca
1. pH Calculation for Aqueous Solution
Example 3: Base:
The weak base solution is treated the same way.
B + H2O
[BH+] [OH-]
kb = [BH+] [OH-] = [BH+] [OH-]
[B]
Cb- [OH-]
Cb = weak base concentration before dissociation
Cb- [OH] = weak base concentration after dissociation at equilibrium
If we assume that the base is very weak and kb very small, then we neglect [OH-]
concentration compare to Cb
kb = [OH-]2
Cb
 = [OH-] = [BH+] at equilibrium
[OH-] = Kb Cb
pOH = - log Kb Cb
kb = weak base contuent
Cb= initial base concentration
[B]= remaining weak base after dissociation
1. pH Calculation for Aqueous Solution

Ka and Kb equal the equilibrium constituent Keq for the dissociation reaction multiplied by
the water concentration (55.56 molar) considering water concentration is constant as long
as acid and base solutions in water are diluted.

Now if we look at the conjugate base A- and the conjugate acid BH+ we will find that both
of them have good basic and acidic qualities based on Bronsted’ s defined of acid and
base. For instance, the base A- reacts with water:
A- + H2O
HA + OHkb- = [HA] [OH-]
. . . . . .. . . . . 
[A-]
and the conjugate acid BH+ reacts with water:
BH+
B + H+
ka- = [B] [H+]
[BH+]
Kb- = conjugate base constant dissociation
Ka- = conjugate acid constant dissociation
1. pH Calculation for Aqueous Solution
And they are concluded from values of ka and kb of the conjugate acid and base, by
multiplying  by  :
ka x kb- = [H+] [A-]
[A-]
x
[HA] [OH-]
[HA]
ka x kb- = [H+] [OH-] = kw
kb- = kw
ka
by using the same method we can conclude that:
ka- = kw
kb
2. Blood pH


The pH of blood at body temperature (37oC) is 7.35 to 7.45. This value represents a
slightly more alkaline solution relative to neutral water than the same value would be at
room temperature.
Since a neutral blood solution at 37oC would have pH 6.8, a blood pH of 7.4 is more
alkaline at 37oC by 0.2 ph units than it would be at 25oC.
Buffer Solutions

A buffer solution consists of a maximum of a week acid and its conjugate base of a
weak base and its conjugate acid at predetermined concentrations or ratios. Which resists
change in pH when a small amount of an acid or base is added or when the solution is
diluted.
A. Buffer Capacity:
The buffer capacity describe how much the buffer solution is able to resist the change in
the pH. And it is determined quantitatively by the amount of moles of the strong base or
acid needed to be added to one liter of buffer solution in order to increase or decrease the
pH –respectively- of the buffer solution by one unit. And if it is at its maximum when the
ratio of the weak acid concentration and its conjugate base is unity.
B.
The Buffering Mechanism:
If we assume that buffer solution consists of a maximum of acetic acid and sodium
acetate, the following reactions will occure:
Bufferd solution
Buffer Solutions
-
-
-

Weak acid ionization:
CH3COOH
CH3COO- + H+
Complete salt ionization:
CH3COONa
Na+ + CH3COOVery small water dissociation:
H2O
H+ + CH3COOHere, we can observe that if an amount of hydrochloride acid was added to this mixture,
then the mixture will suffer a complete ionization to chloride and hydrogen ions. Which
would have caused a sharp decrease in the pH if the mixture above did not exist.
Since the mixture does exist, the hydrogen on will react with the acetate (which we
mentioned as a weak base), and acetate acid will be produced:
H+ + CH3COOCH3COOH
this means we transformed the strong acid (hydrochloric) into weak acid.
Buffer Solutions

On the other hand, if we added strong base such as hydroxide sodium, the produced
hydroxide ions will increase the pH highly if the buffer solution did not exist. But, here it
reacts with the hydrogen ions to produce neutral water:
H+ + OHH2O

The same mechanism is observable when using an alkaline buffer solution, such as
ammonia and ammonium chloride:
NH3 + H2O
NH4+ + OHNH4Cl
NH4+ + Cl-

By adding HCl to this solution, the H+ from this acid will react with [OH-] and will produce
water H2O. Meaning that, the acid HCl has been transformed to water by the buffer
solution. While, when adding NaOH to this buffer solution the following reaction will occur:
NaOH + NH4+
NH3 + Na+ + H2O
meaning, that strong base transformed to a weak base.
3. Calculation for the pH the Buffer
Solution
1. The pH for a weak acid and its salt:
For example, acetic acid with sodium
acetates:
CH3COOH
H+ + CH3COOka = [H+] [CH3COO-]
[CH3COOH]
If we assume that the molar salt
concentration is Cs and that molar acid
concentration is Ca, we will find that:
Ka = [H+] Cs
Ca
ka  [H+] Cs
Ca
[H+] = ka Ca
Cs
pH = pka + log Cs
Ca
2. The pH for a weak base and its salt:
It is calculated the same way
[OH-] = kb Cb
Cs
pOH = pkb + log Cs
Cb
Physiological Buffers



The pH of the blood in a healthy individual remains remarkably constant at 7.35 to 7.45.
This is because the blood contains a number of buffers that protect against pH change
due to the presence of acidic or basic metabolites.
The buffering capacity of blood for handling CO2 is estimated to be distributed among
various buffer systems as follows: hemoglobin and ox hemoglobin , 62%; H2PO4-, 22%;
plasma protein, 11%; bicarbonate, 5%; proteins contain carboxylic and amino groups;
which are weak acids and bases respectively.
Certain diseases cause disturbances in the acid balance of the body. For e.g.; diabetes
may give rise to “acidosis” which can be fatal.
An important diagnostic analysis in the CO2/HCO3- balance in blood. This ratio is related
to the pH of the blood by the Henderson_Hasselbalch equation:
pH = 6.10 + log [HCO3-]
[H2CO3]
Where H2CO3 can be considered to the concentration of dissolved CO2 in the blood; 6.10
is pka of carbonic acid in blood at body at body temperature (37oC). Normally, the
bicarbonate dioxide (CO2) is 1.3 mmal/ L.Accoringly for the blood:
pH = 6.10 + log 26 mmal/ L = 7.40
1.3 mmal/ L
Physiological Buffers

The HCO3- \H2CO3 buffer system is the most important one in buffering blood in the lung
(alveolar blood). As oxygen from inhaled air combines with hemoglobin. The oxygenated
hemoglobin ionizes, releasing a proton. This excess acid is removed by reacting with
HCO3-:
H+ + HCO3H2CO3

But note that the [HCO3- \H2CO3 ] ratio at pH 6.4 is 26/ 1.3 = 20: 1 this is not a very
effective buffering ratio; and significant amounts of HCO3- are converted to H2CO3,,
The pH would have to decrease to maintain the new ratio. But fortunately, the H2CO3,,
Produced is rapidly decomposed to CO2 and H2O by the enzyme decarboxylase , and the
CO2 exhaled by the lungs. Hence, the ratio of HCO3- \H2CO3 remains constant at 20: 1