Application of Statistical and mathematical equations in Chemistry Part 5 Strong Acids and Bases Ph theory Weak Acids and Weak Bases Salts of Weak Acids and Bases theory A buffer solution theory POLYPROTIC ACID IONIZATION

1. Awad Nasser Albalwi‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
Application of Statistical and mathematical
equations in Chemistry
Part 5
Strong Acids and Bases
Ph theory
Weak Acids and Weak Bases
Salts of Weak Acids and Bases theory
A buffer solution theory
POLYPROTIC ACID IONIZATION

2. Awad Nasser Albalwi‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
Strong Acids and Bases
The animation here shows the formation of H3+O ions and OH- ions in an aqueous
system.
Acids and bases that are completely ionized when dissolved in water are called strong
acids and strong bases There are only a few strong acids and bases, and everyone
should know their names and properties. These acids are often used in industry and
everyday life.
The concentrations of acids and bases are often expressed in terms of pH, and as an
educated person, you should have the skill to convert concentrations into pH and
pOH. The pH is an indication of the hydrogen ion concentration, [H+].
Strong Acids
Strong acids are acids that are completely or nearly 100% ionized in
their solutions. Here are some common strong acids: Ionization of a
strong acid HA can be represented by:
HA = H+ + A-
x x
where x is the concentration of H+, [H+]. For a strong acid, [H+] = [A-]
= concentration of acid (= x), if x is much greater than 1x10-
7 (represented as e-7). For a very dilute strong acid solution with
concentration less than 1E-7, the pH is dominated by
the autoionization of water,
H2O = H+ + OH-, Kw = 1e-14 at 298 K.
The pH and pOH Scales
The pH scale is defined as the negative log of the concentration of H+:
pH = -log[H+]
The pOH scale is defined as the negative log of the concentration of OH-, [OH-]:
pOH = -log[OH-]
With this scale, calculating the pOH can be done in the same manner as the pH
scale.
Example 1
Calculate the pH of a solution with 1.2345E-4 M HCl
Solution
The solution of a strong acid is completely ionized. Thus, [H+] = 1.234e-4.
pH = -log(1.234e-4) = 3.909
Ph theory
Strong acids
Type Formula
Hydrogen
halides
HCl HBr HI
Oxyacids
of halogens,
HClO3
HClO4
HBrO3
HBrO4
HIO3
HIO4
Sulfuric acid H2SO4
Nitric acid HNO3
The pH and pOH
scale at 298 K
pH [H+] pOH
1 0.1 13
2 0.01 12
3 0.001 11
4 1e-4 10
5 1e-5 9
6 1e-6 8

3. Awad Nasser Albalwi‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
pH meter provides a value as to how acidic or alkaline a liquid is. The basic principle
of the pH meter is to measure the concentration of hydrogen ions. Acids dissolve in
water forming positively charged hydrogen ions (H+). The greater this concentration
of hydrogen ions, the stronger the acid is. Similarly alkali or bases dissolve in water
forming negatively charged hydrogen ions (OH-). The stronger a base is the higher the
concentration of negatively charged hydrogen ions there are. The amount of these
hydrogen ions present solution is dissolved in some amount of water determines the
pH
pH is defined as the negative logarithm of the H+ concentration.
pH = - log [H+]
A logarithm is an exponent. It is the exponent you have to put on the base number
(which is 10 for common logs) to get the original quantity.
pH = - log [H+], so
-pH = log [H+]
-pH is the exponent you have to put on the base 10 to get the [H+]
10^-pH = [H+]
Weak Acids and Weak Bases
Weak acids and bases don't dissociate completely. Equilibrium exists between a weak
acid, H3O+, water, and the weak acid's anion. For all weak acids the equilibrium lies
predominantly on the left, indicating that a small amount of H3O+ is produced. The
Ka for a weak acid is generally a number less than 1.
Dissociation of a Weak Acid
Here, you will find the H3O+ concentration and the pH of a weak acid when the initial
concentration is known.
Example weak acid problem: What is the H3O+ concentration and pH of a 0.10 M
solution of hypochlorous acid (HOCl)?
Hint: Ka = 3.5 x 10-8

4. Awad Nasser Albalwi‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
First notice that the value of Ka is small. This means that HOCl is a weak acid. Write
out the equilibrium involved:
HOCl + H2O = H3O+ + OCl-
Set Ka equal to the mass-action expression:
Create a table for initial conditions, the change in equilibrium and the final
equilibrium conditions. At the beginning, the concentration of HOCl is 0.10 M. The
concentrations of both H3O+ and OCl- are zero:
Note: water is not included in the calculation since it is the solvent.
For equilibrium to occur, some of the HOCl must dissociate to form H3O+ and OCl-.
Since we do not know how much will dissociate, we'll label the amount of HOCl lost as
-x and the amount of H3O+ and OCl- formed as +x:
Le Chatelier's principle can be used to predict the effect of a change in conditions on a
chemical equilibrium. Le Chatelier's Principle states: if a stress is applied to a system
in a state of dynamic equilibrium, the system will shift in a direction which opposes
the stress. In the problem above, the "stress" is the absence of H3O+ and OCl- which
will cause the equilibrium to shift to the right, relieving the stress by forming
H3O+ and OCl-.
Using a little bit of math, you can create statements to determine the amount of each
ingredient at equilibrium:

5. Awad Nasser Albalwi‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
The above quantities are going be used in the mass-action expression for the
equilibrium of the acid. Remember, the Ka for HOCl is 3.5 x 10-8.
The solution for x is simplified because the x in the statement 0.10 - x can be ignored.
It can be ignored because its value will be extremely small compared to the
concentration of 0.10 M. To determine if x can be ignored, compare the value of the
last decimal place in the acid concentration to the value of the equilibrium constant. A
difference greater than 100 may be ignored.
In the example problem, the concentration is 10-2 and the equilibrium constant is 10-8.
The difference is 106, therefore, x can be clearly ignored. Our equation is now:
Multiply both sides by 0.10:
x2 = 3.5 x 10-9
Take the square root of both sides:
x = [H3O+] = 5.9 x 10-5 = [OCl-] (also)
Take the negative log of H3O+ to find the pH:
pH = -log[H3O+] = -log(5.9 x 10-5) = 4.23
Additionally, the HOCl concentration at equilibrium will be:
[HOCl] = 0.10 - 5.9 x 10-5 = 0.10 M (using 2 significant figures)
Example weak base problem: What is the OH- concentration and pH of a 0.10 M
solution of NH3?
Hint: Kb = 1.8 x 10-5
Notice that the value of Kb is small. We are going to use the same method that we used
for weak acid solutions:

6. Awad Nasser Albalwi‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
We will assume that no NH4+ or OH- will have formed before equilibrium is reached.
A shift to the right is needed to establish equilibrium. The amount of NH3 that is lost(-
x), and the amounts of NH4+ and OH- formed (+x) are not known, we will again assign
the values of -x and +x.
Using a little bit of math, you can create statements to determine the amount of each
ingredient at equilibrium:
Again, we will be using these quantities in the mass-action expression:
Determine if the x in the denominator can be ignored. The last decimal place in 0.10 is
10-2. The value of the constant is 10-5. The difference is 103. Because the difference is
more than 100, the x may be ignored. Our math equation is:
Multiply both sides by 0.10:
x2 = 1.8 x 10-6
Take the square root of both sides:

7. Awad Nasser Albalwi‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
x = [NH4+] = [OH-] = 1.3 x 10-3
To determine the pH, find the pOH by taking the negative log of OH-:
pOH = -log[OH-] = -log(1.3 x 10-3) = 2.87
Now subtract the pOH from 14:
pH = 14.00 - pOH = 14.00 - 2.87 = 11.13
Salts of Weak Acids and Bases theory
A salt is formed when an acid and a base are mixed and the acid releases H+ ions while
the base releases OH- ions. This process is called hydrolysis. The pH of the salt
depends on the strengths of the original acids and bases:
Acid Base Salt pH
strong strong pH = 7
weak strong pH > 7
strong weak pH < 7
weak weak depends on which is stronger
These salts are acidic or basic due to their acidic or basic ions. When weak acids or
weak bases react with water, they make strong conjugate bases or conjugate acids,
respectively, which determines the pH of the salt.
Here is the chemical reaction (net ionic) for the hydrolysis of NaAc:
Ac¯ + H2O <==> HAc + OH¯
2) Here is the Kb expression for Ac¯:
[HAc] [OH¯]
Kb = ----------------
[Ac¯]
Formula
[OH-] = (Kw * Ca / Ka)1/2
[H+] = (Kw * Cb / Kb)1/2
A buffer solution theory

8. Awad Nasser Albalwi‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
A buffer (more precisely, pH buffer or hydrogen ion buffer) is an aqueous
solution consisting of a mixture of a weak acid and its conjugate base. ItspH changes
very little when a small amount of strong acid or base is added to it and thus it is used
to prevent changes in the pH of a solution. Buffer solutions are used as a means of
keeping pH at a nearly constant value in a wide variety of chemical applications.
Formula
pH = pKa + Log ([base]/[acid])
pOH = pKb + Log ([acid]/[base])
Polyprotic Acids and its Salts
Polyprotic acids contain more than one mole ionizable hydronium ions per mole of
acids. They ionize to give more than one H+ ions per molecule. Possible forms of three
polyprotic acids are given below after their dissociation into H+ ions.
H2S, HS-, S2-
H2SO4, HSO4-, SO42-
H3PO4, H2PO4-, HPO42-, PO43-
These acids ionize in several stages, giving out one proton at each stage. The acidity
constants for these acids may be written as K1, K2, K3, ...
Consider H2S,
H2S = H+ + HS-
[H+] [HS-]
K1 = ----------
[H2S]
and
HS- = H+ + S2-
[H+] [S2-]
K2 = -----------
[HS-]
Obviously, for the overall ionization reaction,
H2S = 2 H+ + S2-
[H+]2 [S2-]
Koverall = -----------
[H2S]
= K1 K2

9. Awad Nasser Albalwi‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
Confirm the above obvious result on a sheet of paper to satisfy yourself.
For polyprotic acids, the following is always true:
K1 > K2 > K3 > ...
For most acids, K1/K2 = 1E5 or 100000, and K2/K3 = 1E5, but oxalic acid is different.
For oxalic acid, K1 = 5.6E-2, and K2 = 5.4E-4. The two acidic groups are separated by a
C-C bond in oxalic acid.
Examples 1
Calculate the overall equilibrium constant for oxalic acid
H2C2O4 = 2 H+ + C2O42-.
K1 = 5.6E-2
K2 = 5.4E-5
Solution
The calculation is straight forward:
Koverall = K1 K2
= 3.0E-6
Fractions of Dissociating Species
Alpha Fractions
Introduction
An fraction is the ratio of the equilibrium concentration of one specific form of a
solute divided by the total concentration of all forms of that solute in an equilibrium
mixture. It is thus a number between 0 and 1.
For the example of phosphoric acid, H3PO4:
H3PO4 (aq) H+(aq) + H2PO4-(aq) H+(aq) + HPO42-(aq) H+(aq) + PO43-(aq)
pKa1 = 2.15
pKa2 = 7.20
pKa3 = 12.38
Ctotal = [H3PO4] + [H2PO4-] + [HPO42-] + [PO43-]
0 = [H3PO4] / Ctotal
1 = [H2PO4-] / Ctotal
2 = [HPO42-] / Ctotal
3 = [PO43-] / Ctotal

10. Awad Nasser Albalwi‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬‫ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬
Alpha fractions for a triprotic acid are calculated from the following equation:
[H+]3
0 = --------------------------------------
[H+]3 + Ka1[H+]2 + Ka1Ka2[H+] + Ka1Ka2Ka3
Ka1[H+]2
1 = --------------------------------------
[H+]3 + Ka1[H+]2 + Ka1Ka2[H+] + Ka1Ka2Ka3
Ka1Ka2[H+]
2 = --------------------------------------
[H+]3 + Ka1[H+]2 + Ka1Ka2[H+] + Ka1Ka2Ka3
Ka1Ka2Ka3
3 = --------------------------------------
[H+]3 + Ka1[H+]2 + Ka1Ka2[H+] + Ka1Ka2Ka3
Note that the denominators are the same in all of these equations. Plots of these four
equations are shown below for phosphoric acid. Figure 6.2 on page 186 of Rubinson &
Rubinson shows this plot on a log scale, which better shows the low concentrations.
See Figure 6.2 on page 186 of Rubinson & Rubinson to see this plot on a log scale.
Note that when the pH equals one of the pKa values, the acid to conjugate base ratio
equals 1.
Why would we want to know the fraction of one particular species as a function of
pH?
Example: adenosine diphosphate (ADP) adenosine triphosphate (ATP)
ADP(aq) + HPO42-(aq) + 2 H+(aq) ATP(aq) + H2O

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This reaction has a positive G of 52.4 kJ/mol. Producing ATP is the body's way of
storing energy, e.g., doughnuts ATP. When your body needs energy for
important activities like thinking, ATP converts back to ADP to supply the energy.
Intracellular pH varies from 6.1 in muscle cells to approximately 7 in most other cells.
Looking at the phosphoric acid alpha plot, you can see that 2 varies from 0.07 to 0.4
over this range.
For the body to store energy it needs a reliable supply of HPO42-, which is one of many
reasons why buffer systems are so important in biology. Large changes in pH would
convert HPO42- to H2PO4- or PO43- and you would not be able to convert the chemical
energy of doughnuts to a usable form.
ase])
POLYPROTIC ACID IONIZATION
a class of salts formed by the partial neutralization of diprotic or polyprotic acids
A. Polyprotic acids are acids with two or more acidic hydrogens.
monoprotic: HC2H3O2, HCN, HNO2, HNO3
diprotic: H2SO4, H2SO3, H2S
polyprotic
triprotic: H3PO4, H3BO3
B. Polyprotic acids ionize in steps - one H+ is removed at a time - with a different Ka
for each step.