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Stoichiometry:
• The study of relationship between the relative quantities of substances taking part in a
reaction or forming a compound.
• Stoichiometry provides the basis for a procedure called titration, which is used to
determine the concentrations of acidic and basic solutions.
• For example, in the reaction of sodium hydroxide and hydrogen chloride, 1 mole of
NaOH neutralizes 1 mole of HCl:
NaOH (aq) + HCl (aq)  NaCl (aq) + H2O (l)
IONIC EQUILIBRIUM:
Acid, base and salt when dissolved in water, gets ionized and form corresponding ion to
some extent. Hence, equilibrium is between ion and unionized atom . Such equilibrium
between ion and unionized atom in an aqueous solution is called as ionic equilibrium.
H⁺ + OH⁻ H₂O
CHEMICAL EQUILIBRIUM:
Chemical equilibrium is a state of a chemical reaction at which the rate of a forward
reaction is equal to the rate of the backward reaction.
N₂(g) + 2H₂ (g) 2 NH₃(g)
LE-CHATELIER’S PRINCIPLE
In 1888, Le-chatelier, gave a qualitative rule, which helps chemists in the prediction of the
effect of change of temperature, pressure, and concentration on equilibrium reactions. This
rules can be stated as follows:
a) Effect of concentration:
• In a general reaction at equilibrium, if the concentration of one of the reactants (or both
the reactants) is increased, then the equilibrium will shift towards the products.
• Similarly, if the concentration of one of the reactants (or both the reactants) is
decreased, then the equilibrium will shift towards the reactants.
• Thus, in general forward reaction is favored by higher concentration of reactants, and the
backward reaction is favored by lowering the concentration of the reactants.
b) Effect of pressure:
• Change of pressure is considered only for gaseous reactions at the state of equilibrium.
In a gaseous reaction, at constant temperature and volume, the pressure applied is
directly proportional to the total number of moles present in the reaction mixture.
• Hence, if pressure is increased, the equilibrium is shifted to a direction in which the
number of moles decrease (the volume decreases).
• Similarly, if the pressure is decreased in which the number of moles increase (the
volume increases).
• Thus, in general at constant temperature, increase in pressure will favor a reaction,
which is accompanied by a decrease in volume, and decrease in pressure will favor a
reaction, which is accompanied by an increase in volume.
c) Effect of temperature:
If a chemical reaction at the state of equilibrium is exothermic, then lowering of
temperature will favor the reaction (the equilibrium shifts to the right). Similarly, if a
chemical reaction at a state of equilibrium is endothermic, the increase in temperature will
favor the reaction (the equilibrium shifts to the right).
THEORIES OF ACID AND BASES
There are three theories of acid and base by which one can determine nature of unknown ,
substance either acid and base.
A) ARRHENIUS THEORY OF ACIDS AND BASES:
• Arrhenius, in 1887, suggested the theory of electrolyte dissociated of an aqueous medium
and propose his theory to define acids and bases.
• Thus, according to Arrhenius theory, an acid may be defined as a hydrogen containing
substance , that dissociates to give hydrogen ions (H⁺) When dissolved in water.
• For example:
HCL + Water H⁺(aq) + Cl⁻(aq)
According to Arrhenius theory, a base may be defined as hydroxyl substance, which
dissociates to give hydroxyl ions (OH⁻ ions) when dissolved in water.
• For example:
NaOH + Water Na⁺(aq) + OH⁻(aq)
Neutralisation:
• It may be defined as a process, in which acid and base react completely with each other to
produce salt and water.
Acid + Base Salt + Water
For example:
HCl + NaOH NaCl(aq) + H₂O(aq)
• On the basis of ionic theory, neutralization may be defined as a process in which H⁺ ions of
an acid combine with OH⁻ ions of a base to form unionized water molecules.
HCl + NaOH NaCl + H₂O
(acid) (base) (salt) (water)
Limitations of Arrhenius theory:
• This concept is applicable to define acids and bases, only in aqueous solutions, but not for non- aqueous solutions or
gaseous reactions.
• Some compounds don’t contain hydroxyl group (OH⁻)and are basic in nature. The basic nature of such compounds isn’t
explained by Arrhenius theory of acids and bases.
For example- NH₃, amines.
• According to this theory, H⁺ ions exists freely in an aqueous
solution. But, in an aqueous solution, H⁺ ion is always hydrated to form hydronium ion.
H⁺ + H₂O H₃O ⁺(aq)
• According to this theory, an acid must contain hydrogen ion(H⁺ ion) and a base must contain a hydroxyl ion(OH⁻). But
there are substances whose aqueous solutions are acidic or basic in nature without containing H⁺ or OH⁻ ions.
For example FeCl ₃, CuSO₄ , CO₂. Show acidity in the aqueous solution and NaNO₃ , NH₃. Shows basic nature in the aqueous
solution.
FeCl₃ + 3H₂O Fe(OH)₃ + 3H₂O
B) LOWRY AND BRONSTED CONCEPT OF ACIDS AND BASES:
In 1923, Bronsted and Lowry, modified the definition of acids and bases which deals with
protons exchange and it applies to aqueous as well as non-aqueous solutions.
According to this theory, an acid may be defined as substance (molecule or ion) which
donates protons protons(H⁺ ions).
For example:
HCl + H₂O H₃O⁺ + Cl⁻
(Acid) (Base) (Acid) (Base)
Since HCl donate proton, they are Lowry-Bronsted acids.
According to this theory, a base may be defined as substance which accepts protons(H⁺
ions).
For example:
NH₃ + H₂O NH₄ + OH⁻
(Base) (Acid) (Acid) (Base)
since, NH₃ accepts proton from water, it is a Lowry-Bronsted base.
Conjugate acid and base pairs:
The pairs of substances which can be formed from one another by gain or by loss of proton,
are known as conjugate acid base pairs.
HCN + H₂O H₃O⁺ + CN⁻
(Acid) (Base) (Acid) (Base)
Explanation: consider the following reaction:
HCl + H₂O H₃O⁺ + Cl⁻
(Acid) (Base) (Acid) (Base)
In this example , HCl donates a proton and acts as an acid. Water accepts a proton and so
acts as a base, it is a reverse reaction, hydronium ion donates a proton and so acts as an
acid. The chloride ion accepts a proton and thus behaves as a base. Thus, hydrochloric acid
(HCl) is a conjugate acid of the base chloride ion(Cl⁻) and chloride ion is a conjugate base of
hydronium ion(H₃O⁺) and hydronium ion is a conjugate acid of water.
Some examples of conjugate acid-base pairs are given below:
ACID₁ BASE₂ ACID₂ BASE₁
HCl H₂O H₃O⁺ Cl⁻
HNO₃ H₂O H₃O⁺ NO₃⁻
HSO₄⁻ H₂O H₃O⁺ HSO₄⁻
C) LEWIS CONCEPT OF ACIDS AND BASES:
In 1938, G.N. LEWIS gave a more broad based definition of acids and bases, which don’t depend
upon the presence of protons. According to Lewis, an acid may be defined as a substance which
can accept a pair of electrons.
For example: H⁺ ion, HCl, etc. are Lewis acids
According to Lewis, a base is defined as a substance which can donate a pair of electron to form
a co-ordinate covalent bond.
For example: NH₃, H₂O, etc are Lewis bases.
Explanation : In the following reaction the proton (H⁺) is a Lewis acid and ammonia(:NH₃) is a
Lewis base, because a lone pair of electrons on nitrogen atom of ammonia is donated to the
proton.
Degree of dissociation
The extent of dissociation of an electrolyte can be expressed in terms of the “degree of
dissociation”
Degree of dissociation is the fraction of the total number of moles of an acid or a base or an
electrolyte that split up into its ions in an aqueous solution
It is represented by α.
α = dissociation at equilibrium
total no. of moles present in solution initially
SOME COMMON TERMS
Strength of acids and bases:
STRENGTH OF ACIDS:
The strength of an acid or a base can be expressed in terms of their dissociation constants.
Consider the dissociation of a weak acid HA in water. The equilibrium reaction will be:
HA + H2O H3O+ + A-
Applying the law of mass action, the equilibrium constant K can be given as:
K=[H3O+][A-]
[HA][H2O]
Since water is present in large excess its concentration will be practically constant.
K= [H3O+][A-]
[HA] X Constant
K X constant = [H3O][A-]
[HA]
Here Ka is the constant for a given acid and is called as acidity constant or Dissociation constant
of that acid.
Thus, stronger acids have high acidity constants and weak acids have low values.
STRENGTH OF BASES :
Consider the dissociation of a weak base BOH like ammonium hydroxide in water. The
equilibrium reaction can be written as:
BOH B+ + OH-
By applying law of mass action to above equation, the equilibrium constant K is given as-
Ka = [H+][A-]
[HA]
K=[B+][OH-] or Kb=[B+][OH-]
[BOH] [BOH]
Here Kb is a constant for a given base and is called basicity constant that determines the
strength of the base.
Thus strong base have higher Kb value and weak bases have lower Kb value.
Examples of dissociation constants of some acids and bases:
Acids Formula Ka
Acetic acid CH3COOH 1.75 x 10 -5
Formic acid HCOOH 1.77 x 10 -2
Phosphoric acid H3PO4 7.55 x 10-2
Sulphric acid H2SO4 > 1
Base Formula Kb
Ammonium hydroxide NH4OH 1.81 x 10-5
Sodium hydroxide NaOH > 1
IONIZATION OF WATER
Water is a weak electrolyte and hence it ionizes to a very small extent. It ionizes into
hydronium ions and hydroxyl ions as follows:
H2O H+ + OH- and H2O + H+ H3O+
The net ionization: 2H2O H3O+ + OH- or H2O H+ + OH-
Ionic Product:
Ionic product of water maybe defined as the product of molar concentrations of [H+] and
[OH-] ions in pure water at a given temperature. It is represented as Kw.
Explanation:
The ionization of water can be written as:
2H2O H3O+ + OH-
Applying law of mass action we get K=[H3O+][OH-]
[H2O]2
Water is present in large excess and it dissociated to a very small extent, hence its
concentration will remain practically constant.
K = [H3O+][OH-]
[constant]
K x constant = [H3O+][OH-]
or Kw= [H3O+][OH-]
Kw a constant known as Ionic product of water. For convinence [H3O+] is written as [H+].
Hence,
Kw is a function of temperature and at 298K its value is 1.00 x 10-14
Since one molecule of water ionizes into one H+ ion and one OH- ion,
[ H+ ] = [OH- ] in pure water.
And Kw= [H+][H+] =[ H+]2 ----------- (a)
Therefore, [H+] = Kw
Similarly, [OH-] = Kw ----------by substituting [OH-] in place of [H+]
Kw= [H+][OH+]
And hence [H+] = 1 x 10−14 = 1 x 10-7 mol/dm3
[OH- ]= 1 x 10−14 = 1 x 10-7 mol/dm3
pH and pOH
pH: It maybe defined as the negative logarithm to the base 10 of molar concentration of H+
pH = -log10 [H+ ] or pH = log10
1
[H+ ]
pOH: It maybe defined as the negative logarithm to the base 10 of molar concentration of
OH-
pOH = - log10 [OH- ] or pH = log10
1
[OH+ ]
To prove pH + pOH = 14
The ionic product of water can be written as:
Kw= [H3O+][OH-] -------- (a)
But Kw = 1 x 10-14
Substituting the value of Kw in equation (a)
1 x 10-14 = [H3O+][OH-] -------- Here [H3O+] is taken as [H+] hence
1 x 10-14 = [H+] [OH-]--------- (b)
Taking logarithm on both sides of equation (b)
log [H+] + log[OH- ] = log(1 x 10-14 )
Reversing the sign to above equation:
-log [H+] -log[OH- ] = - log(1 x 10-14 )
-log [H+] -log[OH- ] = - log(1) - log (10-14 ) by rule log (a x b) = log a + log b
and - log (a x b ) = - log a – log b
But log 1 = 0, substituting the value in below eq:
-log [H+] -log[OH- ] = - log(1) - log (10-14 )
-log [H+] -log[OH- ] = - log (10-14 )
By definition : -log [H+] = pH and -log[OH- ] = pOH hence substituting values in above
eq.
pH + pOH = - log10-14
pH + pOH = -(-14log10) (But log 10 = 1)
Hence pH + pOH = -(-14 x 1 )
Therefore,
pH + pOH = 14
pH Scale
Sorensen introduced a measure called pH scale to measure the acidity, alkalinity or neutal
nature of an aqueous solution.
For pure water or neutral solution
[H+] = 1x 10-7 (From the value of ionic procuct of water. Ref Kw)
Taking log of above equation we get-
pH=7
Hence in an aqueous in which [H+] > 1 x 10-7 then pH<7 i.e. the solution is acidic.
If in an aqueous solution [OH- ] > 1 x 10-7 then pH>7 i.e. the solution is basic/ alkaline.
pH scale derivation :
Acidic solution Basic Solution
[H+] mol/dm3 pH [OH+] mol/dm3 pH pOH = 14 - pH
1 x 10 -6 6 1 x 10 -6 6 8
1 x 10 -5 5 1 x 10 -5 5 9
1 x 10 -4 4 1 x 10 -4 4 10
1 x 10 -3 3 1 x 10 -3 3 11
1 x 10 -2 2 1 x 10 -2 2 12
1 x 10 -1 1 1 x 10 -1 1 13
TITRATION
• Titration is a method for determining the concentration of a solution by reacting a
known volume of unknown solution with a solution of known concentration.
• To find the concentration of an acid solution- titrate the acid solution with a
solution of a base of known concentration.
• To find concentration of unknown base – titrate with acid of known
concentration.
• Burette – Solution of known concentration
• Flask - unknown concentration solution
• Indicator – used to determine end point
Aqueous Acid-Base titrations/ Neutralization titrations
• These are titrations in which acid is titrated/neutralized with a base or vice versa .
The end product is salt and water.
• It helps to determine the concentration of either acid or a base.
• Equivalence point: The equivalence of an acid-base titration is the point at which
there are equal amounts (in moles) of H3O+ and OH- in titration flask.
• End point of titration – the point in a titration at which the indicator changes color.
• The indicator should change color sharply at the equivalence point.
• At the end point of the titration,
- all the acid has been neutralized by the alkali
- the solution in the conical flask contain salt and water only.
H3O+ + OH- 2H2O
Equivalence and end Points
Equivalence point
Related to the amount (moles)
End point
Related to the physical sign that is associated with the
condition of chemical equivalence.
The end point is expressed in range, the range should address
to the equivalence point.
In the acid base titrations, pH meter measures the pH of the acid solution in the
beaker/flask, as a solution of a base with a known concentration is added from the
burette.
As the titration progresses, the pH meter is used to monitor change in pH.
1) A measured volume of an acidic or basic solution of unknown concentration is placed in
a beaker/conical flask.(Solution of unknown concentration is called as analyte or titrand.)
2) The electrodes of a pH meter are immersed in this solution, and the initial pH of the
solution is read and recorded.
3) A burette is filled with the titrating solution of known concentration. This is called the
standard solution or titrant.
4) Measured volumes of the standard solution are added slowly and mixed into the solution
in the beaker.
5) The pH is read and recorded after each addition. This process continues until the reaction
reaches the equivalence point, which is the point at which moles of H+ ion from the acid
equal moles of OH- ion from the base. (H+ = OH-)
Titration Procedure
Types of acid base titration curves/Neutralization curves
Types of acid-base titrations Examples
Strong acid and Strong Base
Titration
HCl and NaOH
Strong acid and weak base
titration
HCl and NH3
Weak acid and strong base
titration
CH3COOH and NaOH
Weak acid and weak base
titration
CH3COOH and NH3
Titration Curve
• A titration curve is a plot of pH vs. the amount of
titrant added.
• Typically the titrant is strong (completely)
dissociated acid or base.
• Such curves are useful for :
1. Determining endpoints
2. Dissociation constants of weak acids or bases.
STRONG ACID STRONG BASE TITRATIONS
• Strong acid (HCL) is titrated with Strong Base (NaOH)
• Burette: NaOH (Known conc), Flask/Beaker = HCL Unknown conc)
• Volume of NaOH added to unknown HCl is plotted on X axis
• pH changed is plotted on Y axis
• Graph of pH curve shows the change in pH versus volume of titrant added as the
titration proceeds.
Key points
1. pH increases slowly
2. pH changes quickly
near the equivalence
point.
3. The equivalence point
of a strong acid—
strong base titration =
7.00NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq)
ie H+ = OH-
The titration of a strong base with a strong acid is almost identical.
Burette = HCL known conc
Flask = NaOH unknown
1.Initially the pH measured is low, which reflects high amount of
[H3O+]/[H+] of the strong acid.
2.pH increases gradually as acid is neutralized by the added base.
3.Suddenly the pH rises steeply. This occurs in the immediate vicinity of
the equivalence point. For this type of titration the pH is 7.0 at the
equivalence point.
4.Beyond this steep portion, the pH increases slowly as more base is
added.
Features of the Strong Acid-Strong Base Titration Curve
Strong Acid-Strong Base Titration Curve
This type of titrations can be explained in three parts
1. Before equivalence point
2. At equivalence point
3. After equivalence point
Molarity (M) =
𝑁𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑙𝑖𝑡𝑟𝑒 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 (𝑉)
Moles of titrant = (V titrant)(M titrant
Moles of analyte = (V analyte)(M analyte)
moles of titrant = moles of analyte
(V titrant)(M titrant) = (V analyte)(M analyte)
Consider titration of 100 ml of 1N HCL with 1 N NaOH. Change in pH value can be determined in 3 steps
as:
1. pH value before equivalence:
a) Before proceeding the titration, solution in flask is strong acid, ie 1M/1N HCL
Concentration of H+ in strong acid can be calculated as-
H+ = 1N
pH = - log (H+)
pH = - log (1) since log (1) = 0
So pH = 0
b) After adding 1 ml of base, total volume will be 100 ml of acid+ 1 ml of base = 101 ml
[H+] =
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻
+
𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
[H+] =
[𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻
+
− 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑂𝐻 − ]
𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
[H+] = 𝑚1
𝑣1
𝑜𝑓 𝐻𝐶𝑙 −[𝑚2
𝑣2
𝑜𝑓 𝑁𝑎𝑂𝐻]
𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
= 1𝑥100 −[1 𝑥 1]
100+1
= 100 −[1]
101
= 99
101 =0.98
[H+] = 0.98
( since pH = - log (H+))
pH = - log (0.98)
pH = 0.008
number of H+ ions remaining now
will be 100 – 1= 99 meq or 99/101
eq/ lit
c.) On addition of 50 ml of base
If we add 50 ml base, to 100 ml of acid, then 50 ml of acid is unneutralised in 150ml (100 ml acid +50 ml
base)
[H+] = [𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻
+
− 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑂𝐻 − ]
𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
[H+] = 𝑚1
𝑣1
𝑜𝑓 𝐻𝐶𝑙 −[𝑚2
𝑣2
𝑜𝑓 𝑁𝑎𝑂𝐻]
𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
= 1 𝑥 100 −[1 𝑥 50]
100+50
= 50
150
= 0.33
pH = - log (0.33) pH = 0.48
Similary if we go on adding-
75 ml base added, H+ = 25 x1 /175 = 0.142, pH = 0.8
90 ml base added, H+ = 10 x 1 /190 = 0.052, pH = 1.28
99 ml base added, H+ = 1 x 1 /199 = 0.005, pH = 2.3
99.9 ml base added, H+ = 0.1 x 1 /199.9 = 0.0005 , pH =3.3
2. pH value at equivalence point:
When 100 ml of base is added , pH will change sharply to 7 which is the equivalence
point.
As per definition of equivalence point-
H+ = OH-
pH = pOH (solution is neutral)
Kw= 1 x 10-14 at 25 C hence pKw = 14
pKw = pH + pOH
ie 14 = pH + pOH
14 = pH + pH
14 = 2pH
pH = 7
3. pH after equivalence point:
After steep slope, when more NaOH is added, there is a gradual rise in pH. At equivalence
100 ml of NaOH is required to neutralise 100 ml of HCL. At this point, the volume in the
flask is 200 ml.
A) After equivalence if 1 ml of NaOH is added, the total volume will be 201.
Consider addition of 1 ml excess of alkali-
[OH-] = [𝑚𝑜𝑙𝑒𝑠 𝑜𝑓𝑂𝐻 − 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓𝐻
+
]
𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
[OH-] = 𝑚1
𝑣1
𝑜𝑓𝑁𝑎𝑂𝐻 −[𝑚2
𝑣2
𝑜𝑓 𝐻𝐶𝑙]
𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
[OH-] = 1 𝑥 101 −[1 𝑥 100]
101+100
[OH-] = 101 −[100 ]
201
[OH-] = 1
201
=0.0049
pOH = - log (OH-) so pOH = - log (0.0049)
pOH = 2.3
pKw = pH + pOH ie 14 = pH + pOH
pH = 14 - pOH
pH = 14 - 2.3 = 11.7
B) Similarly after adding 20 ml of excess alkali
pH = 12.96
Sample Calculation: Strong Acid-Strong Base Titration Curve
Consider the titration of 40.0 mL of 0.100 M HCl with 0.100 M NaOH
Region 1. Before the equivalence point, after adding 20.0 mL of 0.100 M NaOH.
(Half way to the equivalence point.)
Amount of H3O+ remaining = Initial moles of H3O+ - Moles of OH- added
baseaddedofvolumeacidofvolumeoriginal
remainingOHof(mol)amount
]O[H 3
3




moles of titrant = moles of analyte
(V titrant)(M titrant) = (V analyte)(M analyte)
Three regions of titration curve exists
Region 2. At the equivalence point, after adding 40.0 mL of 0.100 M NaOH.
Initial moles of H3O+ = 0.0400 L x 0.100 M = 0.00400 M H3O+- Moles of OH- added = 0.0400
L x 0.100 M =0.00400 mol OH-
baseaddedofvolumeacidofvolumeoriginal
remainingOHof(mol)amount
]O[H 3
3




Sample Calculation: Strong Acid-Strong Base Titration Curve
Region 3. After the equivalence point, after adding 50.0 mL of 0.100 M NaOH. (Now calculate
excess OH-)
Total moles of OH- = 0.0500 L x 0.100 M = 0.00500 mol OH- -Moles of H3O+ consumed =
0.0400 L x 0.100 M =0.00400 mol
baseaddedofvolumeacidofvolumeoriginal
remainingOHof(mol)amount
][OH




BUFFERS ?
Definition: It is defined as a solution, which resists change in its pH value caused by
addition of small amount of acid or base or on dilution or keeping it for a certain long time.
It is required to maintain pH of reaction at optimum value.
Types and preparation of buffer solutions:
1. Acidic buffer:
An acidic buffer can be prepared by dissolving Weak acid such as acetic acid (CH3COOH)
and its Salt of strong base such as sodium acetate (CH3COONa) in water.
CH3COOH + CH3COONa Acidic buffer
Acidic buffers have pH less than 7. Other ex. Is Ethanoic acid and sodium ethanoate in
solution. Equal molar conc of both gives pH of 4.76.
You can change the pH of the buffer solution by changing ratio of acid to salt.
2. Basic buffer:
A basic buffer can be prepared by dissolving Weak base such as ammonium hydroxide
(NH4OH) and its Salt of strong acid such as Ammonium chloride (NH4Cl) in water.
NH4OH + NH4Cl Basic buffer
Basic buffers have pH more than 7. ex. ammonia and ammonium chloride in solution.
Equal molar conc of both gives pH of 9.25.
3. Buffer solution of single salt:
It is prepared by dissolving single salt which must be of weak acid and a weak base such as
ammonium acetate in water.
CH3COONH4 + Water Buffer (single salt)
Buffer action:
It is the property of buffer solution to resist change in its pH value on adding a small
amount of acid or a base.
Mechanism of Buffer action
1. MECHANISM OF ACIDIC BUFFER:
Suppose an acid buffer if prepared from a weak acid , CH3COOH and it salt of strong base,
CH3COONa, by dissolving them in water.
Acetic acid being a week acid, ionised feebly and will give only few H+ ions in solution.
CH3COOH CH3COO- + H+
Sodium acetate being a strong electrolyte will completely get dissociated as-
CH3COONa CH3COO- + Na+
The solution will have large no. of CH3COO- ions now.
I. Addition of strong acid like HCl to the buffer:
When a strong acid like HCl is added to the solution, it dissociated completely as follows:
HCl H+ + Cl-
Now H+ ions of HCl wil combine with acetate ions, CH3COO- and from unionised
CH3COOH molecules and as a result the pH of the buffer solution does not change to
considerable extent.
In this way H+ are neutralised by CH3COO- ions and maintains the pH of buffer
constant.
CH3COO- + H+ CH3COOH (unionised)
This resistance to change in pH value of a buffer solution on adding a small amount
of a strong acid is called ‘Reserve Basisity’ due to CH3COO- ions .
II. Addition of strong base like NaOH to the buffer.
When a strong base like NaOH is added to the solution of buffer, it dissociates completely as
NaOH Na+ + OH-
Now OH- of NaOH combine with CH3COOH and for acetate ions , CH3COO- and water.
CH3COOH + OH- CH3COO- + H2O
In this way hydroxyl ions are neutralised by acetic acid and hence pH of the buffer does not
change to a considerable extent even if base is added to it.
This resistance to change in pH value of buffer solution on adding a small amount of strong
base is called as ‘Reserve acidity’ due to acetic acid.
2. MECHANISM OF BASIC BUFFER:
Suppose a basic buffer is prepared from a weak base , NH4OH and its salt of strong
acid,NH4Cl by dissolving them in water.
NH4OH is a weak base , hence it will be feebly ionised and the solution will contain very few
OH- ions.
NH4OH NH4
+ + Cl -
NH4Cl been an electrolyte will be completely dissociated as –
NH4Cl NH4
+ + Cl –
Now the solution has large number of NH4
+ ions.
I. Addition of strong acid like HCl to the buffer:
When a strong acid like HCl is added in the solution, it dissociates almost completely as:
HCl H+ + Cl-
NowH+ of HCl combine with NH4OH and form water and NH4
+ ions.
NH4OH + H+ NH4
+ + H2O
In this way , H+ ions are neutralised by NH4OH and as a result, the pH of buffer
solution does not change to any extent.
This resistance to change in pH value of a buffer solution on adding a small amont
of strong acid is called ‘Reserve basicity’ due to NH4OH.
II. Addition of strong base like NaOH to the buffer.
When a strong base like NaOH is added to the solution of buffer, it dissociates completely as
NaOH Na+ + OH-
Now OH- of NaOH combine with NH4
+ ions and form unionised NH4OH.
NH4
+ + OH- NH4OH
In this way hydroxyl ions are neutralised by NH4
+ ions and hence pH of the buffer does not
change to a considerable extent even if base is added to it.
This resistance to change in pH value of buffer solution on adding a small amount of strong
base is called as ‘Reserve acidity’ due to NH4
+ ions.
BUFFERING
SYSTEMS
BUFFERING pH
ranges at 25 C
Hydrochloric acid
/ Potassium
chloride
1.0- 2.2
Citric acid
/Sodium citrate
3.0 - 6.2
Potassium
chloride /Sodium
hydroxide
12.0-13
Potassium
hydrogen
Pthalate/ Sodium
hydroxide.
4.1-5.9
Properties of buffer:
1. Buffer solution has definite pH value.
2. The pH value of buffer does not change
even if small amount of an acid or base
is added into it.
3. Value of pH does not change even if it is
kept for long time.
4. pH value of buffer solution remains
constant even if diluted with water.
BUFFERS
The dissociation of an acid can be described by an equilibrium expression:
HA + H20 H3O+ + A-
Applying law of mass action, equilibrium constant will be
Ka = [H3O+][A-]
[HA]
Taking the negative log of both sides of the equation gives
-logKa = -log [H3O+][A-]
[HA]
-logKa = -log [H3O+] + (-log [A-] )
[HA]
-logKa = -log [H3O+] - (log [A-] )
[HA]
By definition,
pKa = - logKa and pH = -log[H3O+], so
pka = pH – log [A-]
[HA]
This equation can then be rearranged to give the Henderson-Hasselbalch equation:
HA + H20 H3O+ + A-
This is Henderson-Hasselbalch equation for calculation of pH of buffers.
Ex: CH3COOH + H2O CH3COO- + H3O+
pH = pKa + log [A-] or pH = pKa + log [conjugate base]
[HA] [acid]
HA + H20 H3O+ + A-
Consider reverse equation, here A- is a base and HA is conjugate acid.
Ka = [H3O+][A-] =
𝐾𝑤
𝐾𝑏
-------------------------- Kw = Ka.Kb
[HA]
[H+] = Ka
[𝐻𝐴]
[𝐴−]
=
𝐾𝑤
𝐾𝑏
-log [H+] = - log Ka – log
[𝐻𝐴]
[𝐴−]
OR
-log [H+] = - log
𝐾𝑤
𝐾𝑏
– log
[𝐻𝐴]
[𝐴−]
pH = pKa + log
[𝐴−]
[𝐻𝐴]
and for basic buffer pOH = pKb + log
𝐵𝐻 𝑐𝑜𝑛𝑗 𝑎𝑐𝑖𝑑
𝐵− 𝑏𝑎𝑠𝑒
For buffers formed from mixture of weak base and strong acid ex. NH4OH and HCL
pH = (pKw – pKb )+ log
[𝑨−]
[𝑯𝑨]
-------
from eq pKw = pKa + pKb
Buffer capacity:
• The amount of acid or base that can be added without causing a large change in pH is the buffering capacity of the solution.
• Or the amount of acid or base consumed by 1 litre of buffer to change the pH by one unit.
• The capacity is determined by the concentrations of HA and A- .
• Buffering capacity increases with increase in the concentrations of buffering species
• Higher the concentration, more the acid and base it can tolerate.
The buffer capacity, buffer intensity or buffer index is represented as
β =
𝑑𝐶 𝐵𝑂𝐻
𝑑𝑝𝐻
= -
𝑑𝐶 𝐻𝐴
𝑑𝑝𝐻
Where,
C BOH= number of moles per litre of strong base
C HA = number of moles per litre of strong acid needed to bring a change of about dpH.
Buffer capacity is a positive number.
The larger the value of β , the more resistant the solution is to pH change.
Buffering capacity is governed by
1. ratio of HA to A-.
2. Concentration of HA and A-. Larger the conc. larger is the capacity.
It is maximum when the ratio is one or unity. That is when pH = pKa
pH = pKa + log
1
1
pH = pKa.
This corresponds to the mid point of titration of weak acid strong base.
The buffering capacity
is satisfactory over a pH
range of pKa ± 1 or
pKb ± 1Since adding a strong acid in buffer
decreases the pH value. So dCHA/dpH is
negative.
β decreases rapidly if
concentration ratio of
[A−]/[HA] becomes
larger or small than 1 .
pKa of acid or pKb of base
chosen should lie withing
± 1 unit of desired pH to
get buffer of desired
capacity.
WEAK ACID AND STRONG BASE TITRATION
CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l)
CH3COONa + H2O (l) CH3COOH + NaOH
In ionic form,
CH3COO- + Na+ + H2O CH3COOH + Na+ +OH-
Cancelling common ions on both sides, net reaction is
CH3COO-
(aq) + H2O (l)) CH3COOH + OH-
(aq)
From this equation we can conclude that if weak acid is titrated with strong base then the
pH of the solution will be more than 7, i.e. it will show basic pH. (At equivalence point-
pH>7), since the solution contains OH- ions .
WEAK ACID AND STRONG BASE TITRATION
• At the beginning, the solution contains only a weak acid or a weak base, and the pH is
calculated from the concentration of that and its dissociation constant.
• After various increment of , the solution consists of a series of buffers.
• The pH of each buffer can be calculated from the analytical concentration of the
conjugate base or acid and the residual concentration of the weak acid or base.
• At equivalence point, the solution contains only the conjugate of the weak acid or base
being titrated, and the pH is calculated from the concentration of this product.
• Beyond the equivalence point, the excess of strong acid or base titrate represses the
acidic or basic character of the reaction product to such an extent that the pH is
governed largely by the concentration of the excess titrant.
TITRATION CURVES FOR WEAK ACIDS
• You might think that all titrations must have an equivalence point at pH 7
because that is the point at which concentrations of hydrogen ions and
hydroxide ions are equal and the solution is neutral.
• But some titrations have equivalence points at pH values less than 7, and some
have equivalence points at pH values greater than 7.
• These differences occur because of reactions between the newly formed salts
and water – salt hydrolysis. (Behaviour of salts in water)
• Some salts are basic (weak acid, strong base) and some salts are acidic (strong
acid, weak base).
• The previous slide shows that the equivalence point for the titration of HPr (a
weak acid) with NaOH (a strong base) lies at pH 8.80.
Titration of a Weak Acid with a Strong Base
A. Addition of a strong base to a weak acid forms a Buffer Solution
HA + OH- A- + H2O
B. Important Points
1. pH increases more rapidly at the start than for a strong acid
2. pH levels off near pKa due to HA/A- buffering effect
pH = pKa + log([A-]/[HA]) = pKa + log(1) = pKa (when [A-] = [HA])
3. Curve is steepest near equivalence point. Equivalence Point > 7.0
4. Curve is similar to strong acid—strong base after eq. pt. where OH- is major
For Weak acid and Strong base titration curve
will have four regions:
1. pH before equivalence point
2. pH of Buffer region
3. pH at Equivalence point
4. pH after equivalence point
CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l)
1. pH before equivalence point:
Before any added base, just weak acid (HA) in water
pH determined by Ka
Ka= [H+] [A-]/[HA]
2. pH of buffer region:
With addition of strong base  buffer region
pH determined by Henderson Hasselbach equation
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HA
A
logpKpH a
3. pH at equivalence point
all HA is converted into A-
Weak base with pH determined by Kb
4. pH after equivalence point:
Beyond equivalence point, excess strong base is added to A-
solution.
pH is determined by strong base.
Similar to titration of strong acid with strong base.
Kb = [HA] [OH-]/[A-]
Calculations for Neutralization curve for weak acid and strong base-
Consider titration between 100ml of 0.1 N CH3COOH and 0.1 N NaOH
(KaCH3COOH= 1.75x10-5 pka= 4.74).
Change in pH value is calculated by 4 steps.
1. pH value before equivalence point.
2. pH at buffer region
3. pH value at equivalence point.
4. pH value after equivalence point.
1) pH value before equivalence point:
Before proceeding the titration, the solution in conical flask is 0.1 N CH3COOH.
For weak acid concentration of H+ can also be calculated by formula-
H+ = 𝑲𝒂. 𝑪
= 1. 75 𝑥 10 − 5 𝑥 0.1 hence H+= 0.0013 so pH = 2.87
2. pH at buffer region:
After addition of 50 ml 0.1 N NaOH , 50 ml of NaOH react with 50 ML OF 0.1 N CH3COOH and
50 ml re CH3COOH remains unreacted.
CH3COOH + NaOH ↔ CH3COONa + H2O
This is the phase where buffer if formed and is called as a buffer region. It consist of mixture
of weak acid and salt of weak acid, so pH value is calculated by formula;
pH = pka + log
Conc of [CH3COONa]
Conc of [CH3COOH]
Concentration of CH3COOH =
M1V1CH3COOH
M2V2[CH3COOH]
=
0.1 X 50 ml
150
Concentration of CH3COOH = 3.33 x 10-2 moles
At this stage concentration of CH3COONa = concentration of CH3COOH
so CH3COONa =3.33 x 10-2 moles
pH = pKa + log
Conc of [CH3COONa]
Conc of [CH3COOH]
as, Ka = 1.75 X 10−5
But pKa = - log Ka
So pKa = 4.74
Substituting the values in the equation of pH we get,
pH = 4.74 + log 3.33 x 10−2
3.33 x 10−2
pH = 4.74 + log (1)
= 4.74 + 0
pH = 4.74
3) pH value at equivalence point. (achieved when 100 ml of NaOH is added.)
pH = ½pKw ½ pKa+ ½ log(c)
= ½(14) + ½(4.74)+ ½ log(0.1)
= 7+2.37- 0.5
pH= 8.72
4) After equivalence point solution contain excess of OH- ion causes sharp increase in pH
value, solution becomes alkaline.
1.The initial pH is higher.
2.A gradually rising portion of the curve, called the buffer region,
appears before the steep rise to the equivalence point.
3.The pH at the equivalence point is greater than 7.00.
4.The steep rise interval is less pronounced.
The four Major Differences Between a Strong Acid-Strong Base
Titration Curve and a Weak Acid-Strong Base Titration Curve
HCl (aq) + NH3 (aq) NH4Cl (aq)
NH4Cl + H2O NH4OH + HCL
In ioni form-
NH4
+ + Cl- + H2O NH4OH + H++CL-
NH4
+
(aq) + H2O (l) NH3 (aq) + H+
(aq)
At equivalence point (pH < 7)
WEAK BASE STRONG ACID TITRATIONS
WEAK BASE STRONG ACID TITRATIONS
For Weak base and Strong acid titration curve
will have four regions:
1. pH before equivalence point
Due to base NH3 initialy before titration
Can be determined by Kb value
Kb = [NH4+] [OH-]/ NH4OH
2. pH of Buffer region
By Henderson hasselbalch equation
put pKa= Kw- pKb
3. pH at Equivalence point
All base is converted into its conjugated acid.
Hence Ka is determined
4. pH after equivalence point
Beyond equivalence point, excess strong acid is added.
pH is determined by strong acid.
Similar to titration of strong acid with strong base.
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A
logpKpH a
Calculations for CURvE BETWEEN WEAK BASE AND STRONG ACID
Consider titration of 100ml of 0.1 N aqueous ammonia solution with 0.1N HCl
{Kb(NH4OH) =1.75×10ˉ5, pKb = 4.74}
Change in pH value calculated in 4 steps :
1. pH value before equivalence point.
2. pH at buffer region
3. pH value at equivalence point.
4. pH value after equivalence point.
1) pH value at equivalence point:
a. Before proceeding titration conical flask contain 100ml of 0.1 N NH4OH Solution,
OH -=√𝐾𝑏.C
OH- =√(1.75 × 10
− 5) × (0.1)
OH- = 1.3×10-3 but pOH = - log(OH-) therefore pOH=2.87
Now pKw = pH + pOH
Hence pH = pkw - pOH
= 14 - 2.87
pH = 11.73
2. pH at buffer region:
After addition 50ml 0f 0.1HCl ,it react with ammonia and form buffer solution.
NH4OH + HCl ↔ NH4Cl + H2O
It is a buffer solution consist of mixture of weak base NH4OH and its salt (NH4CL)so pH
value calculated by buffer formula:
pOH = pkb - log
Conc of weak base
conc of its salt
= 4.74 - log
weak base
its salt(conc.of HCl)
As per equation need to calculate conc.of weak base (NH4OH) and its salt (NH4Cl)
Concentration of weak base NH4OH =
m moles of base
total volume
=
M1V1
150
=
0.1×50
150
=
50
150
=3.33×10-3
Conc.of HCl =
volume of HCl × molarity of HCL
Total volume
=
50×0.1
150
= 3.33× 10-3
Conc.of HCl = conc. of salt
put the value in above equation;
pOH = 4.74 - log(1)
= 4.74
As pOH = pkw-pOH
= 14-4.74
pH = 9.26
2. pH value at equivalence point is calculated by formula;
pH =
1
2
pkw -
1
2
pkb -
1
2
log(c)
=
1
2
(14) -
1
2
(4.74) -
1
2
log(0.1)
pH =5.28
3. pH value after equivalence point:
After equivalence point solution contain excess of HCl ,causes sharp decrease in pH value ,
solution become more acidic.
The four Major Differences Between a Weak Acid-Strong
Base Titration Curve and a Weak Base-Strong Acid Titration
Curve
1.The initial pH is above 7.00.
2.A gradually decreasing portion of the curve, called the buffer region, appears
before a steep fall to the equivalence point.
3.The pH at the equivalence point is less than 7.00.
4.Thereafter, the pH decreases slowly as excess strong acid is added.
WEAK ACIDS AND WEAK BASE TITRATIONS
Consider titration 100 ml of 0.1N acetic acid with
0.1N ammonia
CH3COOH + NH3 CH3COO NH4
When salt dissolved in water,
CH3COO NH4 + H2O CH3COOH + NH4OH
Ionic form
CH3COO- + NH4
+ + H2O CH3COOH + NH4OH
No formation of H+ and OH- ions.
Hence pH is 7
At equivalence point,
pH = pKw + 1/2pKa –pKb
pH = 7.0
Change of pH near equivalence point and during whole neutralization
is very gradual .
Hence end point is not detected by ordinary indicator. Mixed indicators
are required.
Types of acids and bases
ACIDS
1. Monoprotic
These acids loose only one proton in the reaction. They probably have only one proton
with them.
Ex- HCL, HNO3, HBr
2. Polyprotic /Polyfunctional acids(Diprotic, triprotic etc)
a) Polyprotic acids are specific acids that are capable of losing more than a single proton per
molecule in acid-base reactions.
b) These acids that have more than one ionizable H+atom per molecule.
c) Protons are lost through several stages (one at each stage).
1. Diprotic acids:
These have two hydrogen ions for donation and act as bronsted acid.
Ex- H2 SO4, H2 CO3, H2S, chromic acid (H2CrO4), and oxalic acid (H2C2O4)
2. Triprotic acids
These can donate three hydrogen ions and act as bronsted acids.
Ex. phosphoric acid (H3PO4) and citric acid C6H8O7
Consider H2SO4 (Diprotic acid)
• pka helps us to determine the ease with which it looses protons.(Strength)
• There is usually a large difference in the ease with which these acids lose the first and second (or
second and third) protons.
• Sulphuric acid will not loose both the protons at a time in water.(It will loose one proton first and
then the next).
• Sulfuric acid is a strong diprotic acid so it will have two dissociation constants as Ka1 and Ka2 since
it looses one proton first and second next.
Hence for two reactions, it will have two Ka constants.
For Triprotic acid, they will have three acidity constants.
Conjugate bases
Ka1
Ka2
Polyprotic acids and their salts
𝑝𝐾𝑎 = − log(
1
𝐾𝑎
)
Buffer calculations for polyprotic acids
The anion on the right hand side of each ionization step can be considered as salt (conjugate base) of the acid.
Hence from the previous equation,
H2PO4
- is the salt of H3PO4
HPO4
-- -- is the salt of H2PO4
-
PO4
--- is salt of HPO4
—
Each of the pairs constitute a buffer system.
Optimum buffering capacity of each pair occurs when pH = pKa
Use Henderson hasselbalch equation
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A
logpKpH a
SALTS OF POLPROTIC ACIDS
The salts of polyprotic acids can be acidic or basic.
Example- Phosphoric acid H3PO4
Its protonated salts = H2PO4
- , HPO4
-- possess both acidic and basic properties. Hence are amphoteric salts.
Its non-protonated salt = PO4
--- is a base and has basic property.
Concentraion of H+ in system calculated by
Common Polyprotic Acids Formula Strong/Weak Acid Number of Ionizable
Hydrogens
Ka1 Ka2 Ka3
Sulfuric acid H2SO4 Strong 2 (diprotic) Very Large 1.1E-2
Sulfurous acid H2SO3 Weak 2 (diprotic) 1.3E-2 6.2E-8
Phosphoric acid H3PO4 Weak 3 (triprotic) 7.1E-3 6.3E-8 4.2E-
13
Carbonic acid H2CO3 Weak 2 (diprotic) 4.4E-7 4.7E-11
Hydrosulfuric acid or Hydrogen
sulfide
H2S Weak 2 (diprotic) 1.0E-7 1E-19
Oxalic acid H2C2O4 Weak 2 (diprotic) 5.4E-2 5.3E-5
BASES
1. Monobasic (Which can accept only one H+) – Ex. NaOH, NH4OH, KOH
2. Polybasic /Polyfunctional bases (Dibasic, tribasic etc)
Polyprotic Bases are bases that can accept more than one H+ ion, or proton in acid-base reactions.
Common Polyprotic
Bases
Formula Strong/Weak Base Diprotic/Triprotic
Base
Phosphate ion PO4
3-
Weak Triprotic
Sulfate ion SO4
2-
Very Weak Diprotic
Carbonate ion CO3
2-
Strong Diprotic
Titration Curve: Polyprotic acids Titration Curve of 0.100M
H2SO3 with 0.100 M NaOH
• Curve for the titration of a weak
polyprotic acid.
• Titrating 40.00mL of 0.1000M H2SO3
with 0.1M NaOH leads to a curve with
two buffer regions and two equivalence
points.
• Because the Ka values are
separated by several orders of
magnitude, in effect the titration
curve looks like two weak acid strong
base curves attached.
• The pH of the first equivalence point is
below 7 because the solution
contains HSO3 which is a stronger
acid than it is a base.
• Ka of HSO3 = 6.5•10-8;
• Kb of HSO3 = 7.1•10-13
FEATURES
1. The loss of each mole of H+
shows up as separate
equivalence point.
2. The pH at the midpoint of
the buffer region is equal to
the pKa of that acid species.
3. The same volume of added
base is required to remove
each mole of H+.
Titration Curve: Poly-Basic
Titration Curve of 0.10 M
Na2CO3 with 0.10 M HCl.
1. CO3 + H+ HCO3 + H2O
2. HCO3 + H+ H2CO3 + H2O
1.Multiple Inflection Points = Multiple Equivalence Points will be seen
2.The volume required to reach each equivalence point will be the same
CO3
2- + H+ HCO3
- Kb1 = KW/Ka2 = 1.8 x 10-4 (pKb1 = 3.74)
HCO3
- + H+ H2CO3 Kb2 = KW/Ka1 = 2.3 x 10-8 (pKb2 = 7.64)
½ Eq. pt 1 Eq. pt 1 ½ Eq. pt 2 Eq. pt 2
pKa2 = 10.26
pKb1 = 3.74
pKa2 = 6.36
pKb1 = 7.64
SUMMARY
Weak base added to
weak acid
Polyfunctional /polyprotic acids
Polyfunctional /polyprotic
bases
TITRATION OF AMINO ACIDS
Each amino acid (except for proline) has:
1. A carboxyl group (-COO-) .
2. An amino group (-NH3
+) .
3. Side chain ("R-group") bonded to the α-carbon atom.
These carboxyl and amino groups are combined in peptide
linkage.
STRUCTURE OF THE AMINO ACIDS
Proteins are complex structure made up of amino acids.
R---C---COOH
NH2
R’
• These are amphoteric substances that contain both acidic groups (COOH) and basic
group (NH2 ).
• Hence they can act as both acid and base.
• In aqueous solutions, these substances tend to undergo internal proton transfer from
the carboxylic acid group to amino group since R NH2 is stronger base than RCOO-
• The result is Zwitter ion.
Since they are amphoteric they can be titrated either with strong acid or strong base
R---CH---COOH
NH2
R---CH---COO-
NH3
+
Zwitter ion
Classification of Amino Acids
They classified according to the side chain:
1. Amino acids with nonpolar side chains.
2. Aromatic R Groups.
3. Amino acids with uncharged polar side chains.
4. Positively Charged (Basic) R Groups.
5. Amino acids with acidic side chains.
ACIDIC AND BASIC PROPERTIES OF AMINO ACIDS
Amino acids in aqueous solution contain weakly acidic α-carboxyl groups and
weakly basic α-amino groups.
Each of the acidic and basic amino acids contains an ionizable group in its side
chain.
Thus, both free and some of the combined amino acids in peptide linkages can
act as buffers.
The concentration of a weak acid (HA) and its conjugate base(A-) is described by
the Henderson-Hasselbalch equation.
For the reaction (HA A- + H+ )
[H+] [A-]
Ka = ───── ------ (1)
[HA]
By solving for the [H+] in the above equation, taking the logarithm of both sides of the
equation, multiplying both sides of the equation by -1, and substituting pH = -log [H+]
and pKa = -log [Ka] we obtain:
[A-]
pH = pKa + log ─── ------ (2)
[HA]
It is the (Henderson-Hasselbalch equation)
Titration Solution of an amino acid
R---CH---COOH
NH2
R---CH---COO-
NH3
+
Zwitter ion
Ka1 Ka2
TTITRATION OF ZWITTER ION OF AMINOACIDS WITH ACIDS AND BASES
Titration of zwitter ion with strong acid and base
• Conjugate acid of the zwitter ion can be considered as weak polyprotic acid.
• It will ionize to form first amphoteric zwitter ion and on second ionization it will give conjugate base of
zwitter ion ie salt of weak acid.
• When zwitter ion of amino acid are titrated with strong acid, buffer region is established containing the
conjugate acid and zwitter ion .ie salt of weak acid
• At half equivalence point pH = pKa1
• At equivalence point pH is determined by Ka,
• For buffer region: pH is determined by Henderson hasselbalch equation.
• When zwitter ion formed is titrated with strong base, buffer region will contain zwitter ion (which will
act as strong acid) and it conjugate base.
• Half of the equivalence point will be pH = pKa2
• Equivalence point can be determined by Kb
Kb =
𝐾𝑤
𝐾𝑎
Since amino acids contain more than one
number of NH2 and COOH groups , titration
will yield stepwise end point like other
polyprotic acids.
Buffers
R---CH---COO-
NH3
+
R---CH---COOH
NH2
Titraton of amino acid as acid with base
pH = pKa1
ka1
Titration curve of glycine
=
=
From acid to zwitter ion,
1. pH determined by ka
2. Buffers at this stage
determined by Henderson
equation.
3.At equivalence point
conjugate base is formed
hence Kb is determined .
With help of kb, Ka is
calculated using eq,
Ka = Kw/Kb
4.Half equivalence point is
pH = pKa
5. Glycine will have 2
dissociation const as Ka1 &
Ka2 due to diprotic nature
ACID-BASE INDICATORS
• Equivalence Point
The point at which the two solutions used in a titration are present in chemically equivalent
amounts is the equivalence point.
• End Point
The point in a titration at which an indicator changes color and we can visually detect it is
called the end point of the indicator.
• The equivalence point of an acid-base titration can be determined by-
1. pH meter
2. Indicators ( Chemical dyes)
A. Finding the equivalence point of a titration
Use a pH meter
•Plot pH versus titrant volume
•Center vertical region = equivalence point
Use an Acid-Base Indicator
•Indicators are chemical species, whose colors are
affected by acidic and basic solutions & are called
acid-base indicators.
•An acid–base indicator is a weak acid or base.
•Acid-base indicators are compounds whose colors
are sensitive to pH or change color based on pH.
Acid base indicators are weak organic
acids /bases that dissociate slightly in
aqueous solutions to form ions.
• The pH range over which an indicator changes color is called its transition interval.
• Consider the above equation
HIn H+ + In-
Where ,
HIn is indicator which is a weak acid (one color)
In- is its conjugate base. (other color)
• HIn and In− will have different colors. One of the “colors” may be colorless.
• The indicators can change colour because their ions have colors that are different from
undissociated molecule when pH changes.
• If a solution is acidic, [H3O+]/[H+] is high. When indicator (weak acid) is added , few [H+] are
dissociated). Hence [H3O+]/[H+] is a common ion and so it suppresses the ionization of the
indicator acid, and we see the color of HA. (i.e indicator will show color of HA).
• Hence in acidic solutions, most of the indicator is HIn (it is undissociated due to common
ion effect)
• In basic solutions, most of the indicator is In– ie dissociated form of conjugate base of
that indicator.( As we proceed for titration with a base, the pH will change to basic side.
Hence the common ion effect will be reduced and the indicator which is a weak acid will
undergo dissociation to some extent to In– , which is a dissociated form or conjugate
base of the indicator molecule.)
• Indicators that change color at pH lower than 7 are stronger acids than the other types
of indicators. They tend to ionize more than the others.
• Indicators that undergo transition in the higher pH or pH more than 7 range are weaker
acids.
• Each indicator its own particular pH or pH ranges over which it changes color.
For Strong acid strong base titration:
• The neutralization of strong acids with strong bases produces a salt solution with a pH of 7.
• Indicators that undergo transition at about pH 7 are used to determine the equivalence point of strong-acid/strong
base titrations. Bromthymol blue is a good choice for the titration of a strong acid with a strong base
For strong acid weak base titrations:
• The equivalence point of a strong-acid/weak-base titration is acidic.( pH less than 7)
• Indicators that change color at pH lower than 7 are used to determine the equivalence point of strong-acid/weak-
base titrations.
For weak acid strong base titrations:
• The equivalence point of a weak-acid/strong-base titration is basic.
• Indicators that change color at pH higher than 7 are used to determine the equivalence point of weak-acid/strong-
base titrations.
Phenolphthalein changes color at the equivalence point of a titration of a weak acid with a strong base.
Types of indicators used for different titrations
For weak acid weak base
titration, since endpoint is
difficult to determine, mixed
type of indicators are used.
1. Choose an indicator that changes color at the equivalence point of titration.
2. Remember that the role of the indicator is to indicate to you, by means of a color change, that
just enough of the titrating solution has been added to neutralize the unknown solution.
3. End Point = when the indicator changes color.
4. If you have chosen the wrong indicator, the end point will be different than the eq. pt.
5. According to the type of titrations, whether SASA, SAWB, WASB or WAWB titrations, consider
the pH at which equivalence point occurs.
6. Titrations whose equivalence point is below 7, indicator should change color at pH below 7.
7. Titrations whose equivalence point is above7, indicator should change color at pH above 7.
8. Titrations whose equivalence point is at 7, indicator should change color at pH 7.
How to select an indicator?
pH of acid base reaction at its equivalent point
Acid Base pH at equivalence point Indicators
Strong Strong = 7 (neutral) Methyl orange
Phenolphthalein
Strong Weak < 7 (acidic) Methyl orange
Weak Strong > 7 (basic) Phenolphthalein
Weak Weak pH depend on Ka and Kb
of acid & base conc.
Mixed indicators
Colours of indicator at different pH
THEORIES FOR ACID BASE INDICATORS
Two theories are postulated for acid base indicator as follows:
1. Ostwald theory
2. Resonance theory
A. OSTWALD THEORY
• The first theory to explain the behaviour of indicators was put forth by W. Ostwald.
• According to this theory , the undissociated indicator acid HIn or base InOH has a color different than its ion.
• For an acid indicator equivalence can be written as :
HIn H+ + In-
• In acid solution, there is depression of ionisation of indicator due to common ion effect. Hence initial concentration of
HIn is greater than the In- and hence the color exhibited will be that of the unionised form of acid.
• As the titration with base proceeds , the alkali medium promotes removal of H+ and hence there is gradual increase in
concentration of ionised form In- and the solution acquires colour of ionised form.
• Apply law of mass action-
KIna =
[H+][In− ]
[𝐻In
]
log H+ = log KIna + log
𝐻In
[In− ]
Taking negative log on both sides
-log H+ = - log KIna - log
𝐻In
[In− ]
= - log KIna + log
In−
[𝐻𝐼n]
where KIna = Dissociation constant of indicator.
The colour of indicator depends upon the ratio of concentration of ionised and unionised form and is directly proportional
to pH.
For base indicator
In OH In
+ + OH-
KInb =
In
+ [OH
−
]
[In OH]
[OH
−
] = KInb
[In OH]
[In
+]
pH = p KIna + log
In−
[𝐻𝐼n]
Now
Kw = [H+] [OH-]
[H+] =
Kw
[OH]
Substituting the value of [OH] in the above equation.
[H+] =
Kw [𝐼𝑛
+
]
KInb[In OH]
Taking log on both sides,
For an acid indicator from dissociation constant
equation,
In−
[𝐻𝐼n]
=
KIn
[H+]
The colour change is affected by H+ concentration.
Since this change is gradual, the colour change of
indicator is also gradual.
To detect the colour change of the indicator , the ratio
of
In−
[𝐻𝐼n]
must be at least
1
10
.
Hence pH at which observable change occurred will be-
pH = p KIn + log
In−
[𝐻𝐼n]
In−
[𝐻𝐼n]
= 0.1
pH = pKIn + log 0.1
pH = pKIn - 1
For basic side, ratio of
In−
[𝐻𝐼n]
= 10/1= 10
pH = pKIn + log 10
pH = pKIn + 1
Therefore general pH range in colour change
observed is pH = pKIn ± 1
This is transition interval of indicator.
pH = pKw – K𝐼nb + log
[In OH]
𝐼 𝑛
+
+
The indicator should be such that pH at equivalence point falls within the transition interval of the indicator.
Titration between pH at end point Commonly used indicator
Weak acid and strong base Alkaline range Thymol blue,
phenolphthalein
thymopthalein
Weak base and strong
acid
Acidic range Methyl orange
Methyl red
Bromocresol green
2. RESONANCE THEORY
All acid- base indicators commonly used are organic compounds.
The difference in colour of same compound in acid and base medium is due to difference
in structure of two forms.
Colour shown by the compound is related to the ability of the compound to absorb visible
light and this capability is related to the electronic structure.
Change in electronic features will result in absorption of different colour components of
light with a resultant colour change.
Types of Indicators
1. Single indicators
2. Mixed indicators
3. Universal or Multiple range indicators
Single indicators
Only one component of indicator
Mixed Indicators
In some cases the pH range is very narrow and the colour change over this range must be very sharp.
Ordinary acid base indicators cannot serve this purpose.
This is achieved by use of suitable mixture of indicators.
These are selected such that their pKIn values are close to each other.
Examples:
Mixture of equal parts of neutral red (0.1% in alcohol) and methylene blue (0.1% in alcohol) gives a sharp
colour change from violet blue to green in passing from acid to alkaline solution at pH 7.
Mixture of phenolphthalein (3 parts of 0.1%) and α-napthopthalein ( 1 part 0.%) passes from pale rose to
violet at pH 8.9
(Titration of phosphoric acid to dibasic stage)
Universal or Multiple range indicators.
By suitably mixing certain indicators the colour change may be made to extend over a considerable portion of
pH range .
Such mixtures are called as universal indicators .
They are not suitable for quantitative analysis.
Can be used for qualitative purpose.
Ex- Dissolve 0.1g of phenolphthalein, 0.2g methyl red, 0.3 g methyl yellow, 0.4 g of bromothymol blue in 500
ml of absolute alcohol and add sufficient sodium hydroxide until the colour is yellow.
Colour change are-
pH 2 = red
pH 4= orange
pH 6= yellow
pH 8 = green
pH 10= Blue
Color Ranges of Indicators
Indicators: Color changes against pH
Application of Neutralization titration :
• For the determination of strong acid Eg .assay of HCL, H₂SO₄ and benzoic acid.
• For determination of strong base Eg assay of Na₂co₃,NaHCO₃.
• For determination of weak acid eg assay of phosphoric acid ,CH₃COOH.
• For determination of weak base eg assay of aminophylline.
• for the determination of salt example ,sodium acetate salt and ammonium chloride .
• To know % purity of acidic and basic substance.
Points to be summarized
• Chemical equillbrium, Ionic equillibrium
• Acid base theories (Arrhenius, lowry bronsted and Lewis theories with examples and limitations.
• Law of Mass action
• Strengths of acids and bases ( Dissociation constants- Ka, Kb, pKa, pKb, relationship of pKa and Ka , pKb
and Kb)
• Strong acids, strong base, weak acid and weak bases examples.
• Degree of dissociation α and equation.
• Ionic product of water (Kw)
• pH and pOH, Scale of pH derivation, pH and pOH formulae
• Le chaterliers principle
• Common ion effect (Mechanism with examples, application)
• Buffers, types, Mechanism, buffer capacity, Henderson hasselbalch equation, pH = pKa
• Solubility product (Ksp)
• Hydrolysis of salts
• Types of Salts (SASB, WASB, WBSA, WAWB), their pH when dissolved in water and examples.
• Titration and its concepts
• Titration curves (SASB, WASB, WBSA, WAWB), equations, significance of each.
• Polyprotic acids and their salts
• Polyfunctional acid base titrations
• Titration of amino acids.
• Acid- base indicators
Some important formulae's for acid base titrations
Kw = Ka. Kb
pKw = pKa + pKb
pKa + pKb = 14
Kw = 1 x 10 -14










][
][
HA
A
logpKpH a
moles of titrant = moles of analyte
(V titrant)(M titrant) = (V analyte)(M analyte)
pH = (pKw – pKb )+ log
[𝐀−]
[𝐇𝐀]
β =
𝒅𝑪 𝑩𝑶𝑯
𝒅𝒑𝑯
= -
𝒅𝑪𝑯𝑨
𝒅𝒑𝑯

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PHARMACEUTICAL ANALYSIS I - ACID BASE TITRATIONS

  • 1. Stoichiometry: • The study of relationship between the relative quantities of substances taking part in a reaction or forming a compound. • Stoichiometry provides the basis for a procedure called titration, which is used to determine the concentrations of acidic and basic solutions. • For example, in the reaction of sodium hydroxide and hydrogen chloride, 1 mole of NaOH neutralizes 1 mole of HCl: NaOH (aq) + HCl (aq)  NaCl (aq) + H2O (l)
  • 2. IONIC EQUILIBRIUM: Acid, base and salt when dissolved in water, gets ionized and form corresponding ion to some extent. Hence, equilibrium is between ion and unionized atom . Such equilibrium between ion and unionized atom in an aqueous solution is called as ionic equilibrium. H⁺ + OH⁻ H₂O CHEMICAL EQUILIBRIUM: Chemical equilibrium is a state of a chemical reaction at which the rate of a forward reaction is equal to the rate of the backward reaction. N₂(g) + 2H₂ (g) 2 NH₃(g)
  • 3. LE-CHATELIER’S PRINCIPLE In 1888, Le-chatelier, gave a qualitative rule, which helps chemists in the prediction of the effect of change of temperature, pressure, and concentration on equilibrium reactions. This rules can be stated as follows: a) Effect of concentration: • In a general reaction at equilibrium, if the concentration of one of the reactants (or both the reactants) is increased, then the equilibrium will shift towards the products. • Similarly, if the concentration of one of the reactants (or both the reactants) is decreased, then the equilibrium will shift towards the reactants. • Thus, in general forward reaction is favored by higher concentration of reactants, and the backward reaction is favored by lowering the concentration of the reactants.
  • 4. b) Effect of pressure: • Change of pressure is considered only for gaseous reactions at the state of equilibrium. In a gaseous reaction, at constant temperature and volume, the pressure applied is directly proportional to the total number of moles present in the reaction mixture. • Hence, if pressure is increased, the equilibrium is shifted to a direction in which the number of moles decrease (the volume decreases). • Similarly, if the pressure is decreased in which the number of moles increase (the volume increases). • Thus, in general at constant temperature, increase in pressure will favor a reaction, which is accompanied by a decrease in volume, and decrease in pressure will favor a reaction, which is accompanied by an increase in volume.
  • 5. c) Effect of temperature: If a chemical reaction at the state of equilibrium is exothermic, then lowering of temperature will favor the reaction (the equilibrium shifts to the right). Similarly, if a chemical reaction at a state of equilibrium is endothermic, the increase in temperature will favor the reaction (the equilibrium shifts to the right).
  • 6. THEORIES OF ACID AND BASES There are three theories of acid and base by which one can determine nature of unknown , substance either acid and base. A) ARRHENIUS THEORY OF ACIDS AND BASES: • Arrhenius, in 1887, suggested the theory of electrolyte dissociated of an aqueous medium and propose his theory to define acids and bases. • Thus, according to Arrhenius theory, an acid may be defined as a hydrogen containing substance , that dissociates to give hydrogen ions (H⁺) When dissolved in water. • For example: HCL + Water H⁺(aq) + Cl⁻(aq) According to Arrhenius theory, a base may be defined as hydroxyl substance, which dissociates to give hydroxyl ions (OH⁻ ions) when dissolved in water.
  • 7. • For example: NaOH + Water Na⁺(aq) + OH⁻(aq) Neutralisation: • It may be defined as a process, in which acid and base react completely with each other to produce salt and water. Acid + Base Salt + Water For example: HCl + NaOH NaCl(aq) + H₂O(aq) • On the basis of ionic theory, neutralization may be defined as a process in which H⁺ ions of an acid combine with OH⁻ ions of a base to form unionized water molecules. HCl + NaOH NaCl + H₂O (acid) (base) (salt) (water)
  • 8. Limitations of Arrhenius theory: • This concept is applicable to define acids and bases, only in aqueous solutions, but not for non- aqueous solutions or gaseous reactions. • Some compounds don’t contain hydroxyl group (OH⁻)and are basic in nature. The basic nature of such compounds isn’t explained by Arrhenius theory of acids and bases. For example- NH₃, amines. • According to this theory, H⁺ ions exists freely in an aqueous solution. But, in an aqueous solution, H⁺ ion is always hydrated to form hydronium ion. H⁺ + H₂O H₃O ⁺(aq) • According to this theory, an acid must contain hydrogen ion(H⁺ ion) and a base must contain a hydroxyl ion(OH⁻). But there are substances whose aqueous solutions are acidic or basic in nature without containing H⁺ or OH⁻ ions. For example FeCl ₃, CuSO₄ , CO₂. Show acidity in the aqueous solution and NaNO₃ , NH₃. Shows basic nature in the aqueous solution. FeCl₃ + 3H₂O Fe(OH)₃ + 3H₂O
  • 9. B) LOWRY AND BRONSTED CONCEPT OF ACIDS AND BASES: In 1923, Bronsted and Lowry, modified the definition of acids and bases which deals with protons exchange and it applies to aqueous as well as non-aqueous solutions. According to this theory, an acid may be defined as substance (molecule or ion) which donates protons protons(H⁺ ions). For example: HCl + H₂O H₃O⁺ + Cl⁻ (Acid) (Base) (Acid) (Base) Since HCl donate proton, they are Lowry-Bronsted acids. According to this theory, a base may be defined as substance which accepts protons(H⁺ ions). For example: NH₃ + H₂O NH₄ + OH⁻ (Base) (Acid) (Acid) (Base) since, NH₃ accepts proton from water, it is a Lowry-Bronsted base.
  • 10. Conjugate acid and base pairs: The pairs of substances which can be formed from one another by gain or by loss of proton, are known as conjugate acid base pairs. HCN + H₂O H₃O⁺ + CN⁻ (Acid) (Base) (Acid) (Base) Explanation: consider the following reaction: HCl + H₂O H₃O⁺ + Cl⁻ (Acid) (Base) (Acid) (Base) In this example , HCl donates a proton and acts as an acid. Water accepts a proton and so acts as a base, it is a reverse reaction, hydronium ion donates a proton and so acts as an acid. The chloride ion accepts a proton and thus behaves as a base. Thus, hydrochloric acid (HCl) is a conjugate acid of the base chloride ion(Cl⁻) and chloride ion is a conjugate base of hydronium ion(H₃O⁺) and hydronium ion is a conjugate acid of water.
  • 11. Some examples of conjugate acid-base pairs are given below: ACID₁ BASE₂ ACID₂ BASE₁ HCl H₂O H₃O⁺ Cl⁻ HNO₃ H₂O H₃O⁺ NO₃⁻ HSO₄⁻ H₂O H₃O⁺ HSO₄⁻ C) LEWIS CONCEPT OF ACIDS AND BASES: In 1938, G.N. LEWIS gave a more broad based definition of acids and bases, which don’t depend upon the presence of protons. According to Lewis, an acid may be defined as a substance which can accept a pair of electrons. For example: H⁺ ion, HCl, etc. are Lewis acids
  • 12. According to Lewis, a base is defined as a substance which can donate a pair of electron to form a co-ordinate covalent bond. For example: NH₃, H₂O, etc are Lewis bases. Explanation : In the following reaction the proton (H⁺) is a Lewis acid and ammonia(:NH₃) is a Lewis base, because a lone pair of electrons on nitrogen atom of ammonia is donated to the proton.
  • 13. Degree of dissociation The extent of dissociation of an electrolyte can be expressed in terms of the “degree of dissociation” Degree of dissociation is the fraction of the total number of moles of an acid or a base or an electrolyte that split up into its ions in an aqueous solution It is represented by α. α = dissociation at equilibrium total no. of moles present in solution initially SOME COMMON TERMS
  • 14. Strength of acids and bases: STRENGTH OF ACIDS: The strength of an acid or a base can be expressed in terms of their dissociation constants. Consider the dissociation of a weak acid HA in water. The equilibrium reaction will be: HA + H2O H3O+ + A- Applying the law of mass action, the equilibrium constant K can be given as: K=[H3O+][A-] [HA][H2O] Since water is present in large excess its concentration will be practically constant. K= [H3O+][A-] [HA] X Constant K X constant = [H3O][A-] [HA]
  • 15. Here Ka is the constant for a given acid and is called as acidity constant or Dissociation constant of that acid. Thus, stronger acids have high acidity constants and weak acids have low values. STRENGTH OF BASES : Consider the dissociation of a weak base BOH like ammonium hydroxide in water. The equilibrium reaction can be written as: BOH B+ + OH- By applying law of mass action to above equation, the equilibrium constant K is given as- Ka = [H+][A-] [HA]
  • 16. K=[B+][OH-] or Kb=[B+][OH-] [BOH] [BOH] Here Kb is a constant for a given base and is called basicity constant that determines the strength of the base. Thus strong base have higher Kb value and weak bases have lower Kb value. Examples of dissociation constants of some acids and bases: Acids Formula Ka Acetic acid CH3COOH 1.75 x 10 -5 Formic acid HCOOH 1.77 x 10 -2 Phosphoric acid H3PO4 7.55 x 10-2 Sulphric acid H2SO4 > 1 Base Formula Kb Ammonium hydroxide NH4OH 1.81 x 10-5 Sodium hydroxide NaOH > 1
  • 17. IONIZATION OF WATER Water is a weak electrolyte and hence it ionizes to a very small extent. It ionizes into hydronium ions and hydroxyl ions as follows: H2O H+ + OH- and H2O + H+ H3O+ The net ionization: 2H2O H3O+ + OH- or H2O H+ + OH- Ionic Product: Ionic product of water maybe defined as the product of molar concentrations of [H+] and [OH-] ions in pure water at a given temperature. It is represented as Kw. Explanation: The ionization of water can be written as: 2H2O H3O+ + OH- Applying law of mass action we get K=[H3O+][OH-] [H2O]2
  • 18. Water is present in large excess and it dissociated to a very small extent, hence its concentration will remain practically constant. K = [H3O+][OH-] [constant] K x constant = [H3O+][OH-] or Kw= [H3O+][OH-] Kw a constant known as Ionic product of water. For convinence [H3O+] is written as [H+]. Hence, Kw is a function of temperature and at 298K its value is 1.00 x 10-14 Since one molecule of water ionizes into one H+ ion and one OH- ion, [ H+ ] = [OH- ] in pure water. And Kw= [H+][H+] =[ H+]2 ----------- (a) Therefore, [H+] = Kw Similarly, [OH-] = Kw ----------by substituting [OH-] in place of [H+] Kw= [H+][OH+]
  • 19. And hence [H+] = 1 x 10−14 = 1 x 10-7 mol/dm3 [OH- ]= 1 x 10−14 = 1 x 10-7 mol/dm3 pH and pOH pH: It maybe defined as the negative logarithm to the base 10 of molar concentration of H+ pH = -log10 [H+ ] or pH = log10 1 [H+ ] pOH: It maybe defined as the negative logarithm to the base 10 of molar concentration of OH- pOH = - log10 [OH- ] or pH = log10 1 [OH+ ]
  • 20. To prove pH + pOH = 14 The ionic product of water can be written as: Kw= [H3O+][OH-] -------- (a) But Kw = 1 x 10-14 Substituting the value of Kw in equation (a) 1 x 10-14 = [H3O+][OH-] -------- Here [H3O+] is taken as [H+] hence 1 x 10-14 = [H+] [OH-]--------- (b) Taking logarithm on both sides of equation (b) log [H+] + log[OH- ] = log(1 x 10-14 ) Reversing the sign to above equation: -log [H+] -log[OH- ] = - log(1 x 10-14 ) -log [H+] -log[OH- ] = - log(1) - log (10-14 ) by rule log (a x b) = log a + log b and - log (a x b ) = - log a – log b
  • 21. But log 1 = 0, substituting the value in below eq: -log [H+] -log[OH- ] = - log(1) - log (10-14 ) -log [H+] -log[OH- ] = - log (10-14 ) By definition : -log [H+] = pH and -log[OH- ] = pOH hence substituting values in above eq. pH + pOH = - log10-14 pH + pOH = -(-14log10) (But log 10 = 1) Hence pH + pOH = -(-14 x 1 ) Therefore, pH + pOH = 14
  • 22. pH Scale Sorensen introduced a measure called pH scale to measure the acidity, alkalinity or neutal nature of an aqueous solution. For pure water or neutral solution [H+] = 1x 10-7 (From the value of ionic procuct of water. Ref Kw) Taking log of above equation we get- pH=7 Hence in an aqueous in which [H+] > 1 x 10-7 then pH<7 i.e. the solution is acidic. If in an aqueous solution [OH- ] > 1 x 10-7 then pH>7 i.e. the solution is basic/ alkaline.
  • 23. pH scale derivation : Acidic solution Basic Solution [H+] mol/dm3 pH [OH+] mol/dm3 pH pOH = 14 - pH 1 x 10 -6 6 1 x 10 -6 6 8 1 x 10 -5 5 1 x 10 -5 5 9 1 x 10 -4 4 1 x 10 -4 4 10 1 x 10 -3 3 1 x 10 -3 3 11 1 x 10 -2 2 1 x 10 -2 2 12 1 x 10 -1 1 1 x 10 -1 1 13
  • 24. TITRATION • Titration is a method for determining the concentration of a solution by reacting a known volume of unknown solution with a solution of known concentration. • To find the concentration of an acid solution- titrate the acid solution with a solution of a base of known concentration. • To find concentration of unknown base – titrate with acid of known concentration. • Burette – Solution of known concentration • Flask - unknown concentration solution • Indicator – used to determine end point
  • 25. Aqueous Acid-Base titrations/ Neutralization titrations • These are titrations in which acid is titrated/neutralized with a base or vice versa . The end product is salt and water. • It helps to determine the concentration of either acid or a base. • Equivalence point: The equivalence of an acid-base titration is the point at which there are equal amounts (in moles) of H3O+ and OH- in titration flask. • End point of titration – the point in a titration at which the indicator changes color. • The indicator should change color sharply at the equivalence point. • At the end point of the titration, - all the acid has been neutralized by the alkali - the solution in the conical flask contain salt and water only. H3O+ + OH- 2H2O
  • 26. Equivalence and end Points Equivalence point Related to the amount (moles) End point Related to the physical sign that is associated with the condition of chemical equivalence. The end point is expressed in range, the range should address to the equivalence point.
  • 27. In the acid base titrations, pH meter measures the pH of the acid solution in the beaker/flask, as a solution of a base with a known concentration is added from the burette. As the titration progresses, the pH meter is used to monitor change in pH.
  • 28. 1) A measured volume of an acidic or basic solution of unknown concentration is placed in a beaker/conical flask.(Solution of unknown concentration is called as analyte or titrand.) 2) The electrodes of a pH meter are immersed in this solution, and the initial pH of the solution is read and recorded. 3) A burette is filled with the titrating solution of known concentration. This is called the standard solution or titrant. 4) Measured volumes of the standard solution are added slowly and mixed into the solution in the beaker. 5) The pH is read and recorded after each addition. This process continues until the reaction reaches the equivalence point, which is the point at which moles of H+ ion from the acid equal moles of OH- ion from the base. (H+ = OH-) Titration Procedure
  • 29. Types of acid base titration curves/Neutralization curves Types of acid-base titrations Examples Strong acid and Strong Base Titration HCl and NaOH Strong acid and weak base titration HCl and NH3 Weak acid and strong base titration CH3COOH and NaOH Weak acid and weak base titration CH3COOH and NH3
  • 30. Titration Curve • A titration curve is a plot of pH vs. the amount of titrant added. • Typically the titrant is strong (completely) dissociated acid or base. • Such curves are useful for : 1. Determining endpoints 2. Dissociation constants of weak acids or bases.
  • 31. STRONG ACID STRONG BASE TITRATIONS • Strong acid (HCL) is titrated with Strong Base (NaOH) • Burette: NaOH (Known conc), Flask/Beaker = HCL Unknown conc) • Volume of NaOH added to unknown HCl is plotted on X axis • pH changed is plotted on Y axis • Graph of pH curve shows the change in pH versus volume of titrant added as the titration proceeds. Key points 1. pH increases slowly 2. pH changes quickly near the equivalence point. 3. The equivalence point of a strong acid— strong base titration = 7.00NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq) ie H+ = OH-
  • 32. The titration of a strong base with a strong acid is almost identical. Burette = HCL known conc Flask = NaOH unknown
  • 33. 1.Initially the pH measured is low, which reflects high amount of [H3O+]/[H+] of the strong acid. 2.pH increases gradually as acid is neutralized by the added base. 3.Suddenly the pH rises steeply. This occurs in the immediate vicinity of the equivalence point. For this type of titration the pH is 7.0 at the equivalence point. 4.Beyond this steep portion, the pH increases slowly as more base is added. Features of the Strong Acid-Strong Base Titration Curve
  • 34. Strong Acid-Strong Base Titration Curve This type of titrations can be explained in three parts 1. Before equivalence point 2. At equivalence point 3. After equivalence point Molarity (M) = 𝑁𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑙𝑖𝑡𝑟𝑒 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 (𝑉) Moles of titrant = (V titrant)(M titrant Moles of analyte = (V analyte)(M analyte) moles of titrant = moles of analyte (V titrant)(M titrant) = (V analyte)(M analyte)
  • 35. Consider titration of 100 ml of 1N HCL with 1 N NaOH. Change in pH value can be determined in 3 steps as: 1. pH value before equivalence: a) Before proceeding the titration, solution in flask is strong acid, ie 1M/1N HCL Concentration of H+ in strong acid can be calculated as- H+ = 1N pH = - log (H+) pH = - log (1) since log (1) = 0 So pH = 0 b) After adding 1 ml of base, total volume will be 100 ml of acid+ 1 ml of base = 101 ml [H+] = 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻 + 𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 [H+] = [𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻 + − 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑂𝐻 − ] 𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
  • 36. [H+] = 𝑚1 𝑣1 𝑜𝑓 𝐻𝐶𝑙 −[𝑚2 𝑣2 𝑜𝑓 𝑁𝑎𝑂𝐻] 𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 = 1𝑥100 −[1 𝑥 1] 100+1 = 100 −[1] 101 = 99 101 =0.98 [H+] = 0.98 ( since pH = - log (H+)) pH = - log (0.98) pH = 0.008 number of H+ ions remaining now will be 100 – 1= 99 meq or 99/101 eq/ lit
  • 37. c.) On addition of 50 ml of base If we add 50 ml base, to 100 ml of acid, then 50 ml of acid is unneutralised in 150ml (100 ml acid +50 ml base) [H+] = [𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻 + − 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑂𝐻 − ] 𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 [H+] = 𝑚1 𝑣1 𝑜𝑓 𝐻𝐶𝑙 −[𝑚2 𝑣2 𝑜𝑓 𝑁𝑎𝑂𝐻] 𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 = 1 𝑥 100 −[1 𝑥 50] 100+50 = 50 150 = 0.33 pH = - log (0.33) pH = 0.48
  • 38. Similary if we go on adding- 75 ml base added, H+ = 25 x1 /175 = 0.142, pH = 0.8 90 ml base added, H+ = 10 x 1 /190 = 0.052, pH = 1.28 99 ml base added, H+ = 1 x 1 /199 = 0.005, pH = 2.3 99.9 ml base added, H+ = 0.1 x 1 /199.9 = 0.0005 , pH =3.3 2. pH value at equivalence point: When 100 ml of base is added , pH will change sharply to 7 which is the equivalence point. As per definition of equivalence point- H+ = OH- pH = pOH (solution is neutral) Kw= 1 x 10-14 at 25 C hence pKw = 14 pKw = pH + pOH
  • 39. ie 14 = pH + pOH 14 = pH + pH 14 = 2pH pH = 7 3. pH after equivalence point: After steep slope, when more NaOH is added, there is a gradual rise in pH. At equivalence 100 ml of NaOH is required to neutralise 100 ml of HCL. At this point, the volume in the flask is 200 ml. A) After equivalence if 1 ml of NaOH is added, the total volume will be 201. Consider addition of 1 ml excess of alkali- [OH-] = [𝑚𝑜𝑙𝑒𝑠 𝑜𝑓𝑂𝐻 − 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓𝐻 + ] 𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 [OH-] = 𝑚1 𝑣1 𝑜𝑓𝑁𝑎𝑂𝐻 −[𝑚2 𝑣2 𝑜𝑓 𝐻𝐶𝑙] 𝑇𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 [OH-] = 1 𝑥 101 −[1 𝑥 100] 101+100
  • 40. [OH-] = 101 −[100 ] 201 [OH-] = 1 201 =0.0049 pOH = - log (OH-) so pOH = - log (0.0049) pOH = 2.3 pKw = pH + pOH ie 14 = pH + pOH pH = 14 - pOH pH = 14 - 2.3 = 11.7 B) Similarly after adding 20 ml of excess alkali pH = 12.96
  • 41. Sample Calculation: Strong Acid-Strong Base Titration Curve Consider the titration of 40.0 mL of 0.100 M HCl with 0.100 M NaOH Region 1. Before the equivalence point, after adding 20.0 mL of 0.100 M NaOH. (Half way to the equivalence point.) Amount of H3O+ remaining = Initial moles of H3O+ - Moles of OH- added baseaddedofvolumeacidofvolumeoriginal remainingOHof(mol)amount ]O[H 3 3     moles of titrant = moles of analyte (V titrant)(M titrant) = (V analyte)(M analyte) Three regions of titration curve exists
  • 42. Region 2. At the equivalence point, after adding 40.0 mL of 0.100 M NaOH. Initial moles of H3O+ = 0.0400 L x 0.100 M = 0.00400 M H3O+- Moles of OH- added = 0.0400 L x 0.100 M =0.00400 mol OH- baseaddedofvolumeacidofvolumeoriginal remainingOHof(mol)amount ]O[H 3 3     Sample Calculation: Strong Acid-Strong Base Titration Curve Region 3. After the equivalence point, after adding 50.0 mL of 0.100 M NaOH. (Now calculate excess OH-) Total moles of OH- = 0.0500 L x 0.100 M = 0.00500 mol OH- -Moles of H3O+ consumed = 0.0400 L x 0.100 M =0.00400 mol baseaddedofvolumeacidofvolumeoriginal remainingOHof(mol)amount ][OH    
  • 44. Definition: It is defined as a solution, which resists change in its pH value caused by addition of small amount of acid or base or on dilution or keeping it for a certain long time. It is required to maintain pH of reaction at optimum value. Types and preparation of buffer solutions: 1. Acidic buffer: An acidic buffer can be prepared by dissolving Weak acid such as acetic acid (CH3COOH) and its Salt of strong base such as sodium acetate (CH3COONa) in water. CH3COOH + CH3COONa Acidic buffer Acidic buffers have pH less than 7. Other ex. Is Ethanoic acid and sodium ethanoate in solution. Equal molar conc of both gives pH of 4.76.
  • 45. You can change the pH of the buffer solution by changing ratio of acid to salt. 2. Basic buffer: A basic buffer can be prepared by dissolving Weak base such as ammonium hydroxide (NH4OH) and its Salt of strong acid such as Ammonium chloride (NH4Cl) in water. NH4OH + NH4Cl Basic buffer Basic buffers have pH more than 7. ex. ammonia and ammonium chloride in solution. Equal molar conc of both gives pH of 9.25. 3. Buffer solution of single salt: It is prepared by dissolving single salt which must be of weak acid and a weak base such as ammonium acetate in water. CH3COONH4 + Water Buffer (single salt)
  • 46. Buffer action: It is the property of buffer solution to resist change in its pH value on adding a small amount of acid or a base. Mechanism of Buffer action 1. MECHANISM OF ACIDIC BUFFER: Suppose an acid buffer if prepared from a weak acid , CH3COOH and it salt of strong base, CH3COONa, by dissolving them in water. Acetic acid being a week acid, ionised feebly and will give only few H+ ions in solution. CH3COOH CH3COO- + H+ Sodium acetate being a strong electrolyte will completely get dissociated as- CH3COONa CH3COO- + Na+ The solution will have large no. of CH3COO- ions now.
  • 47. I. Addition of strong acid like HCl to the buffer: When a strong acid like HCl is added to the solution, it dissociated completely as follows: HCl H+ + Cl- Now H+ ions of HCl wil combine with acetate ions, CH3COO- and from unionised CH3COOH molecules and as a result the pH of the buffer solution does not change to considerable extent. In this way H+ are neutralised by CH3COO- ions and maintains the pH of buffer constant. CH3COO- + H+ CH3COOH (unionised) This resistance to change in pH value of a buffer solution on adding a small amount of a strong acid is called ‘Reserve Basisity’ due to CH3COO- ions .
  • 48. II. Addition of strong base like NaOH to the buffer. When a strong base like NaOH is added to the solution of buffer, it dissociates completely as NaOH Na+ + OH- Now OH- of NaOH combine with CH3COOH and for acetate ions , CH3COO- and water. CH3COOH + OH- CH3COO- + H2O In this way hydroxyl ions are neutralised by acetic acid and hence pH of the buffer does not change to a considerable extent even if base is added to it. This resistance to change in pH value of buffer solution on adding a small amount of strong base is called as ‘Reserve acidity’ due to acetic acid.
  • 49. 2. MECHANISM OF BASIC BUFFER: Suppose a basic buffer is prepared from a weak base , NH4OH and its salt of strong acid,NH4Cl by dissolving them in water. NH4OH is a weak base , hence it will be feebly ionised and the solution will contain very few OH- ions. NH4OH NH4 + + Cl - NH4Cl been an electrolyte will be completely dissociated as – NH4Cl NH4 + + Cl – Now the solution has large number of NH4 + ions.
  • 50. I. Addition of strong acid like HCl to the buffer: When a strong acid like HCl is added in the solution, it dissociates almost completely as: HCl H+ + Cl- NowH+ of HCl combine with NH4OH and form water and NH4 + ions. NH4OH + H+ NH4 + + H2O In this way , H+ ions are neutralised by NH4OH and as a result, the pH of buffer solution does not change to any extent. This resistance to change in pH value of a buffer solution on adding a small amont of strong acid is called ‘Reserve basicity’ due to NH4OH.
  • 51. II. Addition of strong base like NaOH to the buffer. When a strong base like NaOH is added to the solution of buffer, it dissociates completely as NaOH Na+ + OH- Now OH- of NaOH combine with NH4 + ions and form unionised NH4OH. NH4 + + OH- NH4OH In this way hydroxyl ions are neutralised by NH4 + ions and hence pH of the buffer does not change to a considerable extent even if base is added to it. This resistance to change in pH value of buffer solution on adding a small amount of strong base is called as ‘Reserve acidity’ due to NH4 + ions.
  • 52. BUFFERING SYSTEMS BUFFERING pH ranges at 25 C Hydrochloric acid / Potassium chloride 1.0- 2.2 Citric acid /Sodium citrate 3.0 - 6.2 Potassium chloride /Sodium hydroxide 12.0-13 Potassium hydrogen Pthalate/ Sodium hydroxide. 4.1-5.9 Properties of buffer: 1. Buffer solution has definite pH value. 2. The pH value of buffer does not change even if small amount of an acid or base is added into it. 3. Value of pH does not change even if it is kept for long time. 4. pH value of buffer solution remains constant even if diluted with water.
  • 53. BUFFERS The dissociation of an acid can be described by an equilibrium expression: HA + H20 H3O+ + A- Applying law of mass action, equilibrium constant will be Ka = [H3O+][A-] [HA] Taking the negative log of both sides of the equation gives -logKa = -log [H3O+][A-] [HA] -logKa = -log [H3O+] + (-log [A-] ) [HA] -logKa = -log [H3O+] - (log [A-] ) [HA]
  • 54. By definition, pKa = - logKa and pH = -log[H3O+], so pka = pH – log [A-] [HA] This equation can then be rearranged to give the Henderson-Hasselbalch equation: HA + H20 H3O+ + A- This is Henderson-Hasselbalch equation for calculation of pH of buffers. Ex: CH3COOH + H2O CH3COO- + H3O+ pH = pKa + log [A-] or pH = pKa + log [conjugate base] [HA] [acid]
  • 55. HA + H20 H3O+ + A- Consider reverse equation, here A- is a base and HA is conjugate acid. Ka = [H3O+][A-] = 𝐾𝑤 𝐾𝑏 -------------------------- Kw = Ka.Kb [HA] [H+] = Ka [𝐻𝐴] [𝐴−] = 𝐾𝑤 𝐾𝑏 -log [H+] = - log Ka – log [𝐻𝐴] [𝐴−] OR -log [H+] = - log 𝐾𝑤 𝐾𝑏 – log [𝐻𝐴] [𝐴−] pH = pKa + log [𝐴−] [𝐻𝐴] and for basic buffer pOH = pKb + log 𝐵𝐻 𝑐𝑜𝑛𝑗 𝑎𝑐𝑖𝑑 𝐵− 𝑏𝑎𝑠𝑒 For buffers formed from mixture of weak base and strong acid ex. NH4OH and HCL pH = (pKw – pKb )+ log [𝑨−] [𝑯𝑨] ------- from eq pKw = pKa + pKb
  • 56. Buffer capacity: • The amount of acid or base that can be added without causing a large change in pH is the buffering capacity of the solution. • Or the amount of acid or base consumed by 1 litre of buffer to change the pH by one unit. • The capacity is determined by the concentrations of HA and A- . • Buffering capacity increases with increase in the concentrations of buffering species • Higher the concentration, more the acid and base it can tolerate. The buffer capacity, buffer intensity or buffer index is represented as β = 𝑑𝐶 𝐵𝑂𝐻 𝑑𝑝𝐻 = - 𝑑𝐶 𝐻𝐴 𝑑𝑝𝐻 Where, C BOH= number of moles per litre of strong base C HA = number of moles per litre of strong acid needed to bring a change of about dpH. Buffer capacity is a positive number. The larger the value of β , the more resistant the solution is to pH change. Buffering capacity is governed by 1. ratio of HA to A-. 2. Concentration of HA and A-. Larger the conc. larger is the capacity. It is maximum when the ratio is one or unity. That is when pH = pKa pH = pKa + log 1 1 pH = pKa. This corresponds to the mid point of titration of weak acid strong base. The buffering capacity is satisfactory over a pH range of pKa ± 1 or pKb ± 1Since adding a strong acid in buffer decreases the pH value. So dCHA/dpH is negative. β decreases rapidly if concentration ratio of [A−]/[HA] becomes larger or small than 1 . pKa of acid or pKb of base chosen should lie withing ± 1 unit of desired pH to get buffer of desired capacity.
  • 57. WEAK ACID AND STRONG BASE TITRATION CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l) CH3COONa + H2O (l) CH3COOH + NaOH In ionic form, CH3COO- + Na+ + H2O CH3COOH + Na+ +OH- Cancelling common ions on both sides, net reaction is CH3COO- (aq) + H2O (l)) CH3COOH + OH- (aq) From this equation we can conclude that if weak acid is titrated with strong base then the pH of the solution will be more than 7, i.e. it will show basic pH. (At equivalence point- pH>7), since the solution contains OH- ions .
  • 58. WEAK ACID AND STRONG BASE TITRATION
  • 59. • At the beginning, the solution contains only a weak acid or a weak base, and the pH is calculated from the concentration of that and its dissociation constant. • After various increment of , the solution consists of a series of buffers. • The pH of each buffer can be calculated from the analytical concentration of the conjugate base or acid and the residual concentration of the weak acid or base. • At equivalence point, the solution contains only the conjugate of the weak acid or base being titrated, and the pH is calculated from the concentration of this product. • Beyond the equivalence point, the excess of strong acid or base titrate represses the acidic or basic character of the reaction product to such an extent that the pH is governed largely by the concentration of the excess titrant. TITRATION CURVES FOR WEAK ACIDS
  • 60.
  • 61. • You might think that all titrations must have an equivalence point at pH 7 because that is the point at which concentrations of hydrogen ions and hydroxide ions are equal and the solution is neutral. • But some titrations have equivalence points at pH values less than 7, and some have equivalence points at pH values greater than 7. • These differences occur because of reactions between the newly formed salts and water – salt hydrolysis. (Behaviour of salts in water) • Some salts are basic (weak acid, strong base) and some salts are acidic (strong acid, weak base). • The previous slide shows that the equivalence point for the titration of HPr (a weak acid) with NaOH (a strong base) lies at pH 8.80.
  • 62. Titration of a Weak Acid with a Strong Base A. Addition of a strong base to a weak acid forms a Buffer Solution HA + OH- A- + H2O B. Important Points 1. pH increases more rapidly at the start than for a strong acid 2. pH levels off near pKa due to HA/A- buffering effect pH = pKa + log([A-]/[HA]) = pKa + log(1) = pKa (when [A-] = [HA]) 3. Curve is steepest near equivalence point. Equivalence Point > 7.0 4. Curve is similar to strong acid—strong base after eq. pt. where OH- is major
  • 63. For Weak acid and Strong base titration curve will have four regions: 1. pH before equivalence point 2. pH of Buffer region 3. pH at Equivalence point 4. pH after equivalence point CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l) 1. pH before equivalence point: Before any added base, just weak acid (HA) in water pH determined by Ka Ka= [H+] [A-]/[HA] 2. pH of buffer region: With addition of strong base  buffer region pH determined by Henderson Hasselbach equation           ][ ][ HA A logpKpH a 3. pH at equivalence point all HA is converted into A- Weak base with pH determined by Kb 4. pH after equivalence point: Beyond equivalence point, excess strong base is added to A- solution. pH is determined by strong base. Similar to titration of strong acid with strong base. Kb = [HA] [OH-]/[A-]
  • 64. Calculations for Neutralization curve for weak acid and strong base- Consider titration between 100ml of 0.1 N CH3COOH and 0.1 N NaOH (KaCH3COOH= 1.75x10-5 pka= 4.74). Change in pH value is calculated by 4 steps. 1. pH value before equivalence point. 2. pH at buffer region 3. pH value at equivalence point. 4. pH value after equivalence point. 1) pH value before equivalence point: Before proceeding the titration, the solution in conical flask is 0.1 N CH3COOH. For weak acid concentration of H+ can also be calculated by formula- H+ = 𝑲𝒂. 𝑪 = 1. 75 𝑥 10 − 5 𝑥 0.1 hence H+= 0.0013 so pH = 2.87
  • 65. 2. pH at buffer region: After addition of 50 ml 0.1 N NaOH , 50 ml of NaOH react with 50 ML OF 0.1 N CH3COOH and 50 ml re CH3COOH remains unreacted. CH3COOH + NaOH ↔ CH3COONa + H2O This is the phase where buffer if formed and is called as a buffer region. It consist of mixture of weak acid and salt of weak acid, so pH value is calculated by formula; pH = pka + log Conc of [CH3COONa] Conc of [CH3COOH] Concentration of CH3COOH = M1V1CH3COOH M2V2[CH3COOH] = 0.1 X 50 ml 150 Concentration of CH3COOH = 3.33 x 10-2 moles
  • 66. At this stage concentration of CH3COONa = concentration of CH3COOH so CH3COONa =3.33 x 10-2 moles pH = pKa + log Conc of [CH3COONa] Conc of [CH3COOH] as, Ka = 1.75 X 10−5 But pKa = - log Ka So pKa = 4.74 Substituting the values in the equation of pH we get, pH = 4.74 + log 3.33 x 10−2 3.33 x 10−2 pH = 4.74 + log (1) = 4.74 + 0 pH = 4.74
  • 67. 3) pH value at equivalence point. (achieved when 100 ml of NaOH is added.) pH = ½pKw ½ pKa+ ½ log(c) = ½(14) + ½(4.74)+ ½ log(0.1) = 7+2.37- 0.5 pH= 8.72 4) After equivalence point solution contain excess of OH- ion causes sharp increase in pH value, solution becomes alkaline.
  • 68. 1.The initial pH is higher. 2.A gradually rising portion of the curve, called the buffer region, appears before the steep rise to the equivalence point. 3.The pH at the equivalence point is greater than 7.00. 4.The steep rise interval is less pronounced. The four Major Differences Between a Strong Acid-Strong Base Titration Curve and a Weak Acid-Strong Base Titration Curve
  • 69. HCl (aq) + NH3 (aq) NH4Cl (aq) NH4Cl + H2O NH4OH + HCL In ioni form- NH4 + + Cl- + H2O NH4OH + H++CL- NH4 + (aq) + H2O (l) NH3 (aq) + H+ (aq) At equivalence point (pH < 7) WEAK BASE STRONG ACID TITRATIONS
  • 70. WEAK BASE STRONG ACID TITRATIONS
  • 71. For Weak base and Strong acid titration curve will have four regions: 1. pH before equivalence point Due to base NH3 initialy before titration Can be determined by Kb value Kb = [NH4+] [OH-]/ NH4OH 2. pH of Buffer region By Henderson hasselbalch equation put pKa= Kw- pKb 3. pH at Equivalence point All base is converted into its conjugated acid. Hence Ka is determined 4. pH after equivalence point Beyond equivalence point, excess strong acid is added. pH is determined by strong acid. Similar to titration of strong acid with strong base.           ][ ][ HA A logpKpH a
  • 72. Calculations for CURvE BETWEEN WEAK BASE AND STRONG ACID Consider titration of 100ml of 0.1 N aqueous ammonia solution with 0.1N HCl {Kb(NH4OH) =1.75×10ˉ5, pKb = 4.74} Change in pH value calculated in 4 steps : 1. pH value before equivalence point. 2. pH at buffer region 3. pH value at equivalence point. 4. pH value after equivalence point. 1) pH value at equivalence point: a. Before proceeding titration conical flask contain 100ml of 0.1 N NH4OH Solution, OH -=√𝐾𝑏.C OH- =√(1.75 × 10 − 5) × (0.1) OH- = 1.3×10-3 but pOH = - log(OH-) therefore pOH=2.87
  • 73. Now pKw = pH + pOH Hence pH = pkw - pOH = 14 - 2.87 pH = 11.73 2. pH at buffer region: After addition 50ml 0f 0.1HCl ,it react with ammonia and form buffer solution. NH4OH + HCl ↔ NH4Cl + H2O It is a buffer solution consist of mixture of weak base NH4OH and its salt (NH4CL)so pH value calculated by buffer formula:
  • 74. pOH = pkb - log Conc of weak base conc of its salt = 4.74 - log weak base its salt(conc.of HCl) As per equation need to calculate conc.of weak base (NH4OH) and its salt (NH4Cl) Concentration of weak base NH4OH = m moles of base total volume = M1V1 150 = 0.1×50 150 = 50 150 =3.33×10-3
  • 75. Conc.of HCl = volume of HCl × molarity of HCL Total volume = 50×0.1 150 = 3.33× 10-3 Conc.of HCl = conc. of salt put the value in above equation; pOH = 4.74 - log(1) = 4.74 As pOH = pkw-pOH = 14-4.74 pH = 9.26
  • 76. 2. pH value at equivalence point is calculated by formula; pH = 1 2 pkw - 1 2 pkb - 1 2 log(c) = 1 2 (14) - 1 2 (4.74) - 1 2 log(0.1) pH =5.28 3. pH value after equivalence point: After equivalence point solution contain excess of HCl ,causes sharp decrease in pH value , solution become more acidic.
  • 77. The four Major Differences Between a Weak Acid-Strong Base Titration Curve and a Weak Base-Strong Acid Titration Curve 1.The initial pH is above 7.00. 2.A gradually decreasing portion of the curve, called the buffer region, appears before a steep fall to the equivalence point. 3.The pH at the equivalence point is less than 7.00. 4.Thereafter, the pH decreases slowly as excess strong acid is added.
  • 78. WEAK ACIDS AND WEAK BASE TITRATIONS Consider titration 100 ml of 0.1N acetic acid with 0.1N ammonia CH3COOH + NH3 CH3COO NH4 When salt dissolved in water, CH3COO NH4 + H2O CH3COOH + NH4OH Ionic form CH3COO- + NH4 + + H2O CH3COOH + NH4OH No formation of H+ and OH- ions. Hence pH is 7 At equivalence point, pH = pKw + 1/2pKa –pKb pH = 7.0 Change of pH near equivalence point and during whole neutralization is very gradual . Hence end point is not detected by ordinary indicator. Mixed indicators are required.
  • 79. Types of acids and bases ACIDS 1. Monoprotic These acids loose only one proton in the reaction. They probably have only one proton with them. Ex- HCL, HNO3, HBr 2. Polyprotic /Polyfunctional acids(Diprotic, triprotic etc) a) Polyprotic acids are specific acids that are capable of losing more than a single proton per molecule in acid-base reactions. b) These acids that have more than one ionizable H+atom per molecule. c) Protons are lost through several stages (one at each stage). 1. Diprotic acids: These have two hydrogen ions for donation and act as bronsted acid. Ex- H2 SO4, H2 CO3, H2S, chromic acid (H2CrO4), and oxalic acid (H2C2O4) 2. Triprotic acids These can donate three hydrogen ions and act as bronsted acids. Ex. phosphoric acid (H3PO4) and citric acid C6H8O7
  • 80. Consider H2SO4 (Diprotic acid) • pka helps us to determine the ease with which it looses protons.(Strength) • There is usually a large difference in the ease with which these acids lose the first and second (or second and third) protons. • Sulphuric acid will not loose both the protons at a time in water.(It will loose one proton first and then the next). • Sulfuric acid is a strong diprotic acid so it will have two dissociation constants as Ka1 and Ka2 since it looses one proton first and second next. Hence for two reactions, it will have two Ka constants. For Triprotic acid, they will have three acidity constants. Conjugate bases Ka1 Ka2
  • 81. Polyprotic acids and their salts 𝑝𝐾𝑎 = − log( 1 𝐾𝑎 )
  • 82. Buffer calculations for polyprotic acids The anion on the right hand side of each ionization step can be considered as salt (conjugate base) of the acid. Hence from the previous equation, H2PO4 - is the salt of H3PO4 HPO4 -- -- is the salt of H2PO4 - PO4 --- is salt of HPO4 — Each of the pairs constitute a buffer system. Optimum buffering capacity of each pair occurs when pH = pKa Use Henderson hasselbalch equation           ][ ][ HA A logpKpH a
  • 83. SALTS OF POLPROTIC ACIDS The salts of polyprotic acids can be acidic or basic. Example- Phosphoric acid H3PO4 Its protonated salts = H2PO4 - , HPO4 -- possess both acidic and basic properties. Hence are amphoteric salts. Its non-protonated salt = PO4 --- is a base and has basic property.
  • 84. Concentraion of H+ in system calculated by
  • 85. Common Polyprotic Acids Formula Strong/Weak Acid Number of Ionizable Hydrogens Ka1 Ka2 Ka3 Sulfuric acid H2SO4 Strong 2 (diprotic) Very Large 1.1E-2 Sulfurous acid H2SO3 Weak 2 (diprotic) 1.3E-2 6.2E-8 Phosphoric acid H3PO4 Weak 3 (triprotic) 7.1E-3 6.3E-8 4.2E- 13 Carbonic acid H2CO3 Weak 2 (diprotic) 4.4E-7 4.7E-11 Hydrosulfuric acid or Hydrogen sulfide H2S Weak 2 (diprotic) 1.0E-7 1E-19 Oxalic acid H2C2O4 Weak 2 (diprotic) 5.4E-2 5.3E-5
  • 86. BASES 1. Monobasic (Which can accept only one H+) – Ex. NaOH, NH4OH, KOH 2. Polybasic /Polyfunctional bases (Dibasic, tribasic etc) Polyprotic Bases are bases that can accept more than one H+ ion, or proton in acid-base reactions. Common Polyprotic Bases Formula Strong/Weak Base Diprotic/Triprotic Base Phosphate ion PO4 3- Weak Triprotic Sulfate ion SO4 2- Very Weak Diprotic Carbonate ion CO3 2- Strong Diprotic
  • 87. Titration Curve: Polyprotic acids Titration Curve of 0.100M H2SO3 with 0.100 M NaOH • Curve for the titration of a weak polyprotic acid. • Titrating 40.00mL of 0.1000M H2SO3 with 0.1M NaOH leads to a curve with two buffer regions and two equivalence points. • Because the Ka values are separated by several orders of magnitude, in effect the titration curve looks like two weak acid strong base curves attached. • The pH of the first equivalence point is below 7 because the solution contains HSO3 which is a stronger acid than it is a base. • Ka of HSO3 = 6.5•10-8; • Kb of HSO3 = 7.1•10-13
  • 88. FEATURES 1. The loss of each mole of H+ shows up as separate equivalence point. 2. The pH at the midpoint of the buffer region is equal to the pKa of that acid species. 3. The same volume of added base is required to remove each mole of H+.
  • 89. Titration Curve: Poly-Basic Titration Curve of 0.10 M Na2CO3 with 0.10 M HCl. 1. CO3 + H+ HCO3 + H2O 2. HCO3 + H+ H2CO3 + H2O
  • 90. 1.Multiple Inflection Points = Multiple Equivalence Points will be seen 2.The volume required to reach each equivalence point will be the same CO3 2- + H+ HCO3 - Kb1 = KW/Ka2 = 1.8 x 10-4 (pKb1 = 3.74) HCO3 - + H+ H2CO3 Kb2 = KW/Ka1 = 2.3 x 10-8 (pKb2 = 7.64) ½ Eq. pt 1 Eq. pt 1 ½ Eq. pt 2 Eq. pt 2 pKa2 = 10.26 pKb1 = 3.74 pKa2 = 6.36 pKb1 = 7.64
  • 91. SUMMARY Weak base added to weak acid Polyfunctional /polyprotic acids Polyfunctional /polyprotic bases
  • 92. TITRATION OF AMINO ACIDS Each amino acid (except for proline) has: 1. A carboxyl group (-COO-) . 2. An amino group (-NH3 +) . 3. Side chain ("R-group") bonded to the α-carbon atom. These carboxyl and amino groups are combined in peptide linkage. STRUCTURE OF THE AMINO ACIDS Proteins are complex structure made up of amino acids. R---C---COOH NH2 R’
  • 93. • These are amphoteric substances that contain both acidic groups (COOH) and basic group (NH2 ). • Hence they can act as both acid and base. • In aqueous solutions, these substances tend to undergo internal proton transfer from the carboxylic acid group to amino group since R NH2 is stronger base than RCOO- • The result is Zwitter ion. Since they are amphoteric they can be titrated either with strong acid or strong base R---CH---COOH NH2 R---CH---COO- NH3 + Zwitter ion
  • 94. Classification of Amino Acids They classified according to the side chain: 1. Amino acids with nonpolar side chains. 2. Aromatic R Groups. 3. Amino acids with uncharged polar side chains. 4. Positively Charged (Basic) R Groups. 5. Amino acids with acidic side chains.
  • 95.
  • 96. ACIDIC AND BASIC PROPERTIES OF AMINO ACIDS Amino acids in aqueous solution contain weakly acidic α-carboxyl groups and weakly basic α-amino groups. Each of the acidic and basic amino acids contains an ionizable group in its side chain. Thus, both free and some of the combined amino acids in peptide linkages can act as buffers. The concentration of a weak acid (HA) and its conjugate base(A-) is described by the Henderson-Hasselbalch equation.
  • 97. For the reaction (HA A- + H+ ) [H+] [A-] Ka = ───── ------ (1) [HA] By solving for the [H+] in the above equation, taking the logarithm of both sides of the equation, multiplying both sides of the equation by -1, and substituting pH = -log [H+] and pKa = -log [Ka] we obtain: [A-] pH = pKa + log ─── ------ (2) [HA] It is the (Henderson-Hasselbalch equation)
  • 98. Titration Solution of an amino acid R---CH---COOH NH2 R---CH---COO- NH3 + Zwitter ion Ka1 Ka2 TTITRATION OF ZWITTER ION OF AMINOACIDS WITH ACIDS AND BASES
  • 99. Titration of zwitter ion with strong acid and base • Conjugate acid of the zwitter ion can be considered as weak polyprotic acid. • It will ionize to form first amphoteric zwitter ion and on second ionization it will give conjugate base of zwitter ion ie salt of weak acid. • When zwitter ion of amino acid are titrated with strong acid, buffer region is established containing the conjugate acid and zwitter ion .ie salt of weak acid • At half equivalence point pH = pKa1 • At equivalence point pH is determined by Ka, • For buffer region: pH is determined by Henderson hasselbalch equation. • When zwitter ion formed is titrated with strong base, buffer region will contain zwitter ion (which will act as strong acid) and it conjugate base. • Half of the equivalence point will be pH = pKa2 • Equivalence point can be determined by Kb Kb = 𝐾𝑤 𝐾𝑎 Since amino acids contain more than one number of NH2 and COOH groups , titration will yield stepwise end point like other polyprotic acids.
  • 101. Titration curve of glycine = = From acid to zwitter ion, 1. pH determined by ka 2. Buffers at this stage determined by Henderson equation. 3.At equivalence point conjugate base is formed hence Kb is determined . With help of kb, Ka is calculated using eq, Ka = Kw/Kb 4.Half equivalence point is pH = pKa 5. Glycine will have 2 dissociation const as Ka1 & Ka2 due to diprotic nature
  • 102. ACID-BASE INDICATORS • Equivalence Point The point at which the two solutions used in a titration are present in chemically equivalent amounts is the equivalence point. • End Point The point in a titration at which an indicator changes color and we can visually detect it is called the end point of the indicator. • The equivalence point of an acid-base titration can be determined by- 1. pH meter 2. Indicators ( Chemical dyes)
  • 103. A. Finding the equivalence point of a titration Use a pH meter •Plot pH versus titrant volume •Center vertical region = equivalence point Use an Acid-Base Indicator •Indicators are chemical species, whose colors are affected by acidic and basic solutions & are called acid-base indicators. •An acid–base indicator is a weak acid or base. •Acid-base indicators are compounds whose colors are sensitive to pH or change color based on pH. Acid base indicators are weak organic acids /bases that dissociate slightly in aqueous solutions to form ions.
  • 104. • The pH range over which an indicator changes color is called its transition interval. • Consider the above equation HIn H+ + In- Where , HIn is indicator which is a weak acid (one color) In- is its conjugate base. (other color) • HIn and In− will have different colors. One of the “colors” may be colorless. • The indicators can change colour because their ions have colors that are different from undissociated molecule when pH changes. • If a solution is acidic, [H3O+]/[H+] is high. When indicator (weak acid) is added , few [H+] are dissociated). Hence [H3O+]/[H+] is a common ion and so it suppresses the ionization of the indicator acid, and we see the color of HA. (i.e indicator will show color of HA).
  • 105. • Hence in acidic solutions, most of the indicator is HIn (it is undissociated due to common ion effect) • In basic solutions, most of the indicator is In– ie dissociated form of conjugate base of that indicator.( As we proceed for titration with a base, the pH will change to basic side. Hence the common ion effect will be reduced and the indicator which is a weak acid will undergo dissociation to some extent to In– , which is a dissociated form or conjugate base of the indicator molecule.) • Indicators that change color at pH lower than 7 are stronger acids than the other types of indicators. They tend to ionize more than the others. • Indicators that undergo transition in the higher pH or pH more than 7 range are weaker acids. • Each indicator its own particular pH or pH ranges over which it changes color.
  • 106. For Strong acid strong base titration: • The neutralization of strong acids with strong bases produces a salt solution with a pH of 7. • Indicators that undergo transition at about pH 7 are used to determine the equivalence point of strong-acid/strong base titrations. Bromthymol blue is a good choice for the titration of a strong acid with a strong base For strong acid weak base titrations: • The equivalence point of a strong-acid/weak-base titration is acidic.( pH less than 7) • Indicators that change color at pH lower than 7 are used to determine the equivalence point of strong-acid/weak- base titrations. For weak acid strong base titrations: • The equivalence point of a weak-acid/strong-base titration is basic. • Indicators that change color at pH higher than 7 are used to determine the equivalence point of weak-acid/strong- base titrations. Phenolphthalein changes color at the equivalence point of a titration of a weak acid with a strong base. Types of indicators used for different titrations For weak acid weak base titration, since endpoint is difficult to determine, mixed type of indicators are used.
  • 107.
  • 108. 1. Choose an indicator that changes color at the equivalence point of titration. 2. Remember that the role of the indicator is to indicate to you, by means of a color change, that just enough of the titrating solution has been added to neutralize the unknown solution. 3. End Point = when the indicator changes color. 4. If you have chosen the wrong indicator, the end point will be different than the eq. pt. 5. According to the type of titrations, whether SASA, SAWB, WASB or WAWB titrations, consider the pH at which equivalence point occurs. 6. Titrations whose equivalence point is below 7, indicator should change color at pH below 7. 7. Titrations whose equivalence point is above7, indicator should change color at pH above 7. 8. Titrations whose equivalence point is at 7, indicator should change color at pH 7. How to select an indicator?
  • 109. pH of acid base reaction at its equivalent point Acid Base pH at equivalence point Indicators Strong Strong = 7 (neutral) Methyl orange Phenolphthalein Strong Weak < 7 (acidic) Methyl orange Weak Strong > 7 (basic) Phenolphthalein Weak Weak pH depend on Ka and Kb of acid & base conc. Mixed indicators
  • 110. Colours of indicator at different pH
  • 111. THEORIES FOR ACID BASE INDICATORS Two theories are postulated for acid base indicator as follows: 1. Ostwald theory 2. Resonance theory A. OSTWALD THEORY • The first theory to explain the behaviour of indicators was put forth by W. Ostwald. • According to this theory , the undissociated indicator acid HIn or base InOH has a color different than its ion. • For an acid indicator equivalence can be written as : HIn H+ + In- • In acid solution, there is depression of ionisation of indicator due to common ion effect. Hence initial concentration of HIn is greater than the In- and hence the color exhibited will be that of the unionised form of acid. • As the titration with base proceeds , the alkali medium promotes removal of H+ and hence there is gradual increase in concentration of ionised form In- and the solution acquires colour of ionised form. • Apply law of mass action- KIna = [H+][In− ] [𝐻In ]
  • 112. log H+ = log KIna + log 𝐻In [In− ] Taking negative log on both sides -log H+ = - log KIna - log 𝐻In [In− ] = - log KIna + log In− [𝐻𝐼n] where KIna = Dissociation constant of indicator. The colour of indicator depends upon the ratio of concentration of ionised and unionised form and is directly proportional to pH. For base indicator In OH In + + OH- KInb = In + [OH − ] [In OH] [OH − ] = KInb [In OH] [In +] pH = p KIna + log In− [𝐻𝐼n]
  • 113. Now Kw = [H+] [OH-] [H+] = Kw [OH] Substituting the value of [OH] in the above equation. [H+] = Kw [𝐼𝑛 + ] KInb[In OH] Taking log on both sides, For an acid indicator from dissociation constant equation, In− [𝐻𝐼n] = KIn [H+] The colour change is affected by H+ concentration. Since this change is gradual, the colour change of indicator is also gradual. To detect the colour change of the indicator , the ratio of In− [𝐻𝐼n] must be at least 1 10 . Hence pH at which observable change occurred will be- pH = p KIn + log In− [𝐻𝐼n] In− [𝐻𝐼n] = 0.1 pH = pKIn + log 0.1 pH = pKIn - 1 For basic side, ratio of In− [𝐻𝐼n] = 10/1= 10 pH = pKIn + log 10 pH = pKIn + 1 Therefore general pH range in colour change observed is pH = pKIn ± 1 This is transition interval of indicator. pH = pKw – K𝐼nb + log [In OH] 𝐼 𝑛 + +
  • 114. The indicator should be such that pH at equivalence point falls within the transition interval of the indicator. Titration between pH at end point Commonly used indicator Weak acid and strong base Alkaline range Thymol blue, phenolphthalein thymopthalein Weak base and strong acid Acidic range Methyl orange Methyl red Bromocresol green
  • 115. 2. RESONANCE THEORY All acid- base indicators commonly used are organic compounds. The difference in colour of same compound in acid and base medium is due to difference in structure of two forms. Colour shown by the compound is related to the ability of the compound to absorb visible light and this capability is related to the electronic structure. Change in electronic features will result in absorption of different colour components of light with a resultant colour change.
  • 116.
  • 117. Types of Indicators 1. Single indicators 2. Mixed indicators 3. Universal or Multiple range indicators Single indicators Only one component of indicator Mixed Indicators In some cases the pH range is very narrow and the colour change over this range must be very sharp. Ordinary acid base indicators cannot serve this purpose. This is achieved by use of suitable mixture of indicators. These are selected such that their pKIn values are close to each other. Examples: Mixture of equal parts of neutral red (0.1% in alcohol) and methylene blue (0.1% in alcohol) gives a sharp colour change from violet blue to green in passing from acid to alkaline solution at pH 7. Mixture of phenolphthalein (3 parts of 0.1%) and α-napthopthalein ( 1 part 0.%) passes from pale rose to violet at pH 8.9 (Titration of phosphoric acid to dibasic stage)
  • 118. Universal or Multiple range indicators. By suitably mixing certain indicators the colour change may be made to extend over a considerable portion of pH range . Such mixtures are called as universal indicators . They are not suitable for quantitative analysis. Can be used for qualitative purpose. Ex- Dissolve 0.1g of phenolphthalein, 0.2g methyl red, 0.3 g methyl yellow, 0.4 g of bromothymol blue in 500 ml of absolute alcohol and add sufficient sodium hydroxide until the colour is yellow. Colour change are- pH 2 = red pH 4= orange pH 6= yellow pH 8 = green pH 10= Blue
  • 119. Color Ranges of Indicators
  • 120.
  • 121.
  • 123.
  • 124. Application of Neutralization titration : • For the determination of strong acid Eg .assay of HCL, H₂SO₄ and benzoic acid. • For determination of strong base Eg assay of Na₂co₃,NaHCO₃. • For determination of weak acid eg assay of phosphoric acid ,CH₃COOH. • For determination of weak base eg assay of aminophylline. • for the determination of salt example ,sodium acetate salt and ammonium chloride . • To know % purity of acidic and basic substance.
  • 125. Points to be summarized • Chemical equillbrium, Ionic equillibrium • Acid base theories (Arrhenius, lowry bronsted and Lewis theories with examples and limitations. • Law of Mass action • Strengths of acids and bases ( Dissociation constants- Ka, Kb, pKa, pKb, relationship of pKa and Ka , pKb and Kb) • Strong acids, strong base, weak acid and weak bases examples. • Degree of dissociation α and equation. • Ionic product of water (Kw) • pH and pOH, Scale of pH derivation, pH and pOH formulae • Le chaterliers principle • Common ion effect (Mechanism with examples, application) • Buffers, types, Mechanism, buffer capacity, Henderson hasselbalch equation, pH = pKa • Solubility product (Ksp) • Hydrolysis of salts • Types of Salts (SASB, WASB, WBSA, WAWB), their pH when dissolved in water and examples. • Titration and its concepts • Titration curves (SASB, WASB, WBSA, WAWB), equations, significance of each. • Polyprotic acids and their salts • Polyfunctional acid base titrations • Titration of amino acids. • Acid- base indicators
  • 126. Some important formulae's for acid base titrations Kw = Ka. Kb pKw = pKa + pKb pKa + pKb = 14 Kw = 1 x 10 -14           ][ ][ HA A logpKpH a moles of titrant = moles of analyte (V titrant)(M titrant) = (V analyte)(M analyte) pH = (pKw – pKb )+ log [𝐀−] [𝐇𝐀] β = 𝒅𝑪 𝑩𝑶𝑯 𝒅𝒑𝑯 = - 𝒅𝑪𝑯𝑨 𝒅𝒑𝑯