Indicators are substances that change colour depending on whether the solution is acidic or alkaline. An indicator is used to determine the end point in a titration. In acid-base titrations, organic substances (weak acids or weak bases) are generally used as indicators.

1. Theory of indicators
❖Indicators are substances that change colour depending on whether
the solution is acidic or alkaline.
❖An indicator is used to determine the end point in a titration.
❖In acid-base titrations , organic substances (weak acids or weak
bases) are generally used as indicators .

2. Theory of indicators
❖Two theories have been proposed to explain the change of colour of
acid-base indicators with change in pH.
1. Ostwald's theory
2. Quinonoid theory

3. Theory of indicators
Ostwald's theory
❖According to this theory:
The colour change is due to ionisation of the acid-base indicator. The
unionised form has different colour from the ionised form.
❖The ionisation of the indicator is largely affected in acids and bases
as it is either a weak acid or a weak base.

4. Theory of indicators
Ostwald's theory
• In case, the indicator is a weak acid, its ionisation is very much low
in acids due to common H+ ions while it is fairly ionised in alkalis.
• Similarly if the indicator is a weak base, its ionisation is large in
acids and low in alkalis due to common OH- ions.

5. Theory of indicators
Ostwald's theory
❖Considering two important indicators phenolphthalein(a weak acid)
and methyl orange (a weak base),
Ostwald theory can be illustrated as follows:
• Phenolphthalein: can be represented as HPh. It ionises in solution
to a small extent as: HPh H++ Ph-
• Applying law of mass action,
K = [H+][Ph-]/[HPh]

6. Theory of indicators
Ostwald's theory
• The undissociated molecules of phenolphthalein are
colourless while Ph- ions are pink in colour.
• In presence of an acid the ionisation of HPh is practically
negligible as the equilibrium shifts to left hand side due
to high concentration of H+ ions. Thus, the solution would
remain colourless.

7. Theory of indicators
Ostwald's theory
• On addition of alkali, hydrogen ions are removed by OH - ions in
the form of water molecules and the equilibrium shifts to right
hand side. Thus, the concentration of Ph- ions increases in solution
and they impart pink colour to the solution.

8. Theory of indicators
Ostwald's theory
• Methyl orange: It is a very weak base and can be represented as
MeOH. It is ionized in solution to give Me + and OH - ions.
MeOH Me + + OH -
• Applying law of mass action,
K = [Me + ][OH -]/[MeOH]

9. Theory of indicators
Ostwald's theory
• In presence of an acid, OH - ions are removed in the form of water
molecules and the above equilibrium shifts to right hand side.
Thus, sufficient Me + ions are produced which impart red colour to
the solution.

10. Theory of indicators
Ostwald's theory
• On addition of alkali, the concentration of OH - ions increases in
the solution and the equilibrium shifts to left hand side, i.e., the
ionisation of MeOH is practically negligible. Thus, the solution
acquires the colour of unionised methyl orange molecules, i.e.,
yellow.

11. Theory of indicators
Ostwald's theory
• This theory also explains the reason why phenolphthalein is not a
suitable indicator for titrating a weak base against strong acid. The
OH- ions furnished by a weak base are not sufficient to shift the
equilibrium towards right hand side considerably, i.e., pH is not
reached to 8.3. Thus, the solution does not attain pink colour.
• Similarly, it can be explained why methyl orange is not a suitable
indicator for the titration of weak acid with strong base.

12. Theory of indicators
Quinonoid theory:
❖According to this theory:
❖The acid-base indicators exist in two tautomeric forms having
different structures.
❖Two forms are in equilibrium
❖One form is termed benzenoid form and the other quinonoid form.
❖The two forms have different colours.
❖The colour change in due to the interconversion of one tautomeric form into
the other.

13. Theory of indicators
Quinonoid theory:
(c) One form mainly exists in acidic medium and the other in alkaline
medium.
Thus, during titration the medium changes from acidic to alkaline or
vice-versa. The change in pH converts one tautomeric form into other
and thus, the colour change occurs.

14. Theory of indicators
Quinonoid theory:
❖Phenolphthalein has benzenoid form in acidic medium and thus, it is
colourless while it has quinonoid form in alkaline medium which has
pink colour.

15. Theory of indicators
Quinonoid theory:
❖Methyl orange has quinonoid form in acidic solution and benzenoid
form in alkaline solution. The colour of benzenoid form is yellow while
that of quinonoid form is red.

16. The colour change and the pH range of some
common indicators
❖Indictors do not change colour sharply at one particular pH, they
change over a narrow range of pH.

17. The colour change and the pH range of some
common indicators
Indicator pH range Colour change
Methyl orange 3.2-4.5 Pink to yellow
Methyl red 4.4-6.5 Red to yellow
Litmus 5.5-7.5 Red to blue
Phenol red 6.8-8.4 Yellow to red
Phenolphthalein 8.3-10.5 Colourless to pink

18. Henderson Hasselbalch equation
• Let us derive Handerson equation for an indicator
HIn + H2O H3O+ + In-
'Acid form‘ 'Base form'
Conjugate acid-base pair
Kln = [ln][H+3O]/[HIn];
KIn = Ionization constant for indicator
[H3O+] = KIn * [Hln]/lnpH = -log10 [H3O+] = -log10[Kln] -
log10[Hln]/[ln-]
pH = pKIn + log10[ln-]/[Hln] (Handerson equation for indicator)
At equivalence point
[In-] = [HIn] and pH = pKIn

19. Salts
❖A salt is a neutral substance produced from the reaction of an acid and a
base, (salts are compounds that dissociate in water and produce cations
other than H+ and anions other than OH-
❖Composed of the negative ion of an acid and the positive ion of a base.
❖One of the products of a Neutralization Reaction
❖Examples: KCl, MgSO4, NaCl, Na3PO4,

20. Salts
❖When H+ ion of an acid is replaced by a metal ion, a salt is produced
e.g. H2SO4(aq) + 2NaOH(aq) ======➔ Na2SO4(aq) + 2H2O(l)
sodium sulphate (Na2SO4) is the salt formed.
❖Salts are ionic compounds.
❖Note: Ammonia (NH3) is an unusual base - it does not contain a
metal. It forms ammonium salts, containing the ammonium ion, NH4
+
e.g. NH3(aq) + HNO3(aq) →NH4NO3(aq) (ammonium nitrate)

21. Methods of making soluble salts
❖ACID + METAL → SALT + HYDROGEN
❖ACID + BASE → SALT + WATER
❖ACID + CARBONATE → SALT + WATER + CARBON DIOXIDE
❖ACID + ALKALI → SALT + WATER
❖DIRECT COMBINATION

22. Methods of making soluble salts
Method 1 (Acid + Metal)
❖Not suitable for making salts of metals above magnesium, or below
iron/tin in reactivity.
❖e.g.
• Zn + 2HCl -------------------→ ZnCl2 + H2
• Fe + H2SO4 ----------------→ FeSO4 + H2

23. Methods of making soluble salts
Method 2 (Acid + Base)
❖Useful for making salts of less reactive metals, e.g. lead, copper.
❖e.g.
• CuO + H2SO4 ----------------→ CuSO4 + H2O
• MgO + 2HCl ------------------→ MgCl2 + H2O
❖Add excess base to acid.

24. Methods of making soluble salts
Method 3 (Acid + Carbonate)
❖Useful particularly for making salts of more reactive metals, e.g.
calcium, sodium.
❖e.g.
• CaCO3 + 2HCl -------------→ CaCl2 + H2O + CO2.
• Na2CO3 + H2SO4 ------------→ Na2SO4 + H2O + CO2.

25. Methods of making soluble salts
Method 4 (Acid + Alkali)
❖This is useful for making salts of reactive metals, and ammonium salts. It is
different from methods 1-3, as both reactants are in solution. This means
neutralisation must be achieved, by adding exactly the right amount of acid to
neutralise the alkali. This can be worked out by titration
❖e.g.
• NaOH + HCl --------------→ NaCl + H2O
• 2NH4OH + H2SO4 ----------------------→ (NH4)2SO4 + 2H2O

26. Making insoluble salts
❖This involves mixing solutions of two soluble salts that between them
contain the ions that make up the insoluble salt. It is made by two
methods.
• PRECIPITATION
• BaCl2(aq) + MgSO4(aq) → BaSO4(s) + MgCl2(aq)
• DIRECT COMBINATION
• Fe + S ---heat----→ FeS

27. Methods of making insoluble salts:
Precipitation reaction

28. Types of salts
Normal Salts:
❖Normal salts are formed when all the replaceable hydrogen ions in
the acid have been completely replaced by metallic ions.
❖HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
❖H2SO4(aq) + ZnO(aq) → ZnSO4(aq) + H2O(l)
❖Normal salts are neutral to litmus paper.

29. Types of salts
Acid salts:
❖Acid salts are formed when replaceable hydrogen ions in acids are
only partially replaced by a metal. Acid salts are produced only by
acids containing more than one replaceable hydrogen ion. Therefore
an acid with two replaceable ions e.g. H2SO4 will form only one acid
salt, while acid with three replaceable hydrogen ions e.g. H3PO4 will
form two different acid salts.
• H2SO4(aq) + KOH(aq) → KHSO4(aq) + H2O(l)
• H3PO4(aq) + NaOH → NaH2PO4(aq) + H2O(l)
• H3PO4(aq) + 2NaOH(aq) → Na2HPO4(aq) + 2H2O(l)

30. Types of salts
❖An acid salt will turn blue litmus red. In the presence of excess
metallic ions an acid salt will be converted into a normal salt as its
replaceable hydrogen ions become replaced.
• KHSO4(aq) + KOH =========➔ K2SO4(aq) + H2O(l)

31. Types of salts
Basic Salts:
❖Basic salts contain the hydroxide ion, OH-. They are formed when there is
insufficient supply of acid for the complete neutralization of the base. A basic salt
will turn red litmus blue and will react with excess acid to form normal salt.
• Zn(OH)2(s) + HCl(aq) → Zn(OH)Cl(aq) + H2O(l)
• Zn(OH)Cl(aq) + HCl(aq) → ZnCl2(aq) + H2O(l)
• Mg(OH)2(s) + HNO3(aq) → Mg(OH)NO3(aq) + H2O(l)
• Mg(OH)NO3(aq) + HNO3(aq) → Mg(NO3)2(aq) + H2O(l)

32. Types of salts:Hydrated & anhydrous salts
❖Hydrated Salt: Salt that contains
Water of Crystallization is called
Hydrated Salt e.g. CuSO4.5H2O,
Na2CO3.10H2O.
❖Anhydrous Salt: Salt with out
Water of Crystallization is called
Anhydrous Salt. e.g. CuSO4, Na2CO3

33. Types of salts

34. Uses of salts
SALT USE
Ammonium Chloride In torch batteries
Ammonium Nitrate, Ammonium sulphate In fertilizers
Calcium Chloride As drying agent
Iron Sulphate In Iron tablets
Magnesium Sulphate In medicine
Potassium Nitrate In gunpowder etc.
Silver Bromide In photography (a component on
photographic film).
Sodium Chloride Making NaOH, food flavouring (salt)
Sodium Stearate In making soap.
Calcium sulphate Medical uses (plaster of Paris)

35. Buffers
❖A buffer is a solution of a weak acid and its conjugate base that resists
changes in pH in both directions—either up or down, when small amounts
of acids and bases are added.
• They can absorb excess H+ ions or OH- ions, thereby maintaining an
overall steady pH in the solution.
❖ A buffer works best in the middle of its range, where the amount of
undissociated acid is about equal to the amount of the conjugate base.

36. Buffers
❖There are two requirements for a buffer:
• Two substances are needed: an acid capable of reacting with added
OH - ions and a base that can consume added H3O + ions.
• The acid and base must not react with each another

37. Buffers
(a) The pH electrode is indicating the pH
of water that contains a trace of acid (and
bromphenol blue acid–base indicator). The
solution at the left is a buffer solution with a
pH of about 7. (It also contains bromphenol
blue dye.)
(b) When 5 mL of 0.10 M HCl is added to each
solution, the pH of the water drops several
units, whereas the pH of the buffer stays
essentially constant, as implied by the fact that
the indicator color does not change.

38. Uses of buffers
❖Enzyme reactions and cell functions have optimum pH’s for
performance. pH control is important for them to function
optimally.
• Maintaining a constant blood pH

39. Uses of buffers
Blood: pH = 7.35-7.45
Too acidic? Increase respiration rate expelling CO2, driving
reaction to the left and reducing H+ concentration.
Excretory system – excrete more or less bicarbonate
❖Maintaining a constant blood pH

40. How buffers work
❖Equilibrium between acid and base.
❖Example: Acetate buffer
• CH3COOH  CH3COO- + H+
• If more H+ is added to this solution, it simply shifts the equilibrium
to the left, absorbing H+, so the [H+] remains unchanged.
• If H+ is removed (e.g. by adding OH-) then the equilibrium shifts to
the right, releasing H+ to keep the pH constant

41. Limits to the working range of a buffer
❖Consider the previous example:
• CH3COOH  CH3COO- + H+
❖If too much H+ is added, the equilibrium is shifted all the way to the
left, and there is no longer any more CH3COO- to “absorb” H+.
❖At that point the solution no longer resists change in pH; it is useless
as a buffer.
❖A similar argument applies to the upper end of the working range.

42. Chemistry of buffers
❖Lets look at a titration curve
❖Titration is used to determine the concentration of an acid or base by
adding the OTHER and finding an equivalency point.
❖Equivalency point- point in titration at which the amount of titrant
added is just enough to completely neutralize the analyte solution.
❖At the equivalence point in an acid-base titration, moles of base =
moles of acid and the solution only contains salt and water.

43. Chemistry of buffers

44. Chemistry of buffers
❖Suppose you have a KOH solution,
and you want to know its
concentration (molarity).
❖Slowly add an acid (HCl) with a
known concentration (0.1 M) and
find the equivalency point…in this
case it will be at pH = 7… and we use
an indicator that changes color at
that pH determine when that point
has been reached.
❖So, suppose it takes 10ml of 0.1 M
HCl to buffer 50 ml of the KOH.

45. Chemistry of buffers
❖So, suppose it takes 10ml of 0.1 M
HCl to buffer 50 ml of the KOH.
❖The original concentration of the
base = Vol. acid x conc. of acid
Volume of base
= 10 ml x 0.1 M
50 ml
= 0.02 M

46. Chemistry of buffers
Equilibrium constant/Dissociation constant (Ka)
❖Ka = equilibrium constant for H+ ion transfer is also described as the dissociation
constant (the tendancy of an acid to dissociate).
AH → A- (base conjugate) + H+
Ka = [A-] [H+]/ [AH] = [base] [H+] / [acid]
❖Weak acids have low values… contribute few H+ ions…
❖Because we are usually dealing with very small concentrations, log values are
used…
❖The log constant =

47. Chemistry of buffers
❖Since pK is the negative log of K, weak acids have high values … (-2 –
12).
❖HCl = -9.3 – very low ~complete dissociation
• First rearrange the first equation and solve for [H+]
• [H+] = Ka x [acid]/[base]
❖Then take the log of both sides
• log10[H+] = log10Ka + log10 [acid]/[base]
-pH -pKa

48. Chemistry of buffers
❖-pH = -pKa + log10 [acid]/[base]
❖Multiply both sides by –1 to get the Henderson-Hasselbach equation
• pH = pKa - log10 [acid]/[base]

49. Chemistry of buffers
❖What happens when the concentration of the acid and base are equal?
• Example: Prepare a buffer with 0.10M acetic acid and 0.10M acetate
• pH = pKa - log10 [acid]/[base]
• pH = pKa - log10 [0.10]/[0.10]
• pH=pKa
• Thus, the pH where equal concentrations of acid and base are
present is defined as the pKa
❖If there is more of the conjugate base in the solution than acid, for
example, then pH > pKa. Conversely, if there is more acid than conjugate
base in solution, then pH < pKa.
❖A buffer works most effectively at pH values that are + 1 pH unit from the
pKa (the buffer range).