CHAPTER 7.0_IONIC EQUILIBRIA

CHEMISTRY UNIT, KMNS

CHAPTER 7
IONIC EQUILIBRIA

7.1
Acids And Bases

ACIDS AND BASES

ACIDS BASES

• LITMUS PAPER CHANGES COLOUR FROM:

Blue to red Red to blue

• REACTION WITH METAL

Acid + metal → salt + H2(g) No reaction

• REACTION WITH METAL OXIDE

Acid + metal oxide → salt + H2O No reaction

• REACTION WITH CARBONATE, CO32- ION

Acid + CO32- → Salt + H2O + CO2 No reaction

• REACTION WITH ACID

No reaction Base + acid → salt + H2O

ACID-BASE THEORIES

1. ARRHENIUS

Definition:

• Acid : a substance which can produce H+ or H3O+ when dissociates
in water.
Example : HCl (aq)  H+ (aq) + Cl- (aq)

• Base : a substance which can produce OH- when dissociates in water.
Example : NaOH (aq)  Na+ (aq) + OH- (aq)

Limitations :
• Applicable to compounds containing hydrogen and hydroxide.
• Restricted to aqueous solution.

4

2. BRONSTED-LWRY

Definitions:
• Acid : a substance which donates H+. When acid loses H+, conjugate

base is formed.

• Base : a substance which accepts H+. When base accepts H+,
conjugate acid is formed.

Example 1:

CH3COO- + HCN CH3COOH + CN-

Conjugate pair Conjugate pair

Acid Conj. base Base Conj. acid

HCN CN- CH3COO- CH3COOH 5

• Exercise 2: HPO42– + NH4+

H2PO4– + NH3

Conjugate pair Conjugate pair
Acid Conj. base Base Conj. acid

H2PO4– HPO42– NH3 NH4+

• Exercise 3:

CO32– + H2O HCO3– + OH–

Conjugate pair Conjugate pair
Acid Conj. base Base Conj. acid

H2O OH– CO32– HCO3– 6

3. LEWIS

Definition:
• Lewis Acid : a substance which can accept a pair of electrons to form covalent

bond (@ electron-pair acceptor).

Examples:

1) Cations of metals
 Li+, Ca2+, Cu2+ etc

2) Molecules whose central atom does not have a complete octet
 Al in AlCl3, B in BF3

Definition:
• Lewis Base : a substance which can donate a pair of electrons to form covalent

bond (@ electron-pair donor).

Examples:

1) All anions

 OH-, Br-, CN- etc

2) Molecules whose central atom has a lone pair electron to be donated

 N in NH3, O in H2O 7

COMPARISON OF STRONG ACID & WEAK ACID

STRONG ACID:

• Examples: HCl, HNO3, H2SO4
• Degree of dissociation in water,  = 100%
• Example of dissociation equation:

HCl(aq)  H+(aq) + Cl-(aq)

1.0M 1.0M

WEAK ACID:

• Examples: CH3COOH (Ka = 1.8 X 10-5), HCN (Ka = 4.9 X 10-10)
• Degree of dissociation in water,   100%
• Example of dissociation equation:

CH3COOH(aq) CH3COO-(aq) + H+(aq) 8
0.01M  0.01M

COMPARISON OF STRONG BASE & WEAK BASE

STRONG BASE:

• Examples: NaOH, KOH, Mg(OH)2
• Degree of dissociation in water,  = 100%
• Example of dissociation equation:

NaOH(aq)  Na+(aq) + OH-(aq)

2.0M 2.0M

WEAK BASE:

• Examples: NH3 (Kb = 1.8 X 10-5), (C2H5)2NH (Kb = 9.6 X 10-4)
• Degree of dissociation in water,   100%
• Example of dissociation equation:

NH3(aq) + H2O( ) NH4+(aq) + OH-(aq) 9
5 x 10-3M  5 x 10-3M

Examples:

1. Write equations for the acid-base reactions that occurred in the aqueous
solutions for each of the following species

i. NH4+
ii. S2-

Ans: NH4+(aq) + H2O( ) NH3(aq) + H3O+(aq)

Ans: S2-(aq) + H2O( ) HS-(aq) + OH-(aq)

2. At 25 °C, 2.20% of 0.125 M benzoic acid, C6H5COOH ionises. Write the
dissociation equation for benzoic acid in water.

Ans: C6H5COOH(aq) + H2O( ) C6H5COO-(aq) + H3O+(aq)

10

DISSOCIATION CONSTANT OF WATER, KW

Self-ionisation process of H2O

H2O( ) + H2O( ) OH-(aq) + H3O+(aq)

At equilibrium : OH− H3O+
H2O 2
Kc = X 100

Kc x [H2O]2 = [OH-] x [H3O+]

At 25°0C, Kw = [OH-] x [H3O+]

[OH-] = [H3O+] = 1 x 10-7 M
 Kw = (1 x 10-7) x (1 x 10-7)

= 1 x 10-14

pKw = - log Kw = 14 11

DISSOCIATION CONSTANT, K

Water, Kw Weak Acid, Ka Weak Base, Kb

Dissociation H2O( ) + H2O( ) H3O+(aq) + OH-(aq) HA(aq) + H2O( ) H3O+(aq) + A-(aq) B(aq) + H2O( ) BH+(aq) + OH- (aq)
equation @ @

2H2O( ) H3O+(aq) + OH-(aq) HA(aq) H+(aq) + A-(aq)

Expression Kw = [H+] [OH-] Ka = [H+] [OH-] Kb = [BH+] [OH-]
[HA] [B]

- log Kw = pKw - log Ka = pKa - log Kb = pKb

Kw = [H3O+][OH-] = 1 x 10-14 At 25oC, pKa x pKb = pKw = 14 pKb  1
- log Kw = - log [H3O+] – (-log [OH-] ) = - 1x10-14 Kb
Ka x Kb = Kw = 1 x 10-14
pKw = pH + pOH = 14 pKa  1
Ka

pKa , Ka  , acidity  pKb  Kb  , basicity 

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pH SCALE
• pH is defined as : the negative logarithm of [H+] or [H3O+].
• Used to measure the [H+] or [H3O+].
• pH = - log [H+] @ - log [H3O+]
• pOH is defined as : the negative logarithm of [OH-]
• Used to measure the [OH-].
• pOH = - log [OH-]

RELATIONSHIP BETWEEN KW , pH AND pOH

At 25oC, Kw = [H3O+][OH-] = 1 x 10-14

- log Kw = - log [H3O+] – (- log [OH-] ) = - 1x10-14

pKw = pH + pOH = 14

13

Example 1:
Calculate the pH of the following solution:
An aqueous solution contains 0.700g NaOH in 485ml H2O, where NaOH is a
strong base.

Solution:
NaOH (aq)  Na+(aq) + OH-(aq)

nNaOH = mNaOH MNaOH = nNaOH pOH = - log [OH-]
Mr NaOH VNaOH = - log 0.036
= 1.76
= 0.700 = 1.75 x 10-2
40 485 x 10-3 pH + pOH = 1.44
 pH = 12.56
= 1.75 x 10-2 mol = 0.036 M
14
MNaOH  MOH- = 0.036 M

Examples:

2. Calculate the pH of 0.02 M Ba(OH)2 solution.
Solution: Ba(OH)2 (aq)  Ba2+(aq) + 2OH-(aq)

0.02M 2 x 0.02M

pOH = - log [OH-]
= - log (0.04)
= 2.4

pH + pOH = 14
 pH = 12.6

3. Calculate the pH of a 1 x 10–3 M solution of hydrochloric acid and a
1 x 10–3 M solution of calcium hydroxide.

Solution: HCl(aq)  Cl-(aq) + H+(aq)

1 x 10-3M 1 x 10-3M

pH = - log [H+] 15
= - log (1 x 10-3)
=3

4. The pH value of a solution containing 0.245 mol HF in 500mL water is 1.88.
Calculate the acid ionisation constant for HF.

Solution: HF(aq) + H2O( ) F-(aq) + H3O+(aq)

Ka = [F- ] [H3O+ ] MHF = 0.245 mol = 0.49 M
[HF] 500 x 10-3L

= 0.0132 x 0.0132 pH = - log [H3O+]
0.477 1.88 = - log [H3O+]
 [H3O+] = 0.0132 M
= 3.65 x 10-4

[ ]initial/M HF(aq) + H2O(l) F-(aq) + H3O+(aq)
Δ[] 0.49 - 0 0
+x +x
[ ]EQM/M -x - x x
0.49 – x - 0.0132
0.0132
0.477 16

5. At 25 °C, 2.20% of 0.125 M benzoic acid, C6H5COOH ionises.
i. Write the dissociation equation for benzoic acid in water.

Solution: C6H5COOH(aq) + H2O( ) C6H5COO-(aq) + H3O+(aq)

ii. Determine the acid dissociation constant, Ka for benzoic acid.

Solution: Ka = [C6H5COO- ] [H3O+ ]
[C6H5COOH]

C6H5COOH + H2O C6H5COO + H3O+

-

[ ]initial/M 0.125 - 0 0
Δ[] -x - +x +x

[ ]EQM/M 0.125 - x - x x

= 0.122 = 2.75 x 10-3 = 2.75 x 10-3

α = Δ[] x 100% 2.20 = Δ[] x 100 [ ] = 2.75 x 10-3= x = [H3O+]
[ ]initial 0.125 17

Ka = (2.75 x 10-3 ) (2.75 x 10-3 )
0.122

= 6.186 x 10-5

iii. Calculate the pH of the solution.
Solution: pH = - log [H3O+]
= - log 2.75 x 10-3
= 2.56

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6. Write the dissociation equation of base for a 0.5 M aqueous solution of CH3NH2
and calculate the pH. [Kb = 3.7 x 10–4M]

Solution: CH3NH2(aq) + H2O( ) CH3NH3+(aq) + OH-(aq)

Kb = [CH3NH3+ ] [OH- ]
[CH3NH2 ]

[ ]iitial/M CH3NH2 + H2O CH3NH3+ + OH-
Δ[] 0.5 - 0 0
+x +x
[ ]EQM/M -x - x x
0.5 – x -

3.7 x 10-4 = (x)(x) pOH = - log [OH-]
(0.5 - x) = - log (0.134)
= 1.87
x2 + 3.7 x 10-4x – 1.85 x 10-4 = 0

x1 = 0.0134 = [OH-]  pH = 14 – 1.87 19
X2 = - 0.014 (neglected) = 12.13

REMARKS
• If the value of Ka @ Kb is < x 10 -5 : can use approximation.
• If the value of Ka @ Kb is  x 10 -4 : cannot use approximation.

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GUIDELINE IN PROBLEM SOLVING RELATED TO WEAK ACIDS @ BASES
• Write the dissociation equation of the weak acid @ base
 Dissociate partially
• Tabulate the data
 Enter the [ ]Initial of all species
 Determine the Δ in [ ]
 Write the [ ]EQM
• Enter the [ ]EQM into Ka @ Kb expression
• Solve for x by assuming a – x  a (if Ka @ Kb = < 10-5)
• Determine [H+] @ [OH-] & the pH value

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SALT

• Preparation: Acid reacts with base
General equation: Acid + Base  Salt + Water
Example: HCl + NaOH  NaCl + H2O

• Properties:
i. An ionic compound.
ii. Strong electrolyte (dissociate completely in water to form cation and
anion).
Example: NaCl  Na+ + Cl-

• Classification:
 Salt is classified into neutral or basic or acidic through hydrolysis
reaction.
 Hydrolysis reaction is a chemical reaction between anion or cation (of a
salt) with H2O.

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BASIC SALT

Formation : Weak acid + strong base

Example : CH3COOH + NaOH  CH3COONa + H2O
Dissociation equation : CH3COONa  CH3COO- + Na+
Hydrolysis equation :
Anion reacts with H2O
CH3COO - + H2O CH3COOH + OH-

pH > 7 (due to the presence of OH-)

Formation : ACIDIC SALT Formation : NEUTRAL SALT
Example : Strong acid + weak base Strong acid + strong base
Dissociation HCl + NH3  NH4Cl Example : HCl + NaOH  NaCl + H2O
equation : NH4Cl  NH4+ + Cl- Dissociation
Hydrolysis equation : NaCl  Na+ + Cl-
equation : Cation reacts with H2O Hydrolysis
equation : Neither the cation nor anion undergoes
pH NH4+ + H2O NH3 + H3O+ hydrolysis
pH 7

< 7 (due to the presence of H3O+) 23

1. Write the hydrolysis equations for sodium carbonate, Na2CO3 and
ammonium nitrate, NH4NO3 solutions. State whether the solutions are
acidic, basic or neutral.

Ans: Na2CO3  2Na+ + CO32-

Ans: CO32- + H2O H2CO3 + OH- ~ Basic salt

Ans: NH4NO3  NH4+ + NO3-
Ans: NH4+ + H2O NH3 + H3O+ ~ Acidic salt

2. OBr - ion is the conjugate base for a weak acid, HOBr. Write the hydrolysis
equation of NaOBr.

Ans: NaOBr  Na+ + OBr-

Ans: OBr- + H2O HOBr + OH-

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3. Sodium cyanide, NaCN is a salt formed when a strong base, NaOH is reacted
with a weak acid, HCN.
i. Write a balanced equation to show the reaction between NaOH and HCN.
Classify the salt formed.

Solution: HCN + NaOH  NaCN + H2O
NaCN  2Na+ + CN-
Balanced equation:
Salt dissociation equation: CN- + H2O HCN + OH-
Hydrolysis equation: Basic salt.
Classification of salt:

ii. What would be the expected pH of the NaCN solution? Explain the answer
using appropriate equation(s).

Solution:

pH > 7 : Due to the presence of the OH-

CN- + H2O HCN + OH-

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SUMMARY

Type of Type of Salt Species pH Examples
Acid/Base Which Solution
Hydrolyse
Strong Acid + Neutral NaCl, KI,
Strong Base None 7 KNO3, NaSO4

Strong Acid + Acidic Cation <7 NH4Cl,
Weak Base Anion (NH4)2CO3,
Both >7 CuSO4
Weak Acid + Basic
Strong Base Depends CH3COONa,
Neutral @ On Type KCN, NaCN,
Weak Acid + Acidic @ Of Salt Na2S
Weak Base Basic
C2H3O2NH4,
(NH4)2C2O4

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BUFFER SOLUTION

• Definition:
Solution that has the ability to resist changes in pH when a small
amount of strong acid or strong base is added to the solution.

• Type:
i. Acidic buffer
ii. Basic buffer

• Preparation of an acidic buffer:

When weak acid is mixed with its conjugate salt.

Example: Weak acid: CH3COOH CH3COO- + H+

Conjugate salt: CH3COONa  CH3COO- + Na+

• Preparation of basic buffer:

When weak base is mixed with its conjugate salt.

Example: Weak base: NH3 + H2O NH4+ + OH-

Conjugate salt: NH4Cl  NH4+ + Cl- 27

Buffer solution has 2 components:
 A component which is able to neutralise the acids added to the buffer

solution.
 A component which is able to neutralise the bases added to the buffer

solution.

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Example: In acidic buffer solution

i. A component which is able to neutralise the acid added to the buffer solution
is the conjugate base of the weak acid.

ii. A component which is able to neutralise the base added to the buffer solution
is the weak acid.
(Weak acid)

CH3COOH CH3COO- + H+

CH3COONa  CH3COO- + Na+

(Conjugate base of the weak acid)

When acid is added to the acidic buffer solution:  pH of the buffer
H+ + CH3COO-  CH3COOH solution remains
added constant

When base is added to the acidic buffer solution: 29

OH- + CH3COOH  CH3COO- + H2O
added

Example: In basic buffer solution
i. A component which is able to neutralise the acid added to the buffer

solution is the weak base.
ii. A component which is able to neutralise the base added to the buffer

solution is the conjugate acid of the weak base.

(Weak base)

NH3 + H2O NH4+ + OH-

NH4Cl  NH4+ + Cl-

(Conjugate acid of the weak base)

When acid is added to the basic buffer solution:  pH of the buffer
solution remains
H+ + NH3  NH4+ constant
added
30
When base is added to the basic buffer solution:

OH- + NH4+  NH3 + H2O
added

DETERMINATION OF pH of BUFFER SOLUTION

Acidic Buffer Solution

pH = pKa + log [conjugate base] Known as Henderson-Hasselbalch
[weak acid] equation for acidic buffer solution

Where pKa = - log Ka

Basic Buffer Solution

pOH = pKb + log [conjugate acid] Known as Henderson-Hasselbalch
[weak base] equation for basic buffer solution

Where pKb = - log Kb

31

Definition : Solution that has the ability to resist changes in pH when a small amount of strong acid or strong base is added to the solution.

Type : ACIDIC BUFFER BASIC BUFFER
Preparation :
Example : • Weak acid + its conjugate salt • Weak base + its conjugate salt

• CH3COOH CH3COO- + H+ • NH3 + H2O NH4+ + OH-
• Conjugate base : CH3COO- • Conjugate acid : NH4+
• CH3COONa  CH3COO- + Na+ • NH4Cl  NH4+ + Cl-

• When H+ is added into acidic buffer • When H+ is added into basic buffer

H+ + conjugate base  weak H+ + weak  conjugate
added
added of the weak acid acid base acid of the
H+ +
H+ + CH3COO-  CH3COOH added weak base
added
Example : NH3  NH4+
• When OH- is added into acidic buffer
Example : • When OH- is added into basic buffer
Conclusion OH- + weak  conj. base + H2O
OH- + conj. acid of weak + H2O
added acid of the weak added
acid the weak base
OH- +
OH- + CH3COOH  CH3COO- + H2O added base
added
NH4+  NH3 + H2O
• H+ added/OH- added is removed by the conj. base/
weak acid. Thus, pH has no significant Δ. • H+ added/OH- added is removed by the weak
base/conj. acid. Thus, pH has no significant Δ.

32

Example:

A student was asked to prepare a buffer solution of pH 4.6 using 50.00 mL of 0.5 M

benzoic acid, C6H5COOH and sodium benzoate, C6H5COONa.
(Ka C6H5COOH = 6.5 x 10-5)
a) Explain the buffering effect of adding a small amount of NaOH and HCl,

respectively into the buffer solution.

Solution: C6H5COO- + H3O+
Dissociation equation of C6H5COOH: C6H5COOH + H2O

(Weak acid)

Dissociation equation of CH3COONa: C6H5COONa  C6H5COO- + Na+
(Conjugate base of weak acid)

33

Continue:

NaOH  Na+ + OH-
OH- + C6H5COOH  C6H5COO- + H2O
added
OH- added is removed by the weak acid (C6H5COOH).
Thus, pH has no significant change.
HCl  H+ + Cl-

H+ + C6H5COO-  C6H5COOH
added
H+ added is removed by the conjugate base (C6H5COO-).
Thus, pH has no significant change.

34

b) Calculate the mass of sodium benzoate required to prepare the buffer
solution.
(Ka C6H5COOH = 6.5 x 10-5)

Solution: pH = - log Ka + log [C6H5COO-]
[C6H5COOH]

4.6 = - log (6.5 X 10-5) + log [C6H5COO-]

[C6H5COO-] = 1.294 M (0.5)

C6H5COONa  C6H5COO- + Na+

Thus, [C6H5COO-] = 1.294 M = [C6H5COONa] 35
n C6H5COONa = 1.294 x 50 x 10-3
= 0.065 mol x 145 gmol-1
= 9.380 g

Example:

a) 500 mL of 0.10 M hydrazinium chloride solution, N2H5Cl was added to a
500 mL solution of 0. 20 M hydrazine, N2H4. What is the pH of the mixture?
(Kb of N2H4 = 1.70 x 10–7 M, Kw = 1.0 x 10–14 M2)

Solution: n = MV/1000 N2H5Cl  N2H5+ + Cl-

n N2H5Cl = 0.10 x 500 x 10-3 From equation:
= 0.05 mol 1 mol N2H5Cl  1 mol N2H5+

n N2H4 = 0.20 x 500 x 10-3  0.05 mol N2H5Cl  0.05 mol N2H5+
= 0.10 mol [N2H5+] = n/VT

[conjugate acid] = 0.05 / (500 + 500) x 10-3 = 0.05M
pOH = pKb + log [ weak base ]
[N2H4] = 0.10 / (500 + 500) x 10-3 = 0.10M
[N2H5+]
pOH = - log Kb + log [N2H4] (0.05)
pOH = - log 1.70 x 10-7 + log (0.10)

= 6.47  pH = 7.53 36

b) By writing reaction equations, explain what would happen to the pH of
the solution when a small amount of a strong acid or a strong base is
added to the solution mixture in (a).

Solution:

• When H+ is added into basic buffer

H+ + N2H4  N2H5+
added

H+ added is removed by the weak base (N2H4). Thus, pH has no
significant change.

• When OH- is added into acidic buffer 37

OH- + N2H5+  N2H4 + H2O
added

OH- added is removed by the conj. acid (N2H5+). Thus, pH has no
significant change.

Example:

a) Calculate the pH of 250 mL solution containing 0.20 M CH3COOH and
0.30 M CH3COONa.
(Ka of CH3COOH = 1.8 x 10–5 M)

Solution:

pH = - log Ka + log [CH3COO-]
[CH3COOH]

= - log (1.8 X 10-5) + log (0.30)
(0.20)

= 4.92

38

b) Calculate the pH of the resulting solution when 1.0 mL of 1.0 M HCl is added.

Solution:

CH3COONa  CH3COO- + Na+
no CH3COONa = 0.30 x 250 x 10-3 = 0.075 mol = no CH3COO-
no CH3COOH = 0.20 x 250 x 10-3 = 0.05 mol

HCl  H+ + Cl-
n HCl added = 1.0 x 1 x 10-3 = 1 x 10-3 mol = n H+

H+ + CH3COO-  CH3COOH
added

H+ added is removed by the conjugate base (CH3COO-).

39

n initial CH3COO− (aq) + H+ (aq) → CH3COOH (aq)
n addition 0.075 0.050
n change
n final - 1 x 10-3 1 x 10-3 + 1 x 10-3
0.075 – 1 x 10-3 - 1 x 10-3
[ ] final = 0.074 0.050 + 1 x 10-3
0.074 0 = 0.051
0.251 = . 0.051
0 0.251 = .

Using Henderson-Hasselbalch equation : pH = pKa + log CH3COO−
CH3COOH

= − log 1.8 × 10−5 + log 0.295
0.203

= .

40

c) Calculate the pH of the resulting solution when 1.0 mL of 1.0 M NaOH is added.

Solution:

CH3COONa  CH3COO- + Na+
no CH3COONa = 0.30 x 250 x 10-3 = 0.075 mol = no CH3COO-
no CH3COOH = 0.20 x 250 x 10-3 = 0.05 mol

NaOH  Na+ + OH-
n NaOH added = 1.0 x 1 x 10-3 = 1 x 10-3 mol = n OH-

OH- + CH3COOH  CH3COO- + H2O
added
OH- added is removed by the weak acid (CH3COOH).

41

CH3COOH (aq) + OH− (aq) → CH3COO − (aq) + H2O (l)

n initial 0.050 0.075 -

n addition 1 x 10-3 -

n change - 1 x 10-3 - 1 x 10-3 + 1 x 10-3 -
n final -
[ ] final 0.050 – 1 x 10-3 0 0.075 + 1 x 10-3 -
= 0.049 = 0.076
0
0.049 0.076
0.251 = . 0.251 = .

Using Henderson-Hasselbalch equation : pH = pKa + log CH3COO−
CH3COOH

= − log 1.8 × 10−5 + log 0.303
0.195

= .

42

Guideline In Solving Problem Related To Buffer Solution
• For original buffer solution apply Henderson-Hasselbalch equation.

pH = pKa + log [conjugate base] pOH = pKb + log [conjugate acid]
[weak acid] [weak base]

• If acid or base is added to buffer solution

 Step 1 : tabulate the reaction and related data involve

 Assume all acid @ base added is completely used up during the
reaction (acid @ base acts as limiting reactant)

 Step 2 : determine the new pH of buffer solution

 Use Henderson-Hasselbalch equation

43

7.2
Acid-base Titrations

TITRATION

• In an acid-base titration, titrant is slowly added from a
burette to an analyte until the reaction is complete.

Titrant

Analyte

• An indicator is added to determine the end point

~ An indicator is a weak organic acid/base that changes color

when the pH changes. 45

TITRATION

• When the reaction is completed, we have reached the end point of the
titration.

~ End point: the point at which an indicator changes colour.

• When the no. of moles of the acid equals the no. of moles of the base,
the titration has reached its equivalence point.

~ Equivalence point: the point at which the stoichiometrically 46
equivalent amounts of acid and base have reacted.

Tro: Chemistry: A Molecular Approach, 2/e

Past Year PSPM 2018/2019

A sample of 0.214 g of unknown monoprotic weak acid, HA
was dissolved in 25.00 mL of water and titrated with 0.1 M
NaOH. The acid required 27.40 mL of the base to reach the
equivalent point.

(iii) Explain quantitavely the pH of the solution at equivalent point.

Solution :

When reached equivalence point, the solution contains salt
solution @ salt and water.

Since the salt is from weak acid and strong base, the anion A-
undergoes hydrolysis forming OH-

@
A− + H2O ⇌ HA + OH−
Thus, the salt is a basic salt with pH more than 7.

47

TITRATION CURVE

• It is a plot of pH vs volume of titrant added during titration
process.

• Prior to the equivalence point, the analyte is in excess, so
the pH is closest to the pH of the analyte.

• The inflection point of the curve is the equivalence point of
the titration.

• The pH of the equivalence point depends on the pH of the
salt solution
i. equivalence point of neutral salt, pH = 7
ii. equivalence point of acidic salt, pH < 7
iii. equivalence point of basic salt, pH > 7

• Beyond the equivalence point, the titrant is in excess, so
the pH approaches the pH of the titrant.

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Titration Curve: Titration Between Strong Acid – Strong Base
(Titrant: Strong Base; Analyte: Strong Acid)

Neutral salt Beyond
solution formed, equivalence
neither cation nor point, excess strong
anion undergoes base determines
hydrolysis, the pH.
therefore pH = 7
After base has
Before titrant (a been added, but
strong base) is
added, pH is before equivalence
determined by the point, excess strong
strong acid alone.
acid determines
the pH.

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• If the titration is run so that the strong acid is in the
burette/titrant and the base is in the flask/analyte, the
titration curve will be the reflection of the one just shown.

Strong base Strong
Strong acid acid

Strong
base

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