2nd Lecture on Ionic Equilibria Chemistry Part I12th Standard by Rizwana Mohammed

1. The Malegaon High School & Jr. College
Malegaon, (Nashik), 423203
2nd Lecture on Ionic Equilibria
Chemistry Part I, 12th Science
By
Rizwana Mohammad

2. Autoionzation of water:
The ionization equilibrium of water is represented as,
H2O (l) + H2O(l) ⇌ H3O(aq) + OH(aq)
The equilibrium constant given by
𝐾 =
𝐻3
𝑂 [𝑂𝐻 ]
𝐻2
𝑂
𝑙
2 …1
𝐾 𝐻2 𝑂 2 = 𝐻3 𝑂 [𝑂𝐻 ] …2
𝐻2 𝑂 2 = 𝐾′
Therefore from equation 2
𝐾 𝑋 𝐾′ = 𝐻3 𝑂 [𝑂𝐻 ]
𝐾 𝑤 = 𝐻3 𝑂 [𝑂𝐻 ]
Kw = ionic product of water.
The product of molar concentration of hydronium (or hydrogen) ions and
hydroxyl ions at equilibrium in pure water at given temperature is called ionic
product of water.
In pure water,
[H3O ] = [OH ]’= 1 X 10-7 mol/L
Kw = (1.0 X 10-7)(1.0 X 10-7) = 1.0 X10-14

3. pH Scale: Instead of writing concentration of H3O ions mol dm-3, sometimes it is
convenient to express it on the logarithmic scale. This is known as pH Scale
pH : Negative logarithm to the base 10 of the concentration of H ions in solution
in mol dm-3.
p 𝐻 = −log10[𝐻 ]
pOH= Negative logarithm to the base 10 of the molar concentration of OH ions in
solution
p𝑂𝐻 = −log10[𝑂𝐻 ]
Relationship between pH & pOH:
𝐾 𝑤 = 𝐻3 𝑂 𝑂𝐻
1 X 10-14 = [H3O ][OH ]
Taking logarithm on both sides
log10 𝐻3 𝑂 + log10 𝑂𝐻 = −14
−log10 𝐻3 𝑂 + {−log10 𝑂𝐻 } = 14
pH + pOH = 14

4. Acidity, basicity and neutrality of aqueous solution:
1. Neutral Solution:
For pure water or any aqueous neutral solution at 298 k
[H3O ] = [OH ] = 1.0 X 10-7 M
Hence, p 𝐻 = −log10 𝐻 = − log10 1𝑋10 − 7 = 7
2. Acidic Solution:
In acidic solution there is excess of H3O ions, or
[H3O ]>[OH ]
Hence [H3O ] > 1 X 10-7 and pH < 7
3. Basic Solution:
In basic solution excess of OH ions are present
i.e. [H3O ] < [OH ] or
[H3O ] < 1 X 10-7 with pH > 7

5. Hydrolysis of salts:
Types of salts:
There are four types of salts,
1. Salts of strong acid and strong base
e.g. NaCl, Na2SO4, NaNO3, KCl, KNO3
2. Salts of strong acid and weak base.
e.g. NH4Cl, CuSO4, NH4NO3, CuCl2.
3. Salts of weak Acid and strong base.
e.g. CH3COONa, KCN, Na2CO3.
4. Salts weak acid and weak base.
e.g. CH3COONH4, NH4CN.

6. Concept of hydrolysis:
The solution of salt becomes acidic or basic or neutral depends
on type of salt.
The reaction between the ions of salts and the ions of water is
called hydrolysis of salt.
Hydrolysis of salt is defined as, “the reaction in which cations or
anions or both ions of a salt react with ions of water to produce
acidity or alkalinity or neutrality.”
Salts of strong acid and strong base:
When NaCl is dissolved in water, it dissociates as
NaCl(aq)→Na(aq) + Cl(aq)
or Na(aq) + Cl(aq) +H2O → HCl(aq) + NaOH(aq)
SA SB
Possible Products
HCl(aq) + NaOH(aq) + H2O → H3O(aq) + Na(aq) +Cl(aq) + OH(aq)

7. Thus the reactants and products are same.
This implies that neither the cation nor anion of the salt reacts with
water or there is no hydrolysis.
H3O = OH, equality is not disturbed, solution is neutral. It
may be concluded that, “salt of strong acid and strong base does
not undergo hydrolysis.”
Salts of strong acids and weak bases:
CuSO4 is a salt of strong acid H2SO4 and weak base Cu(OH)2
CuSO4 → Cu+2
(aq) + SO4
-2
(aq)
SO4
-2 ions of the salt have no tendency to react with water.
Cu+2 ions with OH ions form unionized Cu(OH)2. The hydrolytic
equilibrium for CuSO4 can be written as
Cu+2
(aq) + 4H2O(l) ⇌ Cu(OH)2(aq) + 2H3O(aq)
Due to presence of excess of H3O ions, resulting solution becomes
acidic.

8. Salts of weak acids and strong bases:
CH3COONa is a salt of weak acid CH3COOH and strong base NaOH,
when dissolved in water it dissociates as:
CH3COONa(aq) → CH3COO(aq) + Na(aq)
Na ions have no tendency to react with water,
CH3COO(aq) reacts with water,
CH3COO(aq) + H2O(l) ⇌ CH3COOH(aq) + OH(aq)
Due to excess of OH ions produced the solution becomes basic.
Salts of weak acids and weak bases:
In this type of salt, acidity, basicity or neutrality depends on Ka
or Kb values.
i. If Ka>Kb, the solution will be acidic
ii. If Ka<Kb, the solution will be basic
iii. If Ka=Kb, the solution will be neutral.

9. i. For Ka>Kb
e.g. NH4F is a salt of weak acid HF and weak base NH4OH
Ka = 7.2 X 10-4, Kb = 1.8 X 10-5
NH4(aq) + F(aq) + H2O ⇌ NH4OH(aq) + HF(aq)
WB WA
The two ions react with water as:
NH4(aq) + 2H2O(l) ⇌ NH4OH(aq) + H3O(aq)
F(aq) + H2O(l) ⇌ HF(aq) + OH(aq)
NH4 ions hydrolyse to a slightly greater extent than F ions,
produces more H3O ions. Therefore the solution is slightly acidic.

10. ii. If Ka<Kb
NH4CN is a salt of weak acid and weak base.
NH4OH (Kb = 1.8 X 10-5)
HCN (Ka = 4.0 X 10-10)
NH4(aq) + CN(aq) + H2O(l) ⇌ NH4OH(aq) + HCN(aq)
WB WA
The two ions of the salt react with water as:
NH4(aq) + H2O(l) ⇌ NH4OH(aq) + H3O(aq)
CN(aq) + H2O(l) ⇌ HCN(aq) + OH(aq)
CN ions hydrolyse to a greater extent than NH4 ions, more OH
ions are produced hence the solution is basic.

11. iii. If Ka=Kb
CH3COONH4 is a salt of weak acid CH3COOH and weak base
NH4OH
Ka = 1.8 X 10-5
Kb = 1.8 X 10-5
CH3COO(aq) + NH4(aq) + H2O(l) ⇌ CH3COOH(aq) + NH4OH(aq)
WA WB
The ions of salt react with water as
CH3COO(aq) + H2O(l) ⇌ CH3COOH(aq) + OH(aq)
NH4(aq) + 2H2O(l) ⇌ NH4OH(aq) + H3O(aq)
As Ka=Kb
therefore the solution is neutral.