IONIC EQUILIBRIUM

L-1
Acid Base Theory
pH Scale
Ionic Product of Water
pH Calculations
Ostwald Dilution Law
Salt Hydrolysis
Buffer Solution
Acid- Base Titration
Theory of Indicator
Solubility & Solubility Product
Factors affecting solubility
 JEE Main Exam

Equilibrium
Percentage Weightage (2019-23)

Physical Chemistry 13.56 %

 JEE Advanced Exam
Physical Chemistry: Relative Weightage (2019-23)

Equilibrium
10.5%
Stoichiometry and Concentration Terms

Thermodynamics and Thermochemistry

Chemical Equilibrium
 Conductors

Thermal Electrical
conductors conductors
Material or substances Substance which can conduct
which can conduct heat. charge (allow passage of charge).

Electronic Electrolytic
conductors conductors
They allow passage of charge Electrolytes conduct electricity
through the physical through the movement of its
movement of electrons. constituent ions.
Substances which dissociate into its constituent ions in their fused / molten
state Or in their solution phase.

Example

NaCl(s) + H2O() ⎯→ Na+(aq) + Cl–(aq)

HCl(aq) ⎯→ H+(aq) + Cl–(aq)

NaOH(aq) ⎯→ Na+(aq) + OH–(aq)

Note

An electrolyte is an acid or a base or a salt.

 1. Strong Electrolytes

Electrolytes

2. Weak Electrolytes
Those ionic conductors which are completely ionized in aqueous solution or
in fused state or in any solution phase are called as strong electrolytes.

For strong electrolytes the value of degree of ionization (dissociation) is

100% i.e. α = 1

Strong electrolytes prefer to remain in ionic form.

⊕ ⊖
HCl(aq) → H (aq) + Cl (aq)

Example (1) Strong acids → H2SO4, HCl, HNO3, HClO4, HBr, HI

(2) Strong bases → KOH, NaOH, Ba(OH)2, CsOH, RbOH
(3) All Soluble Salts → NaCl, KCl, CuSO4
 Those electrolytes which are partially ionized in aqueous solution or in
solution phase are called as weak electrolytes.

For weak electrolytes the value of α < 1.

Weak electrolytes prefer to remain in molecular form.

⊖ ⊕
CH3COOH(aq) ⇌ CH3COO (aq) + H (aq)

Example

Weak acids : HCN, CH3COOH, HCOOH, H2CO3, H3PO3, H3PO2, H3BO3, etc.

Weak bases : NH4OH, Cu(OH)2, Zn(OH)2, Fe(OH)3, Al(OH)3 etc.

Strong electrolytes prefer to remain in ionic form.

Weak electrolytes prefer to remain in molecular form.

All soluble salts are strong electrolytes, no matter, whatever be the

solubility of salt, the amount which is dissolved is 100% ionized.
ACID & BASE THEORY

:
Arrhenius Acids

Substances which give H+ ions on dissolving in water (H+ donor).

eg. HNO3, HClO4, HCl, HI, HBr, H2SO4, H3PO4 etc.
Arrhenius Acids

HX on dissolving in water gives H+(aq) + X–(aq)

Note

H+ combines with H2O and is primarily present as H3O+ .

Some common forms are :

H3O+ H5O2+ H7O3+ H9O4+

Arrhenius Bases

Any substance which releases OH– (hydroxyl) ion in water (OH– ion donor)
e g. NaOH, Ca(OH)2 etc.
 Arrhenius Bases

BOH(aq) on dissolving in water gives B+(aq) + OH–(aq)

Note

The hydroxyl ion also exists in the hydrated form in the aqueous solution
such as H3O2–, H5O3– etc.

H3O2– H5O3–
 :
Brönsted-Lowry Acids

Substances which donate H+ ions are Brönsted Lowry acids or in other

words Brönsted Lowry acids are proton donor/ H+ donor.
 :
Brönsted - Lowry Bases

Substances which accept H+ ions are Brönsted Lowry bases or in other

words Brönsted Lowry bases are proton accepter/H+ acceptor.
 :
Amphiprotic Substances

Substances which can act as an acid (H+ donor) and as well as a base
(H+ acceptor) are known as amphiprotic substances.

Example

H2PO4–, HSO3–, HCO3–, HSO4–, H2PO3–, HPO42–, NH3, H2O etc.

 :
Conjugate acid – base pairs

HX B HB+ X–
(acid) (base) (Conjugate acid) (Conjugate base)

Conjugate pair

Conjugate pair
 :
Acid Base Conjugate acid Conjugate base
HCl + H2O ⇌ H3O+ + Cl–
H2O(l) + NH3(aq) ⇌ NH4+ + OH–(aq)
 :
Conjugate acid - base pair differ by only one proton.

Stronger acid have weaker conjugate base and vice versa.

Acid Conjugate base Base Conjugate acid

HCl Cl– NH3 NH4+

H2SO4 HSO4– H2O H3O+

HSO4– SO42– RNH2 RNH3+

H2O OH–
 Lewis Acids

A Lewis acid is a species which can readily accept an electron pair and
it should therefore possess an empty orbital.

Example

BF3 , AlCl3
 Lewis Bases

A Lewis base is any species possessing an unshared pair of electrons which

can be easily donated (shared).

Example

Molecules with lone pairs : NH3, PH3, H2O, CH3OH

− − − −
Anions : OH , H , NH2− , CH3− , F , Cl
Factors affecting the strength of Lewis Acids

Acidic Strength ∝ Nuclear charge on central atom

Acidic Strength ∝ number of EN atoms attached

Acidic Strength ∝ positive charge on ion

1
Acidic Strength ∝ radius of cation , (for same charge on cation)
 Example

The conjugate base of hydrochloric acid acid is (JEE MAIN 2014)

(A) N3– (B) N3– (C) Cl– (D) HN3–

Solution

Ans. (c)
 Example

The conjugate base of hydrazoic acid is (JEE MAIN 2014)

(A) N3– (B) N3– (C) N2– (D) HN3–

Solution

Ans. (B)
 EXAMPLE

Which of the following is a Lewis acid? (JEE MAIN 2018)

(A) pH3 (B) NF3 (C) NaH (D) B(CH3)3

Solution

Ans.(D)
 Example

Of the given anions, the strongest Bronsted base is (IIT 1981)

(A) ClO– (B) ClO2– (C) ClO3– (D) ClO4–

Solution

Ans. (A)
 Example
The following equilibrium is established when hydrogen chloride is dissolved in
acetic acid.
(IIT 1992)
HCl + CH3COOH ⇌ Cl– + CH3COOH2+
The set that characterizes the conjugate acid-base pairs is
(A) (HCl, CH3COOH) and (CH2COOH2+, Cl–)
(B) (HCl, CH3COOH2+) and (CH2COOH2+, Cl–)
(C) (CH3COOH2+, HCl) and (Cl–, CH3COOH)
(D) (HCl, Cl–) and (CH2COOH2+, CH3COOH)

Solution

Ans. (D)
 Example
Given below are two statements one is labelled as
Assertion A and the other is labelled as Reason R:
Assertion A : The amphoteric nature of water is explained by using Lewis
acid / base concept.
Reason R: Water acts as an acid with NH3 and as a base with H2S.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false
(D) A is false but R is true
(JEE MAIN 2022)
Solution

Ans. (D)
 ⊕ ⊖
HA ⇌ H +A

۩ ⊖
H [A ]
Ka = ∴ Ka = Acid ionization constant.
[HA]

⊕ ⊖
BOH ⇌ B + OH

۩ ⊖
B [OH ] ∴ Kb = Base ionization constant.
Kb = [BOH]
Important results

1 Ka(acid) × Kb(conjugate base) = 10–14

2 Kb(base) × Ka (conjugate acid) = 10–14

3 Ka(HA) × Kb(A–) = 10–14

p

Given by – Sorenson

pH Scale is called Sorenson scale.

pH scale is a measuring scale used to measure strength of acids and

bases in aqueous solution.

The pH of a solution is the negative logarithm to the base 10 of the

concentration of H+ ions which it possess.

pH = –log [H+]

Example [H+] = 10–3

pH = – log 10–3
= + 3 log 10 = 3
p

pOH : It is equal to –log [OH–]

pOH = –log [OH–] [OH–] = 10–pOH

pH has meaning only in an aqueous solution (in water). Many chemicals,

including liquids, do not have pH values.

If there's no water, there's no pH.

Note

There is no pH value for vegetable oil, gasoline, or pure alcohol.

If there is water then there will be both ions H+ and OH– in aqueous solution.
 p Drain
Stomach
Battery Lemon Tomato Milk Blood Tablets Soap Cleaner

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Acidic Neutral Alkaline

Stomach Vinegar Coffee Water Baking Ammonia Bleach
Acid Soda Solution
 KD
H2O H+(aq.) + OH–(aq.)

H + OH–
KD =  KD × [H2O] = [H+] [OH–]
[H2O]

Since, dissociation of water takes place to a very small extent & water is
a pure liquid, [H2O] may be regarded as constant.

[H2O] is incorporated with in the equilibrium constant(K D ) to give a new

constant, Kw which is called the ionic product of water,
(or auto protolysis constant) which increases with the increase in
temperature.
Kw = [H+] [OH–] = KD × [H2O]

The concentration of [H+] = [OH–] = 1.0 × 10-7 M at 298 K.

At 298 K, Kw = (1.0 × 10-7)2 = 10-14 M2
-log Kw = pKw = - log(10-14) = 14
pKw = 14

pH of H2O at 25∘ C = 7 pOH of H2O at 25∘ C = 7

Kw = [H + ] OH − ; pkw = pH + pOH
density
Molarity of pure water or [H2O] =
molar mass
g
1000 L
= g = 55.55 M
18 mole
Thus, KD[H2O] = KW  K w > KD (always)

The ratio of dissociated H2O to the undissociated H2O is

10−7 −9
= 1.8 × 10
55.55

Thus, equilibrium lies mainly towards undissociated water.

Therefore, acidic, neutral, and basic aqueous solutions can be distinguished

by the relative values of [H+] & [OH–] concentrations :

Acidic : [H+] > [OH–]

Basic : [H+] < [OH–]
Neutral : [H+] = [OH–]
Effect of Temperature

endo
H2O exo
H+ + OH–

Ionization of water is endothermic process so,

Effect of Temperature

At 25°C At 90°C

KW 10–14 10–12

pKw 14 12

pK w
pH = = pOH 7 6
2

H + = OH – = Kw 10–7 10–6
 Example

At 90°C, pure water has [H3O+] = 10–6 mole L–1. Find the value of Kw at 90°C .
(A) 10–6 (B) 10–12 (C) 10–14 (D) 10–8
(IIT 1985)
Solution

Ans. (B)
 Example

What is the pH of a 10–4 M OH– solution at 330 K, if Kw at 330 K is 10–13.6 ?

(A) 4 (B) 9.0 (C) 10 (D) 9.6
(JEE MAIN 2013)
Solution

Ans. (D)
 Example

For the following Assertion and Reason, the correct option is

Assertion: The pH of water increases with increase in temperature
Reason: The dissociation of water into H+ and OH– is an exothermic reaction.
(A) Both assertion and reason are true, and the reason is the correct
explanation for the assertion.
(B) Both assertion and reason are false
(C) Both assertion and reason are true, but the reason is not the correct
explanation for the assertion.
(D) Assertion is not true, but reason is true. (JEE MAIN 2020)

Solution

Ans. (B)
p

Strong acid solution

If concentration of H+ ions is greater than 10–6 M, H+ ions coming from

water can be neglected,
So [H+] = normality of strong acid solution

Strong base solution

Calculate the [OH–] which will be equal to normality of the strong base
solution and then use

Kw = [H+] × [OH–] = 10–14, to calculate [H+]

Note

If concentration of H+ ions is less than 10–6 M, H+ ions coming from water

cannot be neglected.
So [H+] = normality of strong acid + H+ ions coming from water in presence
of this strong acid.
(Will Discuss this further in the Chapter)
 Example

pH value of 0.001 M NaOH solution is__________. (JEE MAIN 2022)

Solution

Ans. (11)
p

pH of mixture of two strong acids

If V1 volume of a strong acid solution (I) of normality N1 is mixed with V2

Volume of another strong acid solution (II) of normality N2, then

Number of H+ ions from I-solution = N1V1

Number of H+ ions from II-solution = N2V2

gm-eq = Mole x n-factor n- factor of Acid = Basicity of Acid

Normality = M x n-factor n- factor of Base = Acidity of Base

p

pH of mixture of two strong acids

If final normality is Nf and final volume is Vf, then NfVf = N1V1 + N2V2

Dissociation equilibrium of these acids will not be disturbed as both are

strong acid.
N1 V1 + N2 V2
[H+] = Nf =
V1 + V2

pH of mixture of two strong bases

N1 V1 + N2 V2 10−14
[OH–] = Nf = [H+] =
V1 + V2 OH−
p

pH of mixture of a strong acid and a strong base

Acid Base neutralisation reaction will take place.

The solution will be acidic or basic in nature depending on which

component has been taken in excess.
(Or Which is having more no. of gm eq.)

If V1 volume of a strong acid solution (I) of normality N1 is mixed with V2

volume of a strong base solution (II) of normality N2, then
Number of H+ ions from I-solution = N1V1
Number of OH– ions from II-solution = N2V2
 If N1V1 > N2V2 If N2V2 > N1V1

N1 V1 − N2 V2 N2 V2 − N1 V1
[H+] = N = Mixture of a [OH–] = N =
V1 + V 2 V1 + V2
STRONG ACID
And
Solution will be STRONG BASE Solution will be
acidic in nature basic in nature

10−14 10−14
[H+] = [OH–] =
OH− H+
 Example
The pH of a solution obtained by mixing 50 mL of 1 M HCl and 30 mL of 1 M
NaOH is x × 10–4.
The value of x is _______. (Nearest integer)
[log 2.5 = 0.3979] (JEE MAINS 2021)

Solution

Ans : (6021)
 Example

Following four solutions are prepared by mixing different volumes of NaOH and
HCl of different concentrations. pH of which one of them will be equal to 1 ?
M M
(A) 55 mL 10 HCl + 45 mL 10 NaOH (JEE MAIN 2018)
M M
(B) 75 mL HCl + 25 mL NaOH
5 5
M M
(C) 100 mL HCl + 100 mL NaOH
10 10
M M
(D) 60 mL 10 HCl + 40 mL 10 NaOH

Solution

Ans. (B)
 Example

What will be resultant pH when 200 mL of an aqueous solution of HCl

(pH = 2.0) is mixed with 300 mL of an aqueous solution on NaOH (pH = 12.0) ?
(IIT 1998)
Solution

Ans : (11.3)
Ostwald was the first to apply law of mass action to Ionic Equilibrium.

Ostwald Dilution Law is applicable only for weak electrolytes.

According to Ostwald Dilution Law when solution of a weak electrolyte is

diluted then the degree of ionization of weak electrolyte is increased.

Let AB(aq) ⇌ A+ (aq) + B– (aq)

Initial concentration c 0 0

Degree of ionization () (c – c) (c) (c)

 A+ B– cα cα2
According to law of mass action: K =  K = cα × =
AB c(1 − α) (1 − α)
If,  < 0.05 then (1 – )  1

K
K = c2 or  = (K = Ionization constant, At constant temperature)
c

1 1
 c∝
V α∝  α∝ V α ∝ dilution
c

By Ostwald Dilution Law

Dilution    
At infinite dilution :  = 100%

Thus, for weak electrolytes the degree of ionization is directly proportional

to square root of dilution or inversely proportional to square root of
concentration. This law is known as Ostwald Dilution Law.

Limitation of Ostwald Dilution Law

It is not applicable for strong electrolytes.

When two or more electrolytes furnishes a common ion, then degree
of dissociation of the weak electrolytes furnishing common ion is/are
suppressed.

This is based on Le-Chatelier’s principle.

E.g. H2O at 298 K, [H+] = [OH-] = 10–7 M

H2O ⎯→ H+ + OH-
10–7 M 10–7 M
E.g. 10−7 M HCl + H2O at 298 K

HCl(aq) ⎯→ H+ (aq) + Cl-(aq)

10–7 10–7 + 𝑥 10–7

H2O ⎯→ H+ + OH-
𝑥 + 10−7 𝑥

Kw = [H + ] OH −

Note

Due to common ion effect [H + ] coming from water will be less than 10−7 M.
 Example

Calculate pH of 10-8 M NaOH solution.

Solution

Ans. 7.03
 [HCl] = H+ 10–7 M 10–8 M 10–9 M 10–10 M

pH of solution 6.79 6.97 7 7

[NaOH] = [OH-] 10–7 M 10–8 M 10–9 M 10–10 M

pOH of solution 6.79 6.97 7 7

pH of solution 7.21 7.03 7 7

E.g. 0.1 M CH3COOH (Ka = 10-5)

𝐊𝐚 10−5
= = = 0.01, (1% dissociation)
𝐜 𝟎.𝟏
If x << 1 then x,x2,x3….

If x << 1 then 2x, 3x, ….

Can be neglected in comparison to 1 or greater than 1

1+x≅1 1–x≅1 1 + 2x ≅ 1

1 – 2x ≅ 1 2–x≅2 2 + 3x ≅ 2

1 + x2 ≅ 1 1 – x2 ≅ 1 1 – x3 ≅ 1
If x << 1 then x,x2,x3….

If x << 1 then 2x, 3x, ….

In comparison to 1 or greater than 1 means in addition subtraction only

but in the case of multiplication or division there is no comparison hence
no approximation is needed.

x (2 + 3x) ≅ 2x 𝟐–𝟐𝐱 𝟐
≅
𝐱 𝐱
x2 (2x) = 3x3
 p

For mono basic weak acid (HA)

Given - Ionization constant (Ka), degree of ionization is ().

HA ⇋ H+ + A–
Initial concentration c 0 0
At equilibrium c – c c c

H + A– cα × cα c 2 α2 cα2
Ka = =  Ka = =
HA c– cα c 1– α 1– α

Note

If  ≤ 0.05  (1 –  ) ≈ 1  Ka = c2
 p

For mono basic weak acid (HA)

HA ⇌ H+ + A–
Initial concentration c 0 0
At equilibrium c – c c c

[H+] = c ....(1)

Ka
Ka = c2 or α= ....(2)
c

from Eq. (1) and (2)

Ka
H+ = c × or H+ = Ka × c
c
 p

For mono basic weak acid (HA)

Put the value of [H+] i.e. [H+] = Ka × c

1 Τ2
⇒ pH = – log K a × c = – log K a × c

1
⇒ pH = – logK a + logc
2

1 1
⇒ pH = – log K a – log c
2 2

1
pH = 2 (pK a – log c)
 p

For mono basic weak acid (HA)

If  cannot be ignored (when  > 0.05)

cα2
Ka = ;
1−α

c  + a  − a = 0

−Ka+ K2a + 4cKa

Solve quadratic to obtain ‘’  =
2c
 pH of a mixture of weak acid (monoprotic) and a strong acid solution

E.g. 0.1 M CH3COOH + 0.1 M HCl

⊕ ⊖
HCl (aq) ⎯⎯→ H (aq) + Cl (aq)
0.1M — —
0.1M 0.1M

⊖ ⊕
CH3COOH (aq) CH3COO (aq) + H (aq) ; Ka = 10-5

0.1M — 0.1M(From HCl)

0.1(1–) 0.1 0.1 + 0.1

= 0.1 (1+)
 (0.1) 0.1 (1+) 0.1(1+)
Ka = = = 10-5
0.1 (1−) (1−)

(1 +)
= 10-4 If < 0.05 then (1−) ≃ 1
1 −

0.01% dissociation (Due to common ion effect)

 p

pH of a mixture of two weak acids (both monoprotic)

Both acids will dissociate partially.

Let the acid are HA1 & HA2 and their final concentrations are C1 & C2
respectively, then

HA1 ⇌ H+ + A−
1 HA2 ⇌ H+ + A−2

t=0 C1 0 0 C2 0 0

At eq. C1(1 – 1) C11 + C22 C11 C2(1 – 2) C22 + C11 C22
C1 α1 C1 α1 + C2 α2 C2 α2 C1 α1 + C2 α2
K a1 = K a2 =
C1 (1 −α1 ) C2 (1 −α2 )
 p

pH of a mixture of two weak acids (both monoprotic)

(Since 1, 2 both are small in comparison to unity)

K𝑎1 𝛼
K 𝑎1 C1 𝛼1 + C2 𝛼2 𝛼1 ; K 𝑎2 = C1 𝛼1 + C2 𝛼2 𝛼2 ⟹ K = 𝛼1
𝑎2 2

C1 Ka1 C2 Ka2
H + = C1 𝛼1 + C2 𝛼2 = + ⇒ [H + ]
C1 K𝑎 1 +C2 K𝑎 2 C1 K𝑎 1 +C2 K𝑎 2

= C1 K 𝑎1 + C2 K 𝑎 2
If the dissociation constant of one of the acid is very much greater than that
of the second acid, then contribution from the second acid can be neglected.

So, [H+] = C11 + C22 ≈ C11

 Example

(1) Calculate pH of 10–1 M CH3COOH

(2) Calculate pH of 10–3 M CH3COOH
(KaCH3COOH = 2 x 10–5)

Solution

Ans. (1) 2.85 (2) 3.8

 EXAMPLE

20 mL of 0.1 M H2SO4 solution is added to 30 mL of 0.2 M NH4OH Solution.

The pH of the resultant mixture is (JEE MAIN 2019)
[pKb of NH4OH = 4.7].
(A) 5.2 (B) 9.0 (C) 5.0 (D) 9.4
Solution

Ans. (B)
Important results

1 Ka(acid) × Kb(conjugate base) = 10–14

2 Kb(base) × Ka (conjugate acid) = 10–14

3 Ka(HA) × Kb(A–) = 10–14

 EXAMPLE
Calculate pH of 0.1 M of weak acid HA(aq). Given Kb(A-) = 10-9

Solution

Ans. 3
 EXAMPLE
(1) Calculate pH of 10–1 M CH3COOH
(2) Calculate pH of 10–3 M CH3COOH
(3) Calculate pH of 10–6 M CH3COOH
(KaCH3COOH = 2 x 10–5)

Solution

Ans. (1) 2.85 (2) 3.8 (3) 6

 EXAMPLE

Find pH of 10-4 M C6H5OH (phenol) solution ? (Ka phenol = 10-10)

Solution

Ans. 6.85
 EXAMPLE

Ka for acid HA is 2.5 × 10–8 calculate for its decimolar solution at 25°C.
(1) % dissociation
(2) pH
(3) [OH–]

Solution

Ans. (1) 0.05(2) 4.30 (3) 2 × 10–10 M

 EXAMPLE

Calculate :
(1) Ka for a monobasic acid whose 0.10 M solution has pH of 4.50.

(2) Kb for a monoacidic base whose 0.10 M solution has a pH of 10.50.

Solution

Ans. (1) Ka = 10–8 (2) Kb = 10–6

 EXAMPLE

How many grams of NH4OH (Kb = 2 × 10–5) are present in 500 mL of its
aqueous solution of pOH = 5.

Solution

Ans. 26.25× 10–5

pH of a solution of a polyprotic weak acid

Diprotic acid is the one, which is capable of giving 2 protons per molecule
in water.
Let us take a weak diprotic acid (H2A) in water whose concentration is c M.

H2A ⇌ HA– + H+
t=0 c - -
at eq. c(1–1) (c1− c12) (c1 + c12)

HA– ⇌ A2– + H+
c1 - -
at eq. (c1− c12) c12 (c1+ c12)
 [H + ][HA− ]
K a1 =
[H2 A]

[cα1 (1 + α2 )][α1 (1 − α2 )]
K a1 = ……….(1)
(1 − α1 )

+
[H ][A2− ]
K a2 =
[HA− ]

[cα1 (1+α2 )]α2

K a2 = ……………(2)
(1−α2 )
Using equations (1) and (2)

[H+]total = c1+ c12

Finally, for calculation of pH

If the total [H+]total < 10–6 M, the contribution of H+ from water should be added.

If the total [H+]total > 10–6 M, then [H+] contribution from water can be ignored.
Using this [H+], pH of the solution can be calculated.
 For diprotic acids, K a2 ≪ K a1 and a2 would be even smaller than 1
 1 – 2  1 and 1 + 2  1

cα1 ×α1
Thus, equation (1) can be reduced to K a1 = (1−α1 )

This expression is similar to the expression for a weak monoprotic acid.

Hence, for a diprotic acid (or a polyprotic acid) the [H+] can be calculated
from its first equilibrium constant expression alone provided K a2 ≪ K a1 .

Note

Same as polyprotic weak acids we can conclude for polyacidic weak bases.
 EXAMPLE

Calculate pH and [HS–], [S2–], [Cl–] in a solution which is 0.1 M HCl & 0.1 M
H2S given that Ka1 (H2S) = 10–7, Ka2 (H2S) = 10–14 also calculate a1 & a2.

Solution
Ans. pH = 1
[HS–] = 10–7
[S2–] = 10–20
[Cl–] = 0.1 M
1 = 10–6
2 = 10–13
Solutions of electrolytes are said to be isohydric if the concentration of
the common ion present in them is the same and on mixing such
solutions, there occurs no change in the degree of dissociation of either
of the electrolyte.

Let the isohydric solution is made by HA1 and HA2 acids, then [H+] of both
acids should be equal i.e.

Ka1 C
K a1 C1 = K a2 C2 or = C2
Ka2 1
For two acids of equimolar concentrations.
Strength of acid (I) 𝐾𝑎1
=
Strength of acid (II) 𝐾𝑎2

Strength of base (I) 𝐾𝑏1

Similarly for bases, Strength of base (II) = 𝐾𝑏2

The modern method is to convert Ka as a power of 10 and express acid

strength by power of 10 with sign changed and call this new unit pKa.
Thus, if Ka for acid is equal to 10-4, pKa = 4. So higher pKa value means
lower acid strength, that is, pKa = log Ka

Also, pKb = -log Kb

 Example
Values of dissociation constant, Ka are given as follows: (JEE MAIN 2013)
Acid Ka
HCN 6.2 × 10–10
HF 7.2 × 10–4
HNO2 4.0 × 10–4
Correct order of increasing base strength of the base CN–, F– and NO2– will be
(A) F¯ < CN¯ < NO2¯ (B) NO2¯ < CN¯ < F¯
(C) F¯ < NO2¯ < CN¯ (D) NO2¯< F¯ < CN¯

Solution

Ans. (C)
(Q.) An aqueous solution contains 0.10 M H2S and 0.20 M HCl. If the equilibrium constants for the
formation of HS⊝ from H2S is 1.0 ×10–7 and that of S2⊝ from HS⊝ ions is 1.2×10–13 then the
concentration of S2⊝ ions in aqueous solution is :
(1) 3×10–20 (2) 6×10–21
(3) 5×10–19 (4) 5×10–8

Solution

Answer:(1)
Salts are the ionic compounds formed when its positive part (Cation)
come from a base and its negative part (Anion) come from an acid.

Salts may taste salty, bitter or sweet or tasteless.

Solution of salts may be acidic, basic or neutral.

Fused salts and their aqueous solutions conduct electricity and undergo
electrolysis.

The salts are generally crystalline solids.

 (7) Mixed salts (1) Neutral salts

(2) Acidic salts

Classification
(6) Complex Salts
of
salts (3) Basic salts

(5) Double Salts (4) Salts of weak

acid and weak base
(1) Neutral salt

Salts formed by reaction of strong acids with strong bases.

Solution of such salts are neutral.

Example NaCl, MgCl2, MgSO4 etc.

(2) Acidic salt

Salts formed by reaction of strong acid with weak base.

Aqueous solution of such salts are acidic.

Example NH4Cl, (NH4)2SO4 etc.

(3) Basic salt

Salts formed by reaction of strong base with weak acid.

Aqueous solution of such salts are basic.

Example CH3COONa, HCOONa, (CH3COO)2Ca etc.

(4) Salt of weak acid and weak base

Salts formed by reaction of weak acid with weak base.

Hydrolysis of such salt occur aqueous solution of such salts can be

acidic, basic or neutral.

Example CH3COONH4, NH4F etc.

(5) Mixed Salts
Salts formed by the neutralisation of one acid by two bases or one base
by two acids are called Mixed Salts.

Example Ca(OCl)Cl

(6) Double Salts

A compound of two salts aqueous solution shows the tests for all
constituent ions is called double salt.

Example Mohr salt FeSO4.(NH4)2SO4.6H2O

Potash Alum K2SO4.Al2(SO4)3.24H2O
(7) Complex Salts

A compound whose solution does not give tests for the constituent ions is
called a complex salt.

Example K4[Fe(CN)6]
Salt hydrolysis is defined as the process in which water reacts
with cation or anion or both that may change the concentration of
H+ and OH– ions of water.

Salt hydrolysis is reverse process of neutralization.

Acid + Base ⎯→ Salt + water H = -ve ; Keq= Kneut.

Water + Salt ⎯→ Acid + Base H = +ve ; Keq=Kh=1/Kneut.

Important Terms

Kh = Hydrolysis constant

Kw = Ionic Product of water

Ka = Ionisation constant of acid

Kb = Ionisation constant of base

h = Degree of hydrolysis

C = Concentration of salt (concentration of ions)

When such salts are dissolved in water, they do not undergo hydrolysis.

They simply ionizes and fail to change the H+ ion concentration of

solution and hence the pH of the solution remains same.

pH = 7 at 298 K.

No effect on litmus paper.

Examples

NaCl, CaCl2, BaCl2, KCl, NaNO3, Na2SO4, KClO4, KBr, BaSO4 & Ca(NO3)2
Examples of SASB Types of Salts

Na+ + Cl– + H2O ⎯⎯→ NaOH + HCl

Na+ + Cl– + H2O ⎯⎯→ Na+ + OH– + H+ + Cl–

H2O OH– + H+
Anionic part of salt absorb some H+ ions coming from weakly
dissociated water, which is called anionic hydrolysis.

Such salts give alkaline solutions in water.

pH > 7 at 298 K.

Solution turns red litmus paper into blue.

Examples

NaCN, CH3COONa, HCOONa, KCN, K3PO4, BaCO3, K2CO3

CH3COO– + Na+ + H2O CH3COOH + NaOH

CH3COO– + Na+ + H2O CH3COOH + Na+ + OH–

CH3COO– + H2O CH3COOH + OH–

(1) Relation between Kh, Ka and KW

For weak acid

CH3COOH ⇌ CH3COO– + H+

CH3 COO– H+
Ka =
CH3 COOH …(1)

For Hydrolysis

CH3COO– + H2O ⇌ CH3COOH + OH–

CH3 COOH OH –
Kh = …(2)
CH3 COO–
For water

H2O ⇌ H+ + OH–

KW = [H+] [OH–] …(3)

Now multiply eq. (1) × eq. (2) = eq. (3)

CH3 COOH OH – CH3 COO– H+

– × = H + OH –
CH3 COO CH3 COOH

KW
K h × K a = Kw  Kh = …(4)
Ka
(2) Degree of Hydrolysis (h)

CH3COO– + H2O ⇌ CH3COOH + OH–

t=0 c 0 0
t = teq c – ch ch ch

CH3 COOH OH– ch×ch c2 h2 ch2

Kh = = = =
CH3 COO– c–ch c 1–h 1–h

Since h <<<< 1  (1 – h)  1

Κh Κh
Kh = ch2 or h2 = or h = …(5)
c c

KW
h= …(6)
Ka × c
(3) pH of the solution

CH3COO– + H2O ⇌ CH3COOH + OH–

c 0 0
c – ch ch ch

KW × c
OH – = ch  [OH – ] = …(7)
Ka

Taking – log on both sides

1ൗ
Kw. c 2
– log OH − =– log
Ka
 1ൗ
Kw. c 2
– log OH − =– log
Ka
1
pOH =– logK W + logc– logK a
2
1 1 1 1 1
pOH = pK W – pK a – logc = 7– pK a – log c
2 2 2 2 2

We know that,

 pH + pOH = 14  PH = 14 – poH

1 1
pH = 7 + ½( pKa + logc) ∴ pH = 7 + pK a + logc
2 2
 Example

pKa + pKh = pKw is applicable for :

(A) HCN (B) KCN (C) NH4Cl (D) NH4CN

Solution

Ans. (B)
 Example

Identify the nature of solution :

(1) NaCN(aq) (2) KCl (aq)

Solution

Ans.
1- Basic
2- Neutral
 Example

Calculate :
(i) pH of 0.1 M CH3COONa solution if ionization constant of CH3COO–
is 10–9.
(ii) Calculate Kh & h for hydrolysis of salt.

Solution

Ans. pH = 9
Kh = 10–9
h = 10–4
(Q.) The pH of a 0.02M NH4 Cl solution will be
[given Kb (NH4OH)=10–5 and log2=0.301]
(1) 4.65 (2) 5.35
(3) 4.35 (4) 2.65

Solution

Answer:(2)
 Cationic part of salt combine with OH- ions coming from weakly
dissociated water, which is called cationic hydrolysis.

Such salts give acidic solutions in water.

pH < 7 at 298 K.

Solution turns blue litmus paper into red.

Examples

Mg(NO3)2, MgSO4, (NH4)2SO4, NH4Cl, MgCl2, CuCl2, ZnCl2,

AgCl, AgI, AgNO3
NH3+ + Cl– + H2O NH4OH + HCl

NH4+ + Cl– + H2O NH4OH + H+ + Cl–

NH4+ + H2O NH4OH + H+

(1) Relation between Kh, Kb and KW

For weak base

NH4OH ⇌ NH4+ + OH–

NH4+ ][OH –
Kb = ….(1)
NH4 OH

Hydrolysis constant

B+ + H2O ⇌ BOH + H+

NH4+ + H2O ⇌ NH4OH + H+

NH4 OH][H +
Kh = ….(2)
NH4+
For water

H2O ⇌ H+ + OH–

KW = [H+] [OH–] ….(3)

Now multiply eq. (1) × eq. (2) = eq. (3)

NH4 OH H + NH4+ OH −
+ × = [H + ][OH – ቉
NH4 NH4 OH

KW
Kh × Kb = K w  Kh = ….(4)
Kb
(2) Degree of Hydrolysis (h)

Since h <<<< 1  (1 – h)  1

Kh = ch2

Κh
h2 =
c
Κh
h= ….(5)
c

KW ….(6)
h=
Kb × c
(3) pH of the solution

NH4 + H2O ⇌ NH4OH + H+

t=0 c 0 0
t = teq c – ch ch ch

KW
H + = ch = c
Kb × c

KW × c
H+ = ….(8)
Kb

Taking – log on both sides

KW × c
– log H + = – log
Kb
 1ൗ
KW × c 2
pH = – log
Kb

1
pH = – [log KW + logc – log Kb]
2

1 1 1
pH = – logKW – logc – (–log Kb)
2 2 2

1 1 1
pH = 2 pKw – 2 logc – 2 pKb

1 1
pH = 7 – pKb – logc
2 2

…(9)
1
pH = 7 – 2 (pKb + logc)
 Example

Calculate the equilibrium constant of reaction of strong acid with weak

base (Kb = 10–8)

Solution

Ans. Keq = 106

 Example

Which of the following solution has maximum degree of hydrolysis

(A) 0.1 M NH4Cl (B) 0.01 M NH4Cl
(C) 0.001 M NH4Cl (D) Same in all

Solution

Ans. (C)
 Both cation & anion react respectively with OH- & H+ ions coming
from weakly dissociated water.

Almost Neutral

pH  7

Color of litmus paper depends on pH.

Note If Ka > Kb then final solution will be acidic.

If Kb > Ka then final solution will be basic.

Examples

AgCN, (NH4)2CO3, NH4CN, (NH4)2C2O4 , ZnHPO3, CH3COONH4

CH3COO– + NH4+ + H2O CH3COOH + NH4OH
(1) Relation between Kh, Kw, Ka and Kb

For weak base

NH4OH ⇌ NH4+ + OH–

NH4+ OH –
Kb = ….(1)
NH4 OH

For weak acid

CH3COOH ⇌ CH3COO– + H+

CH3 COO– H+
Ka = ….(2)
CH3 COOH
For Hydrolysis

CH3 COO– + NH4+ + H2O ⇌ CH3COOH + NH4OH

CH3 COOH NH4 OH

Kh = ….(3)
CH3 COO– NH4+

For water

H2O ⇌ H+ + OH–

KW = [H+] [OH–] ….(4)

 CH3 COOH NH4 OH NH4+ OH –
Kh = Kb =
CH3 COO– NH4+ NH4 OH

CH3 COO– H +
Ka = KW = [H+] [OH–]
CH3 COOH

Multiply Eq. (1) × Eq. (2) × Eq. (3) = Eq. (4)

CH3 COOH NH4 OH NH4+ OH – CH3 COO– H+ –] H+

+ × × = [OH
CH3 COO– NH4 NH4 OH CH3 COOH

KW
Kh × Kb × Ka = KW  Kh = ….(5)
Ka × Kb
(2) Degree of hydrolysis (h)

CH3COO– + NH4+ + H2O ⇌ CH3COOH + NH4OH

t=0 c c 0 0
t = teq c – ch c – ch ch ch

CH3 COOH NH4 OH ch × ch c 2 h2

Kh = = =
CH3 COO– NH4+ c– ch 2 c2 1– h 2

h2 h Kh
Kh = 2 Or Kh = Or h=
1– h 1−h 1 + Kh

Note

Hence, the degree of hydrolysis of such salts is independent of the

concentration of salt solution.
(3) pH of the solution

– +
CH3 COO H
CH3COOH ⇌ CH3COO– + H+ ; K a =
CH3 COOH

CH3COO– + NH4+ + H2O ⇌ CH3COOH + NH4OH

t=0 c c 0 0

t = teq c – ch c – ch ch ch

K a × cH3 COOH K a × ch K a × h
H+ = = =
CH3 COO– c– ch 1– h
 [H + ] = K a ×
KW
=
KW ×Ka …(8)
Ka ×Kb Kb

Taking – log on both sides

1ൗ
KW × Ka 2
– log H + = – log
Kb

pH = – ½ [log KW + log Ka – log Kb]

pH = – ½ [log KW] – ½ [log Ka] – ½[ – log Kb ]

pH = + ½ pKW + ½ pKa – ½ pKb

1 1 1
pH = 7 + pK a – pK b  pH = 7 + (pKa – pKb) .…(9)
2 2 2
(Q.) pKa of a weak acid (HA) and pKb of a weak base (BOH) are 3.2 and 3.4, respectively. The pH
of their salt (AB) solution is
(1) 7.2 (2) 6.9
(3) 7.0 (4) 1.0

Solution

Answer:(2)
 SASB
SAWB Salts WASB Salts WAWB Salts
Salts

Neutral Acidic Basic Almost neutral

Solution Solution Solution solution

𝐾𝑤 𝐾𝑤 𝐾𝑤
– 𝐾ℎ = 𝐾ℎ = 𝐾ℎ =
𝐾𝑏 𝐾𝑎 𝐾𝑎 × 𝐾𝑏

𝐾𝑤 𝐾𝑤 𝐾𝑤
– h= h= h=
𝐾𝑏 × 𝑐 𝐾𝑎 . 𝑐 𝐾𝑎 . 𝐾𝑏

𝐾𝑤 × 𝑐 𝐾𝑤 . 𝑐 𝐾𝑤 . 𝐾𝑎
– 𝐻 + = c. h = 𝑂𝐻 – = c. h = 𝐻 + = 𝐾𝑎 . h =
𝐾𝑏 𝐾𝑎 𝐾𝑏

1 1 1 1 1 1
– pH = 7– p𝐾𝑏 – logc pH = 7 + p𝐾𝑎 + logc pH = 7 + p𝐾𝑎 – 𝑝𝐾𝑏
2 2 2 2 2 2
 Example

Which of the following solution has maximum degree of hydrolysis

(A) 0.1 M CH3COONH4 (B) 0.01 M CH3COONH4
(C) 0.001 M CH3COONH4 (D) Same in all

Solution

Ans. (D)
 Example

Which of the following solution has maximum pH?

(A) 0.1 M NaCl (B) 0.1 M NH4Cl
(C) 0.001 M NH4Cl (D) 0.01 M NH4Cl

Solution

Ans. (A)
 Example

Which of the following salt is maximum hydrolysed.

(A) KCl (B) KCN
(C) NH4Cl (D) NH4CN

Solution

Ans. (D)
 Example

The degree of hydrolysis of which of the following salt does not depend
upon the initial concentration of the salt ?
(A) KCN (B) NH4Cl
(C) CH3COONa (D) HCOONH4

Solution

Ans. (D)
NaHCO3, NaHS etc., can undergo ionisation to form H+ ion and can
undergo hydrolysis to form OH– (Na+ ion is not hydrolysed).

(a) NaHCO3 ⎯⎯⎯→ Na+ + HCO3–

Ionisation
Kw
HCO3 + H2O
– CO32– + H3O+ (acid) ; Ka1

Hydrolysis
HCO3– + H2O H2CO3 + OH– (base) ; Ka2

H+ and OH– also react

We can safely assume that both reactions have nearly same degree
of dissociation

H2 CO3 ≈ CO−2
3 ....(1)
Kw H2 CO3 [OH− ]
=
Ka1 HCO−3

1 H2 CO3
⇒ =
Ka1 H+ HCO− 3

CO−2
3 H+
= Ka2 ....(2)
HCO−3

H+ = Ka1 Ka2

pKa1 +pKa2
 pH =
2
(b) Similarly for H2PO4– and HPO42– amphiprotic anions.

pKa1 +pKa2 pKa2 +pKa3

pH (H2 PO−
4)
= = and pH
(HPO2−
4 )
=
2 2
 Example

What is the pH of 0.1M NaHCO3 ? K1 = 5 × 10-7, K2 = 5 × 10-11 for carbonic

acids.

Solution

Ans. 8.3
The solutions which resist change in pH on dilution or with the
addition of small amounts of acid or base are called Buffer
solutions.
 Types of Buffer

Simple Mixed

Weak acid
+ Acidic Buffer Basic Buffer
Weak base
Aqueous solution of the salt of a weak acid(WA) & weak base (WB)
types of salts.

Example

NH4CN, CH3COONH4, AgCN, PhCOONH4

CH3COO– NH4+ + H2O CH3COOH + NH4OH

Case 1 When acid is mixed [H+] :

CH3COO– + H+ CH3COOH

NH4OH + H+ NH4+ + H2O

Case 2 When base is mixed [OH–] :

NH4+ + OH– NH4OH

CH3COOH + OH– CH3COO– + H2O

The solution in which weak acid and its conjugate base are present or
aqueous solution of mixture of weak acid and salt of same weak acid
with any strong base is called acidic buffer solution.

Example

CH3 COOH + CH3 COONa

WA WASB

CH3COOH ⇌ CH3COO– + H+

CH3COONa → CH3COO– + Na+

CH3 COOH + CH3 COO– Na+ (Spectator ion)

WA C Base
The solution in which weak base and its conjugate acid are present
or Aqueous solution of mixture of weak base and salt of same weak
base with any strong acid is called basic buffer solution.

Example

NH4OH + NH4Cl

NH4OH ⇌ NH4+ + OH—

NH4Cl → NH4+ + Cl —

NH4 OH + NH4+ + Cl– (Spectator ion)

W.B. C acid

Note Buffer solutions are actually conjugate acid-base pairs.

Buffer, as we have defined, is a mixture of a conjugate acid-base
pair that can resist changes in pH when small amount of strong acids
or strong bases are added.

When a strong base is added, the acid present in the buffer

neutralizes the hydroxide ions (OH– ions).

When a strong acid is added, the base present in the buffer

neutralizes the hydronium ions (H3O+ ions).
(a) pH of acidic buffer solution

CH3 COOH + CH3 COONa

Acid Salt

CH3COOH ⇌ CH3COO– + H+

CH3COONa ¾® CH3COO– + Na+

CH3 COO– H+
Ka =
CH3 COOH

K a [CH3 COOH] K a [Acid]

Or H+ = =
CH3 COO– Conjugate base
 Ka [CH3 COOH] Ka [Acid]
or H+ = =
CH3 COO– Conjugate base

Taking –log on both sides

Acid
pH = pK a – log
Conjugate base

Or

Conjugate base
pH = pK a + log
Acid

Henderson's equation

Salt Conjugate base

pH = pK a + log or
Acid Acid
[Conjugate base]eq » [Salt] because CH3COO– mainly comes from
salt since dissociation of CH3COOH in presence of CH3COONa is
appreciably decreased.

[Acid]eq initial concentration of acid since it is almost unionised in

presence of salt due to common ion effect.
 Example

100 mL, 0.1 M acetic acid is mixed with 100 mL, 0.3 M sodium acetate.
Calculate the pH of buffer solution ? (pKa of acetic acid is 4.74)

Solution

Ans. 5.22
(b) pH range of acidic buffer solution

It depends on pKa of acid & ratio of salt to acid concentrations.

CH3 COONa
⇒ pH = pK a + log
CH3 COOH

[CH3COOH] : [CH3COONa]

(i) If, 1 : 10
 pH = pKa + log 10/1 = pKa + 1

(ii) If, 10 : 1
 pH = pKa –1
So pH range (pH = pKa ± 1) ^
 Example
s
CH3 COOH + CH3 COO¯Na+ pH = pKa + log
a s a

Solution

(i) add [H+] = x

CH3COO– + H+ ⎯⎯→ CH3COOH

t=0 s x a
t = teq. (s–x) (a + x)

s−x
pH = pKa + log
a+x

pH decreases slightly
Solution

(ii) add [OH–] = y

CH3COOH + OH– ⎯⎯→ CH3COO– + H2O

t=0 a y s
t = teq. (a – y) (s + y)

s+y
pH = pKa + log
a−y

pH increases slightly.
(a) pOH of basic buffer solution

NH4 OH + NH4 Cl
Base Salt

NH4OH ⇌ NH4+ + OH–

NH4Cl → NH4+ + Cl–

NH4+ OH– K b NH4 OH

Kb = or OH – =
NH4 OH NH4+

Taking – log on both sides :

NH4+
pOH = pK b + log
NH4 OH
Henderson’s Equation

Salt Conjugate acid

pOH = pK b + log or
Base Base
(b) pOH range of basic buffer solution

It depends on pKb of base and ratio of salt to base concentrations.

NH4 Cl
pOH = pK b + log
NH4 OH

NH4 OH ∶ NH4 Cl

(i) If, 1 : 10  pOH = pKb + 1

(ii) If, 10 : 1  pOH = pKb – 1

So, pOH range : pOH = pKb ± 1

Example
s
NH4OH + NH4Cl pOH = pKb + log
b
b s

Solution (i) Add small amount of H+(x)

NH4OH + H+ NH4+ + H2O

t=0 b x s
t = teq. (b – x) 0 (s + x)

s+x
pOH = pKb + log
b−x

pOH slightly increases, pH slightly decreases.

Solution

(ii) Add small amount of OH–(y)

NH4+ + OH– NH4OH

t=0 s y b

t = teq. (s – y) 0 (b + y)

s−y
pOH = pKb + log b+y

pOH slightly decreases, pH slightly increases.

Maximum buffer action condition of basic buffer solution

NH4OH : NH4Cl

1 : 1
NH4 Cl
 pOH = pK b + log NH4 OH

pOH = pKb
(Q.) 50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pKb of ammonia
solution is 4.75, the pH of the mixture will be:-
(1) 8.25 (2) 4.75
(3) 9.25 (4) 3.75

Solution

Answer:(3)
(Q.) 3g of acetic acid is added to 250 mL of 0.1 M HCl and the solution made up to 500 mL. To 20
𝟏
mL of this solution mL of 5 M NaOH is added. The pH of the solution is
𝟐

[Given : pKa of acetic acid = 4.75, molar mass of acetic acid = 60 g/mol, log3 = 0.4771]
Neglect any changes in volume
Solution

Answer:(5.23)
 Example

500 ml of 0.5M NH4Cl solution is added into 300 ml of 0.3M NH4OH

solution then find the pH of final mixture. (Given: (pKa)NH4+ = 9.26)

Solution

Ans. 8.82
 It is defined as the number of moles of strong acid (or strong base)
added to one litre of a buffer solution to change its pH by one unit.

Buffer capacity is a quantitative measure of the resistance of a

buffer solution to acid or base.

Larger the value of buffer capacity more resistant is the solution to

pH change.

Number of gm eq. of strong acid or strong base added per litre

Buffer Capacity =
Change in pH of buffer solution
 Let there be a buffer solution of volume 1 L with ‘b’ mole of anion (coming
from salt) and ‘a’ moles of weak acid.

The pH of the buffer would be given by :

𝑏
pH = pKa + log 𝑎 …eqn (1)

On adding x mole of a strong acid (monobasic), the pH changes to pH =

𝑏−𝑥
pKa + log 𝑎+𝑥. …eqn (2)

…eqn (2) -…eqn (1)

𝑏 𝑏−𝑥
 pH = log 𝑎 log 𝑎+𝑥 .
Differentiating with respect to x we get

d∆pH 1 1 b a+b 1 1 a+b

= × × =
dx 2.303 b × a + x a b−x 2 2.303 (a + x)(b − x)
a b−x

Talking the inverse

dx a + x (b − x) ab
= 2.303 ≈ 2.303  𝑥 ≪ 𝑎 ,𝑥 ≪ 𝑏
d∆pH a+b a+b

dx ab
= 2.303 (Buffer capacity)
d∆pH a+b
 Differentiating buffer capacity with respect to ‘b’, the amount of salt present
in the solution and equating it to zero, we get

d dx −1 × b − x + a − b + x
= 2.303 =0
db d∆pH a

a - b + 2x = 0 ; Since x is very small we ignore 2x and we get

a-b=0
 b = a  [Acid] = [Anion of salt]

Note
The buffer shows maximum buffer capacity when the amounts of
acid (or base) and the anion (or cation) from salt are same.
 

max.

[WA = CB] C.B or S

(for acidic buffer solution)

Solution having buffer capacity, when –

[C.B] 1 C. A 1
=1 =1
[WA] [WB]

Therefor for acidic buffer solution : pH = pKa and for basic buffer solution :
pOH = pKb
Solution having max. buffer capacity
 Example
3 g of acetic acid is added to 250 mL of 0.1 M HCl and the solution is made up
1
to 500 mL. To 20 mL of this solution, mL of 5 M NaOH is added. The pH of
2
the solution is _________.
(Given: pKa of acetic acid = 4.75, log 3 = 0.4771) (JEE MAIN 2020)
Solution

Ans. (5.22)
 Example

0.1 mole of CH3NH2 (Kb = 5 × 10–4) is mixed with 0.08 mole of HCl and diluted
to 1 L . What will be the H+ concentration in the solution? (IIT 2005 S)
(A) 8 × 10–2 M (B) 8 × 10–11 M
(C) 1.6 × 10–11 M (D) 8 × 10–5 M

Solution

Ans. (B)
 Example

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl.

If pKb of ammonia solution is 4.75, the pH of the mixture will be:
(A) 3.75 (B) 4.75 (C) 8.25 (D) 9.25
(JEE MAIN 2017)
Solution

Ans. (D)
 Example
In order to prepare a buffer solution of pH 5.74, sodium acetate is added to
acetic acid. If the concentration of acetic acid in the buffer is 1.0M,the
concentration of sodium acetate in the buffer is ________ M.
(Round off to the Nearest Integer).
[Given : pKa (acetic acid) = 4.74] (JEE MAINS 2021)

Solution

Ans : (10)
 Example

50 mL of 0.1 M CH3COOH is being titrated against 0.1 M NaOH. When 25 mL

of NaOH has been added, the pH of the solution will be _________× 10–2.
(Nearest integer) (JEE MAINS 2022)
(Given : pKa (CH3COOH) = 4.76)

Solution

Ans : (476)
 Example
Class XII students were asked to prepare 1L of buffer solution of pH 8.26 by
their chemistry teacher. The amount of ammonium chloride to be dissolved
by the student in 0.2 M ammonia solution to make 1 L of the buffer is
(Given pKb (NH3) = 4.74; Molar mass of NH3 = 17g mol–1 ;
Molar mass of NH4Cl = 53.5 g mol–1) (JEE MAINS 2022)
(A) 53.5 g (B) 72.3 g (C) 107.0 g (D) 126.0 g

Solution

Ans. (C)
 Example
A student needs to prepare a buffer solution of propanoic acid and its sodium
salt with pH = 4.
CH3 CH2 COO–
The ratio of required to make buffer is ……………
CH3 CH2 COOH

Given: Ka(CH3CH2COOH) = 1.3 × 10–5. (JEE MAINS 2022)

(A) 0.03 (B) 0.13 (C) 0.23 (D) 0.33

Solution

Ans. (B)
 Example
How many gram-mole of HCl will be required to prepare 1L of buffer solution
(containing NaCN and HCl) of pH 8.5 using 0.01 gram formula weight of
NaCN? (IIT 1988)
K dissociation (HCN) = 4.1 × 10–10.

Solution

Ans. 8.85 m mole

 Example

A solution is prepared by mixing 0.01 mol each of H2CO3, NaHCO3, Na2CO3,

and NaOH in 100 mL of water. pH of the resulting solution is _________.
(Given : pKa1 and pKa2 of H2CO3 are 6.37 and 10.32, log 2 = 0.30)
(JEE Advanced 2022)
Solution

Ans : (10)
 Example

The pH of blood stream is maintained by a proper balance of H2CO3 and

NaHCO3 concentrations. What volume of 5M NaHCO3 solution should be mixed
with a 10 mL sample of blood which is 2M in H2CO3 in order to maintain a pH
of 7.4 ? Ka for H2CO3 in blood is 7.8 × 10–7. (Antilog(0.4) = 2.51)

Solution (IIT 1993)

Ans. (78.36)
 Example
How many moles of sodium propionate should be added to 1 L of an aqueous
solution containing 0.020 mole of propionic acid to obtain a buffer solution of
pH 4.75? What will be pH if 0.010 mole of HCl is dissolved in the above buffer
solution. Compare the last pH value with the pH of 0.010 molar HCl solution.
Dissociation constant of propionic acid, Ka at 25°C = 1.34 × 10–5. pKa =4.87

Solution [IIT 1981]

Ans. 0.0118 m mole

 Example

20 mL of 0.2 M sodium hydroxide is added to 50 mL of 0.2 M acetic acid to

give 70 mL of the solution. What is the pH of this solution?
Calculate the additional volume of 0.2 M NaOH required to make the pH of
the solution 4.74.
Ka acetic acid is 1.8 × 10–5. (IIT 1982)

Solution

Ans. 5mL, pH=3.52

 Example

Calculate pH of 100ml of 0.1M CH3COOH solution (pKa of CH3COOH = 4.74)

If we add 50ml of 0.1M NaOH to the above solution, then find pH
If we add another 50ml of 0.1M NaOH to the above solution, then find pH
If we add another 50ml of 0.1M NaOH to the above solution, then find pH

Solution

Ans. (2.87, 4.74, 10.02, 12.3)

 Example

Calculate pH of 100ml of 0.1M NH4OH solution (pKb of NH4OH = 4.74)

If we add 50ml of 0.1M HCl to the above solution, then find pH
If we add another 50ml of 0.1M HCl to the above solution, then find pH
If we add another 50ml of 0.1M HCl to the above solution, then find pH

Solution

Ans. (11.13, 9.26, 3.98, 1.7)

 Example

Calculate pH of 100ml of 0.1M H2CO3 solution pKa1 of H2CO3 = 6.37

pKa2 of H2CO3 = 10.32
If we add 50ml of 0.1M NaOH to the above solution, then find pH

Solution
Solution
Maximum concentration of solute which can be dissolved under given set
of conditions.
or
Concentration of dissolved solute in saturated solution.
or
Dissolved concentration of solute at equilibrium.
(Rate of dissolution = Rate of precipitation)

Dissolution
AgCl(s) Ag+(aq.) + Cl–(aq.)
Precipitation
At constant temperature the maximum number of moles of solute which
can be dissolved in a solvent to obtain 1 litre of solution (i.e. saturated
solution) is called solubility.

Number of moles of solute

s(M) =
Volume of solution L

x(gm)
s= mol L–1
Mw × VL

𝒔𝐠Τ = 𝒔𝐌 × 𝐦𝐨𝐥𝐚𝐫𝐦𝐚𝐬𝐬
Types of
Salts

Soluble Solubility > 0.1M

Slightly Soluble 0.01M < Solubility < 0.1M

Sparingly Soluble Solubility < 0.01M

When a sparingly soluble salt such as AgCl is put into water, a very small
amount of AgCl dissolves in water and most of the salt remains
undissolved in its saturated solution.

The salt AgCl is an electrolyte, its dissociation occurs in solution. Hence,

the quantity of AgCl that dissolves in water dissociates into Ag+ and Cl–
ions.

Thus, in the saturated solution of AgCl an equilibrium exists between

undissolved solid AgCl and its ions, Ag+ and Cl– ions.

Solubility of ionic solids in water is very high.

 Dissolution
AgCl(s) Ag+(aq.) + Cl–(aq.)
Precipitation

t=0 ‘a’ mol 0 0

t = teq (a – s) s s

Suppose ‘s’ mole of AgCl are dissolved in 1 L of saturated solution then

we can say that solubility of AgCl is ‘s’ mole/L.

According to law of mass action,

Ag+ Cl−
K=
[AgCl]
According to law of mass action,

Ag+ Cl−
K=
[AgCl]

K[AgCl] = Ag+ Cl−

Ksp = Ag+ Cl−

Solubility product (Ksp) = s x s = s2

At constant temperature product of concentrations of constituent ions in a

saturated solution of substance is called solubility product of that substance.
According to law of mass action,

Value of Ksp depends on temperature & Nature of salt.

𝐾𝑠𝑝2 ∆𝐻𝑠𝑜𝑙 1 1
log = −
𝐾𝑠𝑝1 𝑅 𝑇1 𝑇2
Ksp of AB type salts

Example

AgCl, BaSO4, CH3COONa, CaCO3, NaCN, KCN, NH4CN, NH4Cl etc.

AB(s) ⎯→ A+(aq) + B–(aq)

t=0 a 0 0
t = teq a–s s s

K sp = A+ B –

K sp = 𝑠 2 or s = K sp
Ksp of AB2 or A2B type salts

Example

CaCl2, CaBr2, K2S, (NH4)2SO4, K2SO4, K2CO3 etc.

AB2(s) ⎯→ A2+(aq) + 2B–(aq)

t=0 a 0 0
t = teq a–s s 2s

Ksp = [A+2] [B–]2 = (s) × (2s)2 = 4s3

1ൗ
K sp 3
s=
4
Ksp of AB3 or A3B type salts

Example

FeCl3, AlCl3, K3PO4 etc.

AB3(s) ⎯→ A3+(aq) + 3B–(aq)

t=0 a 0 0
t = teq (a – s) s 3s

Ksp = [A+3] [B–]3 = s × (3s)3 = 27s4

1ൗ
K sp 4
s=
27
Ksp of A2B3 or A3B2 type salts

Example

Al2(SO4)3, Ba3(PO4)2 etc.

A2B3(s) ⎯→ 2A3+(aq) + 3B2–(aq)
t=0 a 0 0
t = teq (a – s) 2s 3s

Ksp = [A+3]2 [B2–]3

Ksp = 2s × 2s × 3s × 3s × 3s = 108 s5
1ൗ
K sp 5
s=
108
General form for salt AxBy

AxBy(s) ⎯→ xAy+(aq) + yBx–(aq)

t=0 a 0 0

t = teq (a – s) xs ys

Ksp = (xs)x . (ys)y

K sp = x x . y y . s x+y
 Example

If the solubility product of PbS is 8 × 10–28, than the solubility of PbS in pure
water at 298 K is x × 10–16 mol L–1. The value of x is _______. (Nearest integer)
[Given 2 = 1.41] (JEE MAINS 2022)

Solution

Ans : (282)
 Example

At 310 K, the solubility of CaF2 in water is 2.34 × 10–3 g/100 mL. The solubility
product of CaF2 is_______ × 10–8 (mol/L)3.
(Given molar mass : CaF2 = 78 mol–1) (JEE MAINS 2022)

Solution

Ans : (0.0108)
 Example

Which salt is most soluble in water ?

(A) CuS(Ksp = 4 × 10–15) (B) FeS (Ksp = 2 × 10–6)
(C) CaCO3 (Ksp = 9 × 10–18) (D) AgCl (Ksp = 5 × 10–21)

Solution

Ans. (B)
 Example
Which of the following ionic solids is maximum soluble in water under the
similar conditions ?
(A) Ag2S (Ksp = 4 × 10–15) (B) ZnS (Ksp = 10–10)
(C) Ag2CO3 (Ksp = 3.2 × 10–11) (D) AlCl3 (Ksp = 2.7 × 10–15)

Solution

Ans. (D)
Ionic product (IP) of an electrolyte is defined in the same way as Ksp. The
only difference is that ionic product expression contains concentrations of
constituent ions at any time whereas the expression of Ksp contains
concentrations of constituent ions at equilibrium for saturated solution only.

Thus, for AgCl  IP = [Ag+]t[Cl–]t

 Ksp = [Ag+]eq[Cl–]eq
Ionic product changes with concentrations of constituent ions but Ksp does
not depend on concentrations of constituent ions, it depends on
temperature only.

AgCl(s) ⎯→ Ag+(aq.) + Cl–(aq.)

Ionic product is applicable to all types of solution i.e. saturated or

unsaturated whereas solubility product is applicable only to saturated
solutions in which there exists a dynamic equilibrium between the
undissolved salt and the ions present in solution .

Therefore, the solubility product is, in fact, the ionic product for a saturated
solution .
To decide whether an ionic compound will precipitate or not, its Ksp is
compared with the value of ionic product. The following three cases arise :

(i) IP < Ksp  Forward  More Salt dissolves

The solution is unsaturated, and precipitation will not occur.

(ii) IP = Ksp  Equilibrium  (r)Dissolution = (r)Precipitation  No more salt dissolves

The solution is saturated and solubility equilibrium exists.

(iii) IP > Ksp  Backward  Precipitation occurs.

The solution is supersaturated, and precipitation will occur.

 Example
The molar solubility of Cd(OH)2 is 1.84 × 10–5 M in water. The expected
solubility of Cd(OH)2 in a buffer solution of pH = 12
2.49
(A) 1.84 × 10–9 M (B) 1.84 × 10−9 M
(C) 6.23 × 10–11 M (D) 2.49 × 10–10 M (JEE MAINS 2019)

Solution

Ans : (D)
 Example
Two salts A2X and MX have the same value of solubility product of 4.0 × 10–12.
𝑠(𝐴2 𝑋)
The ratio of their molar solubilities, i.e., =______.
𝑠(𝑀𝑋)

(Round off to the nearest integer) (JEE MAINS 2021)

Solution

Ans : (50)
 Example
Which of the following sets of concentrations is responsible for the
precipitation of PbI2
(KSP = 1.2 × 10–13 M)
(A) [Pb2+] = 10–5 ; [I–] = 10–5 M (B) [Pb2+] = 10–7 ; [I–] = 10–4 M

(C) [Pb2+] = 10–6 ; [I–] = 10–5 M (D) [Pb2+] = 10–4 ; [I–] = 10–4 M

Solution

Ans. (D)
 Example

At what pH the precipitation of Zn(OH)2 (KSP = 1 × 10–12 M3) will start from the
aqueous solution which is saturated with 0.001 M Zn2+ ion.

Solution

Ans. 9.5
Due to the presence of common ion, solubility of salt will decrease, higher
the concentration of common ion, lesser would be the solubility.

AgCl(s) ⎯→ Ag+(aq.) + Cl-(aq.)

t = teq s s

On addition of ‘c’ M HCl :

HCl(s) ⎯→ H+(aq.) + Cl-(aq.)

c c
 Due to the presence of common ion, solubility of salt will decrease, higher
the concentration of common ion, lesser would be the solubility.

AgCl(s) ⎯→ Ag+(aq.) + Cl-(aq.)

t = teq s s

After addition on c M HCl

AgCl(s) ⎯→ Ag+(aq.) + Cl-(aq.)

s s+c
t = (teq)new (s – x) (s – x) + c

 New solubility (s’) = s – x  Solubility is decreased.

Solubility of sparingly soluble salts always decreases in the presence of
common ion.

According to Le-Chatelier's principle on increasing common ion

concentration equilibrium shifts in backward direction until the equilibrium
is re- established.

The solubility of substance decreases but Ksp remains same because it is

an equilibrium constant which depends only on temperature.
 Example

Find out the solubility of AgCl in the presence of ‘c’ molar NaCl solution ?

Solution

Ksp
Ans. c
 Example
The solubility of AgCl will be maximum in which of the following ?
(A) 0.01 M KCl (B) 0.01 M HCl (JEE MAINS 2022)
(C) 0.01 M AgNO3 (D) Deionised water

Solution

Ans : (D)
 Example
The solubility product of PbI2 is 8.0 × 10–9. The solubility of lead iodide in 0.1
molar solution of lead nitrate is x × 10–6 mol/L. The value of x is ________.
(Rounded off to the nearest integer)
[Given : 2 = 1.4] (JEE MAINS 2014,2019,2022)

Solution

Ans : (141)
(Q.) 3g of acetic acid is added to 250 mL of 0.1 M HCl and the solution made up to 500 mL. To 20
𝟏
mL of this solution mL of 5 M NaOH is added. The pH of the solution is
𝟐

[Given : pKa of acetic acid = 4.75, molar mass of acetic acid = 60 g/mol, log3 = 0.4771]
Neglect any changes in volume
Solution

Answer:(5.23)
 Example
Concentration of H2SO4 and Na2SO4 in a solution is 1 M and 1.8 × 10–2 M,
respectively. Molar solubility of PbSO4 in the same solution is x × 10– Y M
(expressed in scientific notation).
The value of Y is _________.
(Given : Solubility product of PbSO4 (Ksp) = 1.6 × 10–8. for H2SO4, Ka1 is very
large and Ka2 = 1.2 × 10–2) (JEE Advanced 2022)

Solution

Ans : (6)
AgCl(s) ⎯→ Ag+(aq) + Cl–(aq) ; Ksp1
(x + y) x

AgBr(s) ⎯→ Ag+(aq) + Br–(aq) ; Ksp2

(x + y) y

x = solubility of AgCl

y = solubility of AgBr
 Ksp1 = x (x + y) ………(1) for AgCl

Ksp2 = y (x + y) ………(2) for AgBr

Add : Ksp1 + Ksp2 = (x + y)2

x + y = Ksp1+Ksp2

Ksp1
x=
Ksp1+Ksp2
Ksp2
y=
Ksp1+Ksp2
 Example

Calculate [F–] in a solution saturated with respect of both MgF2 and SrF2.
Ksp(MgF2) = 9.5 × 10–9, Ksp(SrF2) = 4 × 10–9.

Solution

Ans. 3 × 10–3 M
Effect of complex formation on solubility of sparingly soluble salt

Solubility of AgCl(Ksp=10–10) in ‘C’ (mol/L) NH3(aq.) solution

+
Given ∶ K f Ag NH3 2 = 1012

(1) AgCl Ag + aq + Cl− aq. : K sp = 10–10

s ………… (1)
s s
=s–x
+
Ag + + 2NH3 Ag NH3 2: K f = 1012
………… (2)
s c –
s–x c – 2x xs
L.R = c – 2s
Reaction almost goes to completion Approximation should be taken x  s

Add equation (1) & (2)

+
AgCl s + 2NH3 Ag NH3 2 + Cl– : K sp × K f
C–2s s s

s 2
K sp × K f =
c– 2s

Note

Complex formation increases the solubility

 Example
⊕ ⊕
If Ag + NH3 ⇌ [Ag(NH3)] ; K1=1.6 × 103
⊕ ⊕
[Ag(NH3)] + NH3 ⇌ [Ag(NH3)2] ; K2 = 6.8 × 103.
⊕
The formation constant of [Ag(NH3)2] is :
(A) 6.08 × 10–6 (B) 6.8 × 10–6 (C) 1.6 × 103 (D) 1.088 × 107
(JEE Advanced 2006)
Solution

Ans. (D)
 Example
Consider the following equilibrium
AgCl ↓ + 2NH3 ⇌ Ag(NH3 )2 + + Cl−
White precipitate of AgCl appears on adding which of the following ?
(A) NH3 (B) aqueous NaCl
(C) aqueous HNO3 (D) aqueous NH4Cl (JEE MAINS 2014)

Solution

Ans : (B)
 Example

Solubility of AgCN (Ksp) in water :

Given : (Ka)HCN = 10-10

(1) AgCN s Ag + aq. + CN – aq. ; K sp = s s − x

s — —
— s s
= (s–x)
Hydrolysis of anion

– Kw
CN – + H2 O HCN + OH : ∴ Kh =
Ka
= Kb CN–

s — —
s-x x x = 10–5

Kw x2
∴ Kh = =
Ka s−x

Note

Due to hydrolysis solubility of Sparingly soluble salt increases.

 Example

Solubility of AgCN (Ksp) in acidic solution and (Ka)HCN = 10–10

Given : ‘C’ mol of H+ is added

(1) AgCN s Ag + + CN – ; K sp = s s − x
s s
= (s–x)
1
(2) CN – + H + HCN ; = 1010
(from acid) Ka
s c —
s-x c-x x
= c–s =s  Approx : x  s x = 10–5
Add eq.(1) & (2)

AgCN s + H+ Ag + aq. + HCN aq.

(from acid)
c-s s s

1 s2
K sp × =
Ka c– s

Note

Conclusion in acidic solution solubility increases.

Special NOTE

If H+ is given by buffer solution then [H]+ = (c–s)  c

1 s2 K sp
then eq. (1) should be – K sp × = ⇒S= ×c
Ka c Ka

[H+]  pH  solubility 
 Example
The solubility of AgCN in a buffer solution of pH = 3 is x. The value of x is :
[Assume : No cyano complex is formed;
Ksp (AgCN) = 2.2 × 10–16 and Ka(HCN) = 6.2 × 10–10]
(A) 1.9 × 10–5 (B) 1.6 × 10–6
(C) 2.2 × 10–6 (D) 0.625 × 10–6 (JEE MAINS 2021)

Solution

Ans : (A)