1. INTRODUCTION
Most of the chemical reactions occur in solutions. When a given substance is dissolved in the
solvents like water. In an ionic solution the substance splits up into ions whereas in molecular
solution, the substance remains as s1uch. Both the solutions can be represented as
XY(s) aq XY(aq) X (aq) Y (aq)
 
  (for ionic solution)
XY(s) aq XY(aq)
 (for molecular solution)
The substances are classified regarding this concept is;
1. Electrolytes and
2. Non-electrolytes
1. Electrolyte (conducting)
Those substance whose aqueous solution or molten form can conduct electricity.
They are further classified into
(a) Strong electrolyte: Those substance whose aqueous solution or molten form conduct
electricity to a greater extent. They almost completely ionised in water.
e.g. NaCl, H2SO4, HCl, NaOH, NH4Cl
Since strong electrolyte completely ionised in aqueous solution so their ionisation is
represented as
HCl aq H (aq) Cl (aq)
 
 
 
(b) Weak electrolyte: Those substance whose aqueous solution or molten form conducts
electricity to a lesser extent. They do not completely ionised in water i.e. partly ionised.
They behaves as poor conductor of electricity.
e.g. when CH3COOH is dissolved in water, it is ionised partly and an equilibrium is setup
between the ions and the unionised electrolyte.
3 3
CH COOH aq CH COO (aq) H (aq)
 
 
2. Non-electrolyte (non-conducting)
Those substance whose aqueous solution or molten form does not conduct electricity to any
extent. They are bad conductor of electricity.
e.g. aqueous solution of sugar, urea, etc. do not conduct electricity.
DEGREE OF IONISATION ‘’
It may be defined as “fraction of total number of molecules which dissociates into ion”. It is
represented by .
Number of molecules dissociated as ions
Total number of molecules dissolved
 
Number of moles dissociated
Total number of moles dissolved

Ostwald dilution law (Ionisation of weak electrolytes)
It is the law of mass action as applied to weak electrolytes like CH3COOH, NH4OH, HCN, etc.
Consider a binary electrolyte AB which dissociates into its ions, there exist a dynamic equilibrium
between ions and unionised molecule of the electrolyte as

2. 2
AB aq A (aq) B (aq) ...........(1)
Initial conc.
when t 0 C 0 0
Conc. at
equilibrium C - C C C
 
 

  
Apply the law of mass action to equilibrium (1)
[A (aq)] [B (aq)]
K
[AB]
 

C C
[C C ]
 

 
2
C
1


 
For weak electrolyte, 1
, 1 1
      i.e.  < 5%
2
K C
 
or
K
C
  ……… (2)
If one mole of electrolyte is dissolved in V litre of solution i.e.
1
C
V

KV
  ………… (3)
Thus “degree of dissociation of a weak electrolyte is proportional to the square root of
dilution”. This is called Ostwald law.
Note:
If  > 5% then we can not neglect  from denominator.
2
C
K
(1 )


 
 can be calculate by quadratic equation (ax2 + bx + c = 0)
2
C (1 )K 0
    
2
b b 4ac
2a
  
 
Here b (1 )
   
a = C and c = 0
COMMON ION EFFECT
Considering the dissociation of weak electrolyte AB with a salt containing a common ion [A+] or
[B–].
AB A B
 

AC A C
 

 
Common ion
Salt AC is completely ionised, as a result concentration of A+ ions in equilibrium increases. By
Le-chateliers principle, the addition of A+ ion to an equilibrium, having common ion makes the
equilibrium in backward direction i.e. dissociation of AB electrolyte is suppressed. This is called
common ion effect.
In other words “The degree of dissociation of weak electrolyte is suppressed by the addition of
strong electrolyte containing a common ion is called common ion effect”.

3. 3
Example: To the solution of weak acid CH3COOH if its salt CH3COONa is added. The
dissociation of CH3COOH is suppressed.
3 3
CH COOH CH COO H
 

3 3
CH COONa CH COO Na
 

 
(salt)
Common ion
Illustration 1. Give reason that acetic acid is less acidic in sodium acetate solution than in
sodium chloride solution.
Solution: Ionisation of acetic acid is suppressed in sodium acetate due to common ion
effect. While in sodium chloride, no such effect is noticed.
THEORY OF ACID AND BASE
Classical concept of Acids and Bases: According to this concept.
Acid: Acid is a substance which in aqueous solution have the following property.
(i) has a sour test.
(ii) react with active metal to give H2.
(iii) conduct electricity.
(iv) turns blue litmus red.
(v) react with base and loose its property.
Base: Base is a substance which in aqueous solution gives the following property.
(i) has a bitter test.
(ii) have soapy touch.
(iii) conduct electricity.
(iv) turns red litmus blue.
(v) react with acids and loose its property.
The above definition is based upon certain operations (tests). This definition is replaced by
conceptual definition.
Arrhenious concept: According to Arrhenious
Acid: An acid is a substance which gives hydrogen ion (H+) in water.
Example HCl is an acid as it gives H+ in water.
HCl aq. H (aq) Cl (aq)
 
 
 
Example: Substance like H2SO4, HCl, HNO3 etc. when dissolved in water dissociate completely
are called strong acids.
Substance like CH3COOH, HCN, H3PO4, H2CO3 etc. when dissolved in water dissociate to a
lesser extent called weak acids.
3 3
CH COOH aq CH COO (aq) H (aq)
 
 
Base: A base is a substance which gives hydroxyl ion (OH–) in water.
Example: Substance like NaOH, KOH etc. when dissolved in water dissociate completely are
called strong bases.
NaOH aq Na (aq) OH (aq)
 
 
 
Neutralisation reaction involves the combination of H+ ion from acid and OH– ion from base.
2
H (aq) Cl (aq) Na (aq) OH (aq) Na (aq) Cl (aq) H O( )
     
   
  
i.e. 2
H (aq) OH (aq) H O( )
 
 

Limitations of Arrhenious concept
(i) Nature of H+
and OH–
ion
In aqueous solutions H+ and OH– ions can not exist as such but exist as hydrated ions as
H+(aq) and OH–(aq).

4. 4
(ii) It is failed to explain the acidic and basic character of certain substances which do not
contain H+ and OH– ions.
e.g. CO2, SO2, SO3 etc. are acids and NH3, Na2CO3, CaO etc. are bases.
(iii) It failed to explain the reaction between an acid and base in absence of water.
e.g. 3 4
CaO(s) SO (g) CaSO (s)
 

3 4
NH (g) HCl(g) NH Cl(s)
 

(iv) This theory is limited to aqueous solution only.
Bronsted Lowry concept (proton donor acceptor concept):
According to Bronsted Lowry,
Acid: Acid is a substance that donates proton (H+).
Base: Base is a substance that accepts proton (H+).
An acid after loosing H+, the residual part becomes a conjugated base. Similarly a base after
accepting H+ becomes a conjugate acid.
Acid H conjugated base


Base H conjugated acid


3 4
HCl NH NH Cl
 
 
Acid
Base Conjugate
Conjugate
acid
base
If an acid is strong then its conjugated base will be weak and vice versa.
Water can act both as acid as well as base. Hence they are amphiprotic. Similarly 3 2 4
HCO , H PO ,
 
2
4 2 3 3
HPO , H PO , HSO
  
are amphiprotic or amphoteric.
e.g. 2
3 2 3 3
HCO H O CO H O
  
 
3 2 2 3
HCO H O H CO OH
 
 
Advantages of Bronsted Lowry concept
(i) This theory includes the ionic species which act as cation or anion.
(ii) This theory explains the acid-base reaction involving in non aqueous solution.
(iii) It can explain the nature of basic oxide like Na2O, CaO which do not contain H+ and OH–
ions.
Limitations
(i) It can not explain the acidic character of substance like BF3, AlCl3 etc. which do not have any
H+ but known to be acids.
(ii) It can not explain the reaction between the acidic oxides like CO2, SO2, SO3 etc. and basic
oxides like CaO, BaO, MgO etc. taking place in absence of solvent.
3 4
CaO SO CaSO
 

Illustration 2. Which is the stronger base towards a proton -
2
NH or -
2
PH and why?
Solution: Bond energy (N—H  P—H) ionisation suggests that 2
NH
will be stronger base.
This is constituent with the relative strengths of the respective conjugate acids:
NH3  PH3.

5. 5
Illustration 3. H3BO3 is ……………….acid
(A) Monobasic (B) Dibasic
(C) Tribasic (D) None
Solution: H3BO3 is monobasic acid
H3BO3 + H2O  B(OH)4
– + H+
Hence, (A) is correct.
Illustration 4. Which of the following is the strongest acid?
(A) HClO4 (B) HBrO4
(C) HIO4 (D) HNO3
Solution: The acidic character of oxy acids decreases down the group and increases along
the period. Also acidity increases with increase in oxidation number of central
atom.
Hence, (A) is correct.
Lewis concept of acids and base: According to this concept
Acid is a substance which accept a pair of electrons.
Base is a substance which donate a pair of electrons.
In simple way; Acid – electron pair acceptor and Base – electron pair donor
All electrophiles are Lewis acids e.g. AlCl3, BF3, H+ etc.
All nucleophiles are Lewis base e.g. 3
NH ,

H2O, R–NH2, R–OH, F–, OH–, Br– etc.
Example: BF3 is an acid as it can accept an electron pair and NH3 is a base as it donate an
electron pair.
The neutralisation reaction involves the donation of electron pair from base to acid resulting the
formation of coordinate bond.
3 3 3 3
BF NH F B NH

 
 

Type of Lewis acids
(i) Molecules having incomplete octet of central atom e.g. BF3, AlCl3, MgCl2, BCl3 etc.
(ii) Simple cations Ca2+, Ag+, H+, Fe3+ etc.
(iii) Molecules having vacant d-orbitals of central atom e.g. SnCl4, PF5, SiF4, RX3, TiCl4 etc.
(iv) Molecules containing multiple bond between the two atoms of different electronegativity.
O C O O C O
 
   
Type of Lewis bases
(i) Neutral molecules like H2O, NH3, R–OH, R–NH2 etc. having atleast one lone pair of electrons.
(ii) All negative ions like OH–, Cl–, Br–, 2
NH
etc.
Advantages
Lewis concept is most general idea out of all the concept which explain acidic, basic nature of
oxides, and other substances which do not contain H+ and OH– ions etc. But it has following
limitations.
Limitations of Lewis concept
(i) Lewis concept can not explain the strength of acids and bases.
(ii) In acid base reaction it involves the formation of coordinate bond which is not always true
e.g. HCl and H2SO4 do not contain coordinate bond.
(iii) Acid-base reactions are very fast but formation of coordinate bond is slow.

6. 6
Illustration 5. Which salt undergoes hydrolysis?
(A) CH3COONa (B) KNO3
(C) NaCl (D) K2SO4
Solution: Salt of strong acid and strong base does not undergo hydrolysis.
Hence, (A) is correct.
Illustration 6. The conjugate acid of 3-
4
PO is
(A) H3PO4 (B) H2
-
4
PO
(C) 2-
4
HPO (D) 2-
3
HPO
Solution: Conjugate acid and base differs by one proton hence the conjugate acid of PO4
–3
is HPO3
–2
Hence, (D) is correct.
Illustration 7. The conjugate base of hydrazoic acid is
(A) NH3 (B) N3H
(C) N3
– (D) N2
–
Solution: N3H H


 N3
–
Hence, (C) is correct.
Exercise 1.
SnO2 is an amphoteric oxide, explain.
Exercise 2.
Why -3
4
PO ion is not amphiprotic?
RELATIVE STRENGTH OF WEAK ACIDS AND BASES
The relative strength of weak acids and bases are generally determined by their dissociation
constant Ka and Kb respectively.
For weak acid CH3COOH:
a
K
3 3
CH COOH aq CH COO (aq) H (aq)
Initially C 0 0
At equilibrium C - C C C
 
 
  
a
C C
K
C C
  

 
2
C
1


 
(1 1
   assuming  is very small).
2
a
K C
  
Where Ka is equilibrium constant of weak acid which depends on temperature. Greater the value
of Ka, more is the strength of an acid.
Similarly for base NH4OH
2
b
K C
 
where Kb is the equilibrium constant for weak base.
For two acids having equi-molar concentrations
1
2
a
a
K
Strength of acid I
Strength of acid II K

Similarly for two base

7. 7
1
2
b
b
K
Strength of base I
Strength of base II K

SELF IONIZATION OF WATER (IONIC PRODUCT OF WATER)
Pure water is a weak electrolyte and itself ionises as
2 2 3
H O H O H O OH
 
 
or simply
2
H O H OH
 

Applying law of mass action
2
[H ] [OH ]
K
[H O]
 

or 3
2
2
[H O ] [OH ]
K
[H O]
 

where K is called dissociation constant.
Since the ionisation of water is very small it means very few H2O molecule are dissociated into
H and OH– ions i.e.
[H2O] = constant
1000
55.55 mole/litre
18
 
At 25oC, the value of K is
14
10
55.55

14
2
10
K [H O] [H ] [OH ] 55.55
55.55

 
   
or 14
w
K [H ] [OH ] 10
  
 
Kw is called ionic product of water.
pH SCALE
Sorenson gave the best way of expressing the acidic or basic nature of solution. This is a
logarithmic scale in which hydrogen ion concentration ranges from 10–14 to 1 mole/litre.
pH may be defined as logarithm of reciprocal of [H3O+].
or negative logarithm of hydronium ion concentration.
pH = –log [H3O+]
pH = – log [H+]
The pH range is taken from 0 to 14. The acidity or alkalinity of a solution can be represented on
pH scale as
pH value 0, 1
, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
, 12, 13, 14
strongly
acidic
weakly
acidic
neutral weakly
alkaline
strongly
alkaline
In similar manner, we can define pOH scale as
pOH = – log [OH–]
Both pH and pOH scale are related as
[H+] [OH–] = 10–14 at 25oC
–log [H+] – log [OH–] = 14
pH + pOH = 14
Limitations of pH scale
(i) pH value of solutions do not give immediate idea of their relative strength.
(ii) pH value can be negative too.

8. 8
(iii) A 10–8 M solution of acid can not have pH = 8.
Since the value of pH of acid may close to 7 but <7.
Illustration 8. Which of the following concentration has the largest degree of dissociation for a
weak acid?
(A) 1.0 M (B) 0.5 M
(C) 0.10 M (D) 0.01 M
Solution: (D)
Illustration 9. Which of the following solutions can be titrated with HCl as well as NaOH?
(A) Glycine (B) Pyruvic acid
(C) Triethyl amine (D) Aniline
Solution: (A)
Illustration 10. The dissociation constant of monobasic acids A, B and C are 104, 106 and 1010
respectively. The concentration of each monobasic acid is 0.1 M. Which of the
following has been arranged in increasing order of pH?
(A) A < B < C (B) C < A < B
(C) B < C < A (D) B < A < C
Solution: (A)
Illustration 11. Among the following, which causes the greatest change in pH on addition to
50 mL of a 0.2 M malonic acid solution?
(A) Addition of 25 mL of 0.02 M malonic acid
(B) Addition of 25 mL of 0.02 M NaOH solution
(C) Addition of 25 mL of 0.02 M HCl solution
(D) Addition of 50 mL of 0.2 M acetic acid
Solution: (B)
Determination of pH of acids and bases
(A) Strong Acid
A strong acid is defined as a substance which completely dissociates to give all the maximum
possible H+ ions that it is capable of giving.
For example
HCl  H+ + Cl–
Therefore [HCl] = [H+]
If we have to find the pH of 10–4 M HCl,
[HCl] = [H+]
pH = –log 10–4 = 4.
Now to find the pH of 10–7 M HCl, by applying the above logic, the pH would come out to be
7. This is not possible since the solution of HCl is acidic and it should have pH less than 7.
The error that is made here is that, [H+] which comes from acid is being considered, whereas
the H+ from acid and water should be taken into account.
Assuming that a is the amount of H+ (or OH–) coming from water. In the case of pure water
a  a = 10–14 ---------------------- (1)
In the case of 10–7 M HCl, the total [H+] will be (10–7+a) if a is the amount of H+ coming from
water.

9. 9
Since [OH–] = a.
[H+]T  [OH–]T = (10–7 + a)  a = 10–14 ---------------------- (2)
Where [H+]T is the total H+ coming from both acid and water.
Therefore a (the [H+] coming from water in the presence of 10–7 M HCl) can be calculated
from equation (2) and therefore the total [H+] would be equal to [H+]T = a+10–7 and
pH = –log [H+]T
Note: In order to figure out when to take the contribution of water, it should be noted that
when [H+]A > 10–6 M, water contribution need not be taken and when [H+]A < 10–6 M, water
contribution should be taken.
(B) Weak Acids
By definition weak acids are acids which dissociate weakly or feebly in water. For example
CH3COOH CH3COO– + H+. The equilibrium constant is represented as Ka.
3
a
3
[CH COO ][H ]
K
CH COOH
 

If we start with `C' moles of acetic acid and  is the degree of dissociation which is defined as
the number of moles of acetic acid which dissociates from one mole of acetic acid, then the
no. of moles dissociated would be `C'
CH3COOH CH3COO– + H+
Initial C 0 0
At eqb. C–C C C
 
3
a
3
CH COO H
K
CH COOH
 
   
   
 =
2
C C C
C(1 ) 1
  

   
Generally, for weak acids `' is very small and to a reasonable approximation we can neglect
 in comparison to 1.
Ka = C2;
 C
K
C
K
C
C
]
H
[
Since
;
C
K
a
a
A
a







 
Where [H+]A is the [H+] coming from the weak acid. Again if [H+]A <10–6 M, we must take into
account H+ coming from water.
Let us follow the above discussion and calculate the pH of 10–6 M CH3COOH with
Ka = 1.8  10–5. Following the above procedure
[H+]A = 5 6
1.8 10 10
 
  = 4.24  10–6
 pH = –log 4.24  10–6
= 6–log 4.24 = 6–0.6274 = 5.37
(C) Strong Acid + Weak Acid
Let us assume that we have a strong acid HA and a weak acid HB. Let the concentration of
HA and HB be C1 and C2 respectively and let the dissociation constant of HB be Ka.
HA  H+ + A–
HB H+ + B–
C2(1–) C2+C1 C2
[ is the degree of dissociation of HB in the presence of HA
[HB]=C2 (1–); [B–] = C2; [H+]T = C2 + C1
Therefore Ka = T
[H ] [B ]
[HB]
 
=
 
 
2 1 2
2
C C C
C 1
  
 
Solving this equation we can get the value of  and then determine the [H+] and find the pH.

10. 10
(D) Two Weak Acids
If we have two weak acids, HA and HB with concentrations C1 & C2 and dissociation
constants Ka1 and Ka2, then
HA H+ + A–,
C1(1–1) C11+C22 C11
HB H+ + B–
C2(1–2) C11+C22 C22
[1 and 2 are the degree of dissociation of
HA & HB in presence of each other]
[HA] = C1 (1–1); [A] = C11; [HB] = C2 (1–2);
[B–] = C22 and [H+] = C11 + C22
 
 
 
1
1 1 2 2 1 1
a
1 1
H A C C C
K
HA C 1
 
       
   
 
 
 
 
 
2
1 1 2 2 2 2
a
2 2
H B C C C
K
HB C 1
 
       
   
 
 
The above two equations can be solved for 1 and 2 from which the H+ concentration can be
calculated and the pH can be found out.
Note: The pH of bases will be calculated just like we do for acids, with the difference that
instead of H+ ions we would have OH– ions, instead of Ka, we would have Kb and instead of
pH, we would first calculate pOH. Then pH can be calculated as pH = 14–pOH.
Illustration 12. 0.2 (M) solution of monobasic acid is dissociated to 0.95% Calculate its
dissociation constant. Given  = 0.0095; C = 0.2 mole lit–1
Solution: Ka = C2 = 0.2  (0.0095)2 = 1.8  10–5
Illustration 13. Calculate pH for (i) 0.01 (N) Ca(OH)2, (ii) 102(M) HCl
Solution: (i) Ca(OH)2  Ca2+ + 2OH–
10–2N 0 0
0 10–2 2  10–2
 [OH] = 10–2  2 gm equivalent / lit = 10–2 gm mole / lit
pOH = 2, pH = 12
(ii) HCl H+ + Cl–
[H+] = 102 (M) or pH = –2.
This is not possible because pH range is 0–14.
pH = – log10
H
a 
Where H
a  = activity = molar concentration  activity coefficient
 Unless and unit activity coefficient is given it is not possible to calculate pH
of solution.
Illustration 14. If the H+ ion concentration of a solution is increased to ten times its initial value,
its pH will
(A) increase by one (B) remains changed
(C) decrease by one (D) increase by ten
Solution: (C)
Illustration 15. What will be the H+ ion concentration in a solution prepared by mixing
50.0 mL of 0.20 M NaCl, 25 mL of 0.10 M NaOH and 25.0 mL of 0.30 M HCl?
(A) 0.5 M (B) 0.05 M
(C) 0.02 M (D) 0.10 M
Solution: (B)

11. 11
Illustration 16. Determine the pH of the solution that results from the addition of 20.00 mL of
0.01 M Ca(OH)2 to 30.00 mL of 0.01 M HCl
(A) 11.30 (B) 10.53
(C) 2.70 (D) 8.35
Solution: (A)
Illustration 17. To 250.0 mL of M/50 H2SO4, 4 g of solid NaOH is added and the resulting
solution is diluted to 1.0 L. The pH of the resulting solution is
(A) 12.00 (B) 11.25
(C) 11.95 (D) 12.15
Solution: (D)
Illustration 18. A given weak acid (0.10 M) had pKa = 6. The pH of the solution is
(A) 2.5 (B) 3.5
(C) 4.5 (D) 6.5
Solution: (B)
Illustration 19. The pH of 1010 M H2SO4 will be almost
(A) four (B) seven
(C) six (D) zero
Solution: (B)
Exercise 3.
Saccharin (Ka = 2  1012
) is a weak acid represented by formula HSaC. A 4  104
mole
amount of saccharin is dissolved in 200 cc water of pH 3. Assuming no change in
volume, calculate the concentration of SaC1
ions in the resulting solution at
equilibrium.
Exercise 4.
Justify the statement that water behaves like an acid and also like a base on the basis
of protonic concept.
Exercise 5.
(i) What is the pH of 10–8
(N) HCl?
(ii) Calculate pH of basic solutions having [OH] as
(a) [OH–
] = 0.05 (M)
(b) [–
OH] = 5 (M)
DETERMINATION OF pH DUE TO HYDROLYSIS
When a salt is dissolved in a solvent, it first dissociates into its constituent ions. This process is
called dissolution. Now, if these ions chemically react with water, the process is called hydrolysis.
The salts that undergo hydrolysis after dissolution are:
(a) Salts of weak acids + strong bases
(b) Salts of weak bases + strong acids
(c) Salts of weak acids + weak bases
(a) Salt of a Weak Acid and Strong Base
Let us take a certain amount of weak acid (CH3COOH) and add to it the same amount
(equivalents) of a strong base (NaOH). They will react to produce CH3COONa.
CH3COOH + NaOH  CH3COONa + H2O
CH3COONa being a strong electrolyte, completely dissociates into its constituent ions.
CH3COONa  CH3COO– + Na+

12. 12
Now, the ions produced would react with H2O. This process is called hydrolysis
Na+ + CH3COO– + H2O CH3COOH + NaOH
We know that NaOH is a strong base and therefore it would be completely dissociated to give
Na+ and OH– ions.
Na+ + CH3COO– + H2O CH3COOH + OH– + Na+
Canceling Na+ on both the sides,
 CH3COO– + H2O CH3COOH + OH–
We can note here that ions coming from strong bases do not get hydrolysed. We should
note here that the solution will be basic. This is because the amount of CH3COOH produced
and OH– produced are equal. But CH3COOH will not completely dissociate to give H+ ions.
Therefore [OH–] ions will be greater than [H+] ions.
Since the reaction is at equilibrium,
KC =
CH COOH OH
3
CH COO
3
 
 
   
   
 
 
 


This equilibrium constant Kc is given a new symbol, Kh.

 
3
h
3
CH COOH OH
K
[CH COO ]


 
 

If we multiply and divide the above equation by [H+] of the solution, then
Kh =
 
3
3
CH COOH OH H
[CH COO ] [H ]
 
 
   
    =
3
3
H OH
[CH COO ] [H ]
[CH COOH]
 
 
   
   
Kh = w
a
K
K
 Kh =
 
3
3
CH COOH OH
[CH COO ]


 
  = w
a
K
K
CH3COO–+H2O CH3COOH + OH–
Initial: C 0 0
At eq: C(1–) C C
Where  is the degree of hydrolysis of CH3COO– ion.

 
C C
C 1
  
 
= w
a
K
K
If  is very much less than 1,
C2 = w
a
K
K
,  = w
a
K
K C
As [OH–] = C, [OH–] = C  w w
a a
K K C
K C K

[H+] =
K K
K w a
w =
- C
[
OH ]
or pH = – log [H+] = –log W a
K K
C
 

 
 
 
=
1
2
W a
1 1
pK pK logC
2 2
 
(b) Salt of a Weak Base and a Strong Acid
Let the acid be HCl and the base be NH4OH.
Therefore the salt would be NH4Cl.
NH4Cl completely dissociates into 4
NH
and Cl– ions.

13. 13
NH4Cl  4
NH
+ Cl–
Cl– + 4
NH
+H2O NH4OH + HCl
HCl being a strong acid dissociates completely to give H+ ions and Cl– ions.
 Cl– + 4
NH
+ H2O NH4OH + H+ + Cl–
4
NH
+ H2O NH4OH + H+
In this hydrolysis, NH4OH and H+ are being produced. This implies that the solution is acidic.
To calculate pH,
4
NH
+ H2O NH4OH + H+
Initial: C 0 0
At eqb: C(1–) C C
Where  is the degree of hydrolysis of 4
NH
.
Kh =
 
4
4
NH OH H
[NH ]


 
 
Multiplying and dividing by OH– and rearranging,
Kh =
 
4
4
NH OH H OH
[NH ] [OH ]
 
 
   
    =
4
4
H OH
[NH ] [OH ]
[NH OH]
 
 
   
    = w
b
K
K
 Kh = w
b
K
K
Now, substituting the concentrations,
Kh =
 
 
2
4 w
b
4
NH OH H K C C C
K C 1 1
[NH ]


    
    
   
If   0.1, then, C2 = w
b
K
K
,  = w
b
K
K C
Since [H+] = C, [H+] = C w
b
K
K C
= w
b
K C
K
or pH =
1
2
W b
1 1
pK pK logC
2 2
 
(c) Salt of a Weak Acid and Weak Base
Let the weak acid be CH3COOH and the weak base be NH4OH. Therefore, the salt of these
is CH3COONH4.
The salt completely dissociates.
CH3COONH4  CH3COO– + 4
NH
The ions get hydrolysed according to the reaction.
CH3COO– + 4
NH
+ H2O NH4OH + CH3COOH
Initial: C C 0 0
At equilibrium: C(1–) C(1–) C C
Kh =
   
4 3
4 3
NH OH CH COOH
[NH ] [CH COO ]
 

14. 14
Multiplying and dividing by H+ & OH– and rearranging,
Kh =
   
4 3
4 3
NH OH CH COOH H OH
[NH ] [CH COO ] [H ] [OH ]
 
   
   
   
=
3 4
3 4
H OH
[CH COO ] [H ] [NH ] [OH ]
[CH COOH] [NH OH]
 
   
   
   

= w
a b
K
K K
 Kh = w
a b
K
K K
Substituting the concentration terms,
Kh =
   
 
   
2 2
w
2
2
a b
C
K C C
K K C 1 C 1 C 1 1

   
  
     
   
 
 w
a b
K
1 K K


 
There is an important issue that needs clarification before we move on further. In this case,
we can see that both the ions (i.e., cation and anion) get hydrolysed to produce a weak acid
and a weak base (hence, we can’t predict whether the solution is acidic, basic or neutral). We
have considered the degree of hydrolysis of both the ions to be the same. Now we present an
explanation as to why this is incorrect and then state reasons for the validity of this
assumption.
Actually the hydrolysis reaction given earlier,
CH3COO– + 4
NH
+ H2O CH3COOH + NH4OH
is made up of the following three reactions,
CH3COO– + H2O CH3COOH + OH–
4
NH
+ H2O NH4OH + H+
H+ + OH– H2O
In fact the equilibrium between NH4OH, 4
NH
and OH– also exists.
Now, we calculate the pH of the solution as,
CH3COOH CH3COO– + H+
C C(1–)
Ka =
 
3
3
CH COO H
CH COOH
 
   
    =
 
C 1 [H ]
C

 

 [H+] = Ka 
1

 
 
 
 
Substituting
1

 
as W
a b
K
K K
[H+] = Ka
1

 
 
 
 
= Ka  w
a b
K
K K
= w a
b
K K
K
or pH =
1
2
W a b
1 1
pK pK pK
2 2
 

15. 15
Illustration 20. Calculate the pH at the equivalence point between the titration of 0.1 M,
25 ml CH3COOH with 0.05 M NaOH solution. Ka (CH3COOH) = 1.8  10–5
Solution: Volume of NaOH required to reach equivalence point
=
05
.
0
25
1
.
0 
= 50 mL.
 Concentration of salt formed =
millimoles of acid
total volume
=
25 0.1 0.1
75 3


Since [H+] =
14 5
w a
K K 10 1.8 10
0.1
C
3
 
  

 pH = 8.63
Illustration 21. When 0.2 M CH3COOH is neutralised with 0.2 M NaOH in 0.5 litre of water the
resulting solution is slightly alkaline. Calculate pH of resulting solution
Ka(CH3COOH) = 1.8  10–5.
Solution:
 
1
3 2 3
Cx
C 1 x Cx
CH COO H O CH COOH OH
 

 
2
2
h
Cx
K Cx , x 1
1 x
 

14
10
w
h 5
a
K 1 10
K 5.5 10
K 1.8 10




   

So, 2 10
h
K Cx 5.5 10
  
x2 = 55  10–10

14 14
9
w
6
K 10 10
H 1.3477 10 M
Cx 7.42 10
OH
 
 


      
  
 
 
pH = log[H+] = log(1.3477  109
) = 8.87
Illustration 22. Calculate amount of ammonium chloride required to dissolve in 500 ml water to
have pH = 4.5 (Kb for NH4OH = 1.8  105).
Solution: [H+] = 10pH
= 104.5
= 3.162  105
M
 
4 2 4
Ch
C 1 h Ch
NH H O NH OH H
 

 
Kh = Ch2

10
5
h h
5
K K 5.5 10
h 1.74 10
Ch 3.162 10
H





    

 
 

5
5
H 3.162 10
C 1.8 mol/L
h 1.74 10
 

  
 
  

500 ml of H2O contains
1.8
0.9 mole
2

Mass in gm = 0.9  53.5 = 48.15 gm

16. 16
Illustration 23. Calcium lactate Ca(Lac)2 is salt of weak organic acid. A saturated solution of
Ca(Lac)2 contains 0.13 mole of this salt in 0.5 litre solution. The pOH of this
solution is 5.60. Assuming a complete dissociation of salt, calculate Ka of lactic
acid.
Solution: 2 1
2
2 2 0.13
0.13 2
Ca(Lac) Ca 2Lac
 
 


 
1 1
2
0.52 x x x
Lac H O HLac OH
 

 
2 2
h
x x
K
0.52 x 0.52
 

as x is small
[OH1
] = 1056
= 2.5  106
= x
6 6
12
h
2.51 10 2.5 10
K 12.12 10
0.52
 

  
  
14
4
w
a 12
h
K 10
K 8.26 10
K 12.12 10



   

Exercise 6.
Calculate pH of an aqueous solution of 1 M CH4COONH4 using complete dissociation
pKa = 3.8
pKb (NH3) = 4.8
Exercise 7.
Explain the pH of an aqueous solution of sodium acetate is more than seven.
Exercise 8.
Why aqueous solution of borax is alkaline to litmus?
Exercise 9.
(i) Calculate pH of mixture containing 50 ml 0.05 (M) NH4OH and 50 ml 0.05 (M) CH3CO2H
 
a b
K K
p = p = 4.74
(ii) Calculate for 0.01 (N) solution of CH3CO2Na
(a) hydrolysis constant,
(b) degree of hydrolysis,
(c) pH given Ka (CH3COOH) =1.9 10–5
and Kb (NH4OH) = 1.9  105
BUFFER SOLUTIONS
A buffer solution is a solution which resists a change in its pH when such a change is caused by
the addition of a small amount of acid or base. This does not mean that the pH of the buffer
solution does not change (we make this assumption while doing numerical problems). It only
means that the change in pH would be less than the pH that would have changed for a solution
that is not a buffer.
There are two types of buffer solutions:
(i) Acidic buffer (ii) Basic buffer
(i) Acidic buffer: It is the solution of a weak acid and it salts with strong base.
e.g. (CH3COOH + CH3COONa), (Phthalic acid + Potassium acid phthalate),
(Boric acid + Borax)
(ii) Basic buffer: It is the solution of a weak base and its salt with strong acid.
e.g. (NH4OH + HCl), (Glycine + Glycine hydrochloride) etc.
(iii) Salt buffer: Aqueous solution of the salt of weak acid and weak base e.g. CH3COONH4.

17. 17
Explanation of acidic buffer
Since acidic buffer is the mixture of weak acid and its salt with strong base e.g. CH3COOH and
CH3COONa.
They have the following equilibria in solution.
3 3
CH COOH CH COO H
 
 (Feebly ionised)
3 3
CH COONa CH COO Na
 

  (Completely ionised)
2
H O H OH
 
 (Very feebly ionised)
The salt is completely ionised. Due to common ion effect CH3COO– ion from salt (CH3COONa)
suppresses the ionisation of CH3COOH. Thus acid buffer is a mixture of more unionised
CH3COOH molecule and more CH3COO– ion that obtained from ionisation of salt alone.
Addition of acid: If a drop of HCl is added, H+ ion from HCl combine with CH3COO– ion to form
feebly ionised CH3COOH molecule.
HCl H Cl
 

 
3 3
from HCl
CH COO H CH COOH
 

Addition of base
If a drop of NaOH is added, OH– ion from NaOH combine with CH3COOH molecule to form feebly
ionised H2O molecule.
NaOH Na OH
 

 
3 3 2
from NaOH
CH COOH OH CH COO H O
 
 
 
Thus small addition of H+ and OH– to buffer solution makes no appreciable change in pH of the
buffer.
Calculation of pH of acidic buffer
It consist of a weak acid which ionises as
a
K
HA H A
 

where Ka is ionisation constant.
a
a
K [HA]
[H ][A ]
K [H ]
[HA] [A ]
 


   ………… (1)
In preparation of buffer salt highly ionised salt BA is also added which ionise completely.
BA B A
 

 
Due to common ion effect, A– ions from salt BA suppresses the ionisation of weak acid.
Thus, [BA] [A ]


equilibrium initial
[HA] [HA] [acid]
 
and [A–] at equilibrium = [BA] = [salt]
a
[acid]
[H ] K
[salt]


Taking log and reversing the sign.
10 10 a
[acid]
log [H ] log K log
[salt]

   

18. 18
10 a
[salt]
pH log K log
[base]
  
a
[salt]
pH pK log
[base]
 
This is called Hendersons equation.
Basic Buffers
Having pH towards base side. It is a mixture of weak base and its salt with strong acid.
e.g. NH4OH + NH4Cl
Explanation:
4 4
NH OH NH OH
 

4 4
NH Cl NH Cl
 

 
The salt is almost completely ionised. Due to common ion effect 4
NH
ion from NH4Cl suppresses
the ionisation of NH4OH. Thus a basic buffer is a mixture of more unionised NH4OH molecule and
more 4
NH
ions that are obtained from ionisation of NH4Cl alone.
Addition of acid:
If a drop of HCl is added the H+ from HCl combine with NH4OH to form feebly ionised H2O
molecules.
HCl H Cl
 

 
4 4 2
feebly ionised
NH OH H NH H O
 
 
Addition of base:
If a drop of NaOH is added OH– ion from NaOH combine with 4
NH
ion forming feebly ionised
NH4OH molecules.
NaOH Na OH
 

 
4 4
NH OH NH OH
 

Hence small addition of acid or base to basic buffer solution makes no appreciable change in pH.
Calculation of pH of basic buffers
It consist of a weak base which ionises as
b
K
BOH B OH
 
 , where Kb is ionisation constant of weak base.
b
b
K [BOH]
[B ][OH ]
K [OH ]
[BOH] [B ]
 


  
In preparation of basic buffers a highly ionised salt BA is used. It ionised as
BA B A
 

 
Due to common ion effect, B+ ions from salt BA suppresses the ionisation of BOH. It can be
assumed that all B+ ions in basic buffers are from salt BA.
i.e. [BA] = [B+]
or [BOH] at equilibrium [base]

equilibrium
[B ] [BA] [salt]

 
b
[salt]
[OH ] K
[base]


Taking log and reversing the signs,

19. 19
10 10 b 10
[base]
log [OH ] log K log
[salt]

   
b 10
[salt]
pOH pK log
[base]
 
This is called Hendersons equation.
Illustration 24. CH3COOH (50 ml, 0.1M) is titrated against 0.1M NaOH solution. Calculate the
pH at the addition of 0 ml, 10 ml, 20 ml, 25 ml, 40 ml, 50 ml and 60 ml of NaOH.
Ka of CH3COOH is 2  10–5.
Solution: (i) When 0 ml of NaOH is added, the pH is due to acetic acid,
 [H+] = a
K C
 = 5
2 10 0.1

  = 6
2 10

pH = – log 6
2 10
 = –
1
2
[log 2–6] = 3–0.15 = 2.85
(ii) When 10 ml of NaOH is added, it reacts with CH3COOH to produce salt and
water. The solution is then a buffer.
pH = pKa + log
 
 
Conjugate base
Acid
= 4.699 + log
10 0.1
50 0.1 10 0.1

  
= 4.699 + log
1
4
= 4.0969
(iii) When 20 ml of NaOH is added.
pH = pKa + log
20 0.1
50 0.1 20 0.1

  
= 4.699 + log
2
3
= 4.5229
(iv) When 25 ml of NaOH is added,
pH = 4.699 + log
25 0.1
50 0.1 25 0.1

  
= 4.699
(v) When 40 ml of NaOH is added,
pH = 4.699 + log
40 0.1
50 0.1 40 0.1

  
= 4.699 + log 4 = 5.3011
(vi) When 50 ml of NaOH is added,
Here, if we use the buffer equation, pH would be = 
But we can’t use the buffer equation as there is no acid. Therefore we used
the hydrolysis equation.
 [H+] = W a
K K
C
C =
0.1
2
[  Total Volume is 100 ml and millimoles of salt is 50  0.1]
 [H+] =
14 5
10 2 10 2
0.1
 
  
= pH = 8.699
(vii) When 60 ml of NaOH is added, the excess OH– ion from NaOH would
suppress the hydrolysis of CH3COO– ion. So we can ignore the contribution
of OH– ion from the hydrolysis of CH3COO– ion.
 [OH–] =
0.1 10
110

[10 ml of OH– ion is in excess] =
1
110
pOH = 2.0414
 pH = 14 – 2.0414 = 11.9586

20. 20
Illustration 25. How much water should be added to 10.00 g of acetic acid to give a hydrogen
ion concentration equal to 1.0  103 M (given pKa = 4.74).
(A) 130 ml (B) 230 ml
(C) 30 ml (D) 330 ml
Solution: (D)
Illustration 26. The pH of an HCl solution is 2.0. Sufficient water is added to it, to make the pH of
the new solution equal to 5.0. The decrease in the hydrogen ion concentration is
(A) 10 times (B) 7 times
(C) 1000 times (D) 100 times
Solution: (C)
Illustration 27. Calculate the pH of a solution prepared by mixing 20 mL of a strong acid solution
of pH 3.0 and 3.0 mL of a strong base of pH 10.0.
(A) 2.5 (B) 3.5
(C) 4.5 (D) 6.5
Solution: (C)
Exercise 10.
Suppose it is required to make a buffer solution of pH = 4, using acetic acid and
sodium acetate. How much of sodium acetate is to be added to one litre of
N
10
acetic
acid? Ka = 1.8  105
.
Exercise 11.
How many g moles of HCl will be required to prepare one litre of buffer solution
(containing NaCN and HCN) of pH = 8.5 using 0.01 gm formula mass of NaCN.
Ka(HCN) = 4.1  1010
.
Exercise 12.
Calculate change in pH of 1 litre buffer solution containing 0.10 mole each of NH3 and
NH4Cl upon addition of (a) 0.02 mole of dissolved gaseous HCl, (b) 0.02 mole of
dissolved NaOH.
Exercise 13.
Explain the following:
Equal volume of 0.1 M of NaCN + 0.05 M of HCl acts as buffer.
OSTWALD THEORY OF INDICATORS
An indicator generally a weak organic acid or weak organic bases is a substance which is used to
determine the end point in a titration. They change their colours within a certain pH range
generally the colour change is due to shifting of indicator equilibrium, eg, for phenolphthalein
(HPh).
HPh H+ + Ph
Colourless pink
and shifting of this equilibrium from left to right produces pink colours
KIn =
 
H Ph
HPh
 
   
    and pH = pKIn +
 
Ph
HPh

 
 

21. 21
At equilibrium point [Ph
] = [HPh]
 pH = pKIn. also KIn = [H+]
Where KIn is the ionization constant of the indicator
SOLUBILITY PRODUCT
Consider a binary electrolyte AxBy with solubility S mole/litre.
y x
x y
A B xA yB
xS yS
 

y x x y
eq
x y
[x ] [B ]
K
[A B ]
 

y x x y
eq x y
K [A B ] [x ] [B ]
 

y x x y
sp
K [x ] [B ]
 
 = xxyyS(x+y)
Ksp is called solubility product.
Solubility product may be defines as “the product of the concentration of ions in a saturated
solution of an electrolyte at a given temperature”.
From solubility product, we may conclude that;
Case I: If, [ionic product] < Ksp
Then, the solution is unsaturated i.e. more solute go into the solution.
Case II: If, [ionic product] = Ksp
The solution is just saturated i.e. no more solute can be dissolved.
Case III: If, [ionic product] > Ksp
The solution is supersaturated i.e. precipitation takes place.
Simultaneous solubility
Solubility of two sparingly soluble salt having common ion dissolved in same solution is called
simultaneous solubility. Let S1 and S2 be the simultaneous solubility of AgBr and AgSCN
respectively.
sp 1 2 1
1 2 1
AgBr(s) Ag (aq) Br (aq); K [Ag ][Br ] (S S )S .......(1)
(S S ) S
   
   

sp 1 2 2
1 2 2
AgSCN(s) Ag (aq) SCN (aq); K [Ag ][SCN ] (S S )S .......(2)
(S S ) S
   
   

Thus simultaneous solubility of AgBr and AgSCN can be calculated by the above two expression.
Exercise 14.
The following solutions were mixed: 500 ml of 0.01M AgNO3 and 500 ml. of solution
that was both 0.01M in NaCl and 0.01M NaBr. Calculate [Ag+
], [Cl–
] and [Br–
]
Ksp(AgCl) = 1.0  10–10
Ksp (AgBr) = 5  10–13
Fractional Precipitation
It is the technique used to separate two or more ions from a solution by adding a selective
reagent that precipitates first one ion and then the second.
Considering a solution of 0.1 M Ba2+ and 0.1 M Sr2+. If K2Cr2O4 is added to this solution as a
precipitating agent. Ksp of BaCrO4 is 1.2 × 10–10
and Ksp of SrCrO4 is 3.5 × 10–5
Concentration of 2
4
CrO 
required to precipitated BaCrO4
10
sp 9
2
K 1.2 10
1.2 10 M
0.1
[Ba ]




   
Concentration of 2
4
CrO 
required to ppt. SrCrO4 =
5
4
3.8 10
3.5 10
0.1



 

22. 22
Since the required concentration of 2
4
CrO 
is low so BaCrO4 will ppt. first.
On addition of chromate ions to the solution, BaCrO4 starts precipitating in the solution when
concentration reaches 1.2 × 10–9 M. On adding further chromate ions to the solution when
concentration reaches upto 3.5 × 10–4, then SrCrO4 also start precipitating in the solution.
[Ba2+] left when SrCrO4 starts precipitating
10
sp 4
2 4
4
K BaCrO 1.2 10
[CrO ] 3.5 10

 

 

= 3.4 × 10–7 M
7
3.4 10
% (Ba ) left 100
0.1

 
  = 0.00034%
Calculation of solubilities of salts
We shall now discuss the solubilities of different types of salts under various conditions.
(i) Solubilities of AgCl (salt of a strong acid and strong base) in water
AgCl would dissolve in water as,
AgCl (s)  Ag+ (aq) + Cl– (aq)
At saturation point,
AgCl (s) Ag+(aq) + Cl– (aq)
If the solubility of the salt is x moles / l
 [Ag+] = xM, [Cl–] = xM
 x2 = Ksp
or x = sp
K
(ii) Solubility of AgCl in a solution that is having 0.1M in AgNO3
AgCl would dissolve and finally reach saturation.
AgCl(s) Ag+(aq) + Cl– (aq)
The Ksp of AgCl is approximately 10–10 . If AgCl were to be dissolved in water (pure), its
solubility would have been 10–5M (previous section). In the presence of 0.1M AgNO3 its
solubility will decreases due to common ion effect. This means that [Ag+] from AgCl would be
less than 10–5 M. Hence, we can ignore the contribution of Ag+ from AgCl.
If the solubility of AgCl is x  moles / l in the presence of 0.1M AgNO3, then
[Ag+] = 0.1 M, [Cl–] = x  M
 x =
sp
K
0.1
= 10–9 moles / l
(iii) Solubility of CH3COOAg (salt of weak acid and strong base) in water
CH3COOAg dissolves and reaches saturation. Since it is a salt of weak acid and strong base,
it would hydrolyse. If the solubility of the salt is x moles/ l then
CH3COOAg (s) CH3COO– (aq) + Ag+ (aq)
At eqb: x –y x
CH3COO– (aq) + H2O CH3COOH (aq) + H+ (aq)
At eqb: x –y y y
Where y is the amount of CH3COO– ion that is hydrolysed.
 (x –y) x = Ksp
 
2
y
y

x
= w
a
K
K
Knowing the values of Ksp and Ka, solubility of the salt can be calculated.
(iv) Solubility of CH3COOAg (salt of a weak acid and strong base) in an acid buffer of pH = 4
(assuming that the buffer does not have any common ion by CH3COOAg):
CH3COOAg would dissolve and reach equilibrium. It would then be hydrolysed. If the
solubility of the salt is x  M in this solution, then

23. 23
At eq; CH3COOAg (s) CH3COO– (aq) + Ag+ (aq)
x –y x 
CH3COO– (aq) + H+ CH3COOH (aq)
x – y 10–4 y
Since the solution is a buffer, the pH will be maintained.
( x  – y) x  = Ksp
 
4
y
10 y


 
 
x
=
a
1
K
Since in presence of basic buffer, the degree of hydrolysis will be suppressed by already
existing –OH ions, therefore the approximated formula which can be used is
x2 = sp
K (CH3COOAg) (neglecting y)
Knowing Ksp and Ka, the solubility can be calculated.
(v) Solubility of CH3COOAg in an buffer solution of pH = 9
Following the same logic as give in the earlier section,
CH3COOAg (s) CH3COO– (aq) + Ag+ (aq)
At eqb: x –y x 
CH3COO–(aq) + H2O CH3COOH + OH–
At eqb: x  – y y 10–5
Where x M is the solubility of the salt and y the extent to which it is hydrolysed.
( x  – y) x  = Ksp
5
y 10
y )


 

x
(
= w
a
K
K
Knowing, Ksp and Ka, the solubility can be calculated.
(vi) Solubility of AgCl in an aqueous solution containing NH3
Let the amount of NH3 initially be ‘a’ M. If the solubility of the salt is x moles/ l, then
AgCl (s) Ag+ (aq) + Cl– (aq)
At eq: x –y x
Ag+(aq) + 2NH3 (aq) Ag(NH3)2
+(aq)
x–y a–2y y
Where y is the amount of Ag+ which has reacted with NH3.
( x –y) x = Ksp
   
2
y
y a 2y
 
x
= Kf (formation constant of Ag(NH3)2
+)
Knowing Ksp and Kf, the solubility can be calculated.
Illustration 28. The solubility of BaSO4 in water is 2.3  10–4 gm/100 mL. Calculate the
percentage loss in weight when 0.2 gm of BaSO4 is washed with (a) 1lt of water
(b) 1lt of 0.01N Na2SO4.
Solution: (a) Solubility is in general expressed in gm/lt,
so solubility of BaSO4 = 2.3  10–3 g /lt
Loss in weight of BaSO4 = amount of BaSO4 soluble
 %loss =
3
2.3 10
.2



 100 = 1.15%
(b) Now 0.01 N N Na2SO4  0.01 N 2
4
NSO 
ions
 0.005 M 2
4
SO 
ions
Now presence of 2
4
SO 
prior to washing BaSO4 will suppress the solubility of
BaSO4 (due to common ion effect). The suppression will be governed by Ksp
value of BaSO4. So first calculate Ksp of BaSO4.

24. 24
Solubility of BaSO4 in fresh water = 2  10–3 g/lt

3
2.3 10
233.4


= Mol/lt = 9.85  10–6 M
Ksp = [Ba2+] 2
4
SO 
 
  = (9.85  10–6)2 = 9.71  10–11
Now let x be solubility in mol/lt
 [Ba2+] in solution = x mol/lt and 2
4
SO 
 
  in solution
= (x + 0.005) mol/lt
Ionic product = [Ba2+] 2
4
SO 
 
  = (x) (x + 0.005)
Ksp = Ionic Product at equilibrium (saturation)
 9.71  10–11 = (x) (x + 0.005)  x2 + 0.005 x –9.71  10–11 = 0
 x =
11
9.71 10
0.005


= 1.94  10–8 mol/lt = 1.94  10–8  233.4 g/lt
 4.53  10–6 gm of BaSO4 are washed away
 percentage loss =
6
4.53 10 100
0.2

 
= 2.26  10–3%
Illustration 29. Two weak solutions are isohydric when their
(A) hydrogen ion concentrations are the same before mixing
(B) hydrogen ion concentrations are the same before and after mixing
(C) degrees of dissociation are the same
(D) chemical properties are the same
Solution: (A)
Exercise 15.
20 ml of 0.001 M AgNO3 solution is added to one litre of 0.002 M K2CrO4 solution. Will
there be any precipitation? Ksp Ag2CrO4 = 2.4  1012
.
Exercise 16.
Explain, a solution of FeCl3 in water gives a brown precipitate on standing.
Exercise 17.
Explain the following:
(i) Mg(OH)2 is soluble in NH4Cl and not in NaCl.
(ii) MgSO4 gives a precipitate with NH4OH, but not with NH4Cl and NH4OH.
Exercise 18.
Explain the following:
(i) CaC2O4 is insoluble in CH3COOH but soluble in dilute HCl.
(ii) AgCl is less soluble in AgNO3 solution than in pure water.
Exercise 19.
Explain the following:
(i) An excess of H2SO4 is added in gravimetric estimation of Ba2+
as BaSO4.
(ii) Little acetic acid is added while preparing the solution of CuSO4.
Exercise 20.
Explain the following:
(i) CoS is precipitated from acid solution of a cobalt salt by H2S in presence of
sodium acetate.
(ii) Little H2SO4 is added while preparing the solution of Mohr’s salt.

25. 25
ANSWERS TO EXERCISES
Exercise 1:
2 2 3 2
SnO 2NaOH Na SnO H O
 
 
SnO2 is acidic oxide since soluble in base.
2 4 2
SnO 4HCl SnCl 2H O
 
 
SnO2 is basic oxide since it is soluble in acid hence amphoteric oxide.
Exercise 2:
An amphiprotic ion is one which can donate proton as well as can accept. Proton 3
4
PO
ion can accept proton but cannot donate any proton. Hence 3
4
PO
is not amphiprotic.
Exercise 3:
Concentration of saccharin
4
3
4 10
1000 2 10 M
200



   
[H+] = 10pH
= 103
M
3
3
10 x
2 10 x
HSac H Sac


 

 

 
  
 
3
a 3
10 x Sac
H Sac
K
HSac 2 10 x
 
 

    
   
 
 
 
 
12 3
1
3
2 10 2 10 x
SaC
10 x
 


   
  
  
Since x is very small, it can be neglected
12 3
1 12
3
2 10 2 10
SeC 4 10 M
10
 
 

  
    
 
Exercise 4:
Water ionises as: 2 2 3
H O H O H O OH
 
 
With strong acid water behaves as a base and accept the proton given by the acid e.g.
2 3
HCl H O Cl H O
 
 
While with strong base, water behaves as an acid by liberating a proton e.g.
2 3 4
H O NH NH OH
 
 
Such solvents are known amphiprotic solvents.
Exercise 5:
(i) pH = 6.9586
(ii) (a) pH = 12.6989
(b) pH = near to 14
Exercise 6:
a b
pK pK
pH 7
2 2
  
1 1
pH 7 3.8 4.8 7 1.9 2.4 6.5
2 2
        
pH = 6.5

26. 26
Exercise 7:
Sodium acetate on hydrolysis forms sodium hydroxide and acetic acid. Since former is a
strong base and later is weak acid hence the pH of the aqueous solution is more than
seven.
3 2 3
CH COONa H O CH COOH NaOH
 
Exercise 8:
On hydrolysis, borax gives strong base (NaOH) and weak acid (H3BO3) to bring in
alkaline nature.
2 4 7 2 3 3
Na B O 7H O 2NaOH 4H BO
 
Exercise 9:
(i) 7
(ii) (a) 5.26  10–10
(b) 2.29  10–4
(c) 8.36
Exercise 10:
Using Henderson’s equation
 
  a
Salt
pH log logK
Acid
 
4 = log[salt]  log(0.1)  log1.8  105
log[Salt] = (4  1  5 + 0.2552)
 [Salt] = 0.018 g mole/litre
 Amount of salt = 0.018  82 = 1.476 gm
Exercise 11:
a a a
NaCN HCl NaCl HCN
  
[NaCN] = (0.01  a), [HCN] = a

 
 
a
NaCN
pH pK log
HCN
 
10
0.01 a
8.5 log log4.1 10
a


  

0.01 a
log 8.5 0.617 10
a

  
 a = 0.0089 mole
Exercise 12:
pOH of NH3 and NH4Cl buffer
 
 
5
b
salt 0.1
logK log log1.8 10 log 4.75
base 0.1

       
pH = (14  4.75) = 9.25
First case: 3 4
NH HCl NH Cl
 
[Salt] = (0.1 + 0.02) = 0.12 M
[Base] = (0.1 + 0.02) = 0.08 M

27. 27
b
0.12
pOH logK log 4.75 0.176 4.926
0.08
     
pH = (14 – 9.926) = 9.074
 
pH 9.25 9.074 0.176 unit
   
Second case: 4 3
NH OH NaOH NH NaCl
  
[Salt] = (0.1  0.02) = 0.08 M
[Base] = (0.1 + 0.02) = 0.12 M
 
b
0.08
pOH logK log 4.75 0.176 4.576
0.12
     
pH = (14  4.574) = 9.426
 
pH 9.426 9.25 0.176 unit
   
Exercise 13:
On combination of NaCN + HCl
NaCN HCl NaCl HCN
Millimoles before reaction 0.1 V 0.05 V 0 0
Millimoles after reaction 0.05V 0 0.05V 0.05V
 
 
 
Thus solution contains millimoles of NaCN(0.05 V) and HCN (0.05 V) in 2 V mL to act as
buffer.
Exercise 14:
[Cl–] = .005 M
[Ag+] = 2  10–8
[Br–] = 2.5  10–5
Exercise 15:
Number of moles of Ag+ in 20 ml 5
20
0.001 2 10
100

   
Number of moles 2
4
CrO 
in one litre = 0.002 = 2  103
After mixing, total volume of the solution = 1000 + 20 = 1.02 L
5
5
2 10
Ag 1.96 10
1.02

 

    
 
3
2 3
4
2 10
CrO 1.96 10
1.02

 

    
 
Ionic product    
2
2 2 5 3 13
4
Ag CrO 1.96 10 1.96 10 7.53 10
    
   
     
   
There will be no precipitation of Ag2CrO4 as ionic product is less than Ksp.
Exercise 16:
FeCl3 hydrolysed in water thus Fe3+ ions and OH– ions form; which combine to form
brown precipitate of Fe(OH)3.
3
3
Fe OH Fe(OH)
 
 

Exercise 17:
(i) Addition of NH4Cl to Mg(OH)2 brings in interaction:
2 4 2 4
Mg(OH) 2NH Cl MgCl 2NH OH
 
 
The NH4OH being weak base reduces the OH– in solution and thus product of [Mg2+]
and [OH–]2 remains lower than Ksp of Mg(OH)2 to give no precipitation of Mg(OH)2.
On the other hand interaction of NaCl with Mg(OH)2 gives strong alkali NaOH and the
product of [Mg2+] and [OH–]2 exceeds their Ksp to show precipitation.

28. 28
(ii) No doubt NH4OH is weak base but it provides appreciable OH– ion to exceed the
product of ionic concentration of Mg2+ and OH– than their Ksp and thus MgSO4 is
precipitated out as Mg(OH)2. On the other hand the dissociation of NH4OH is
suppressed in presence of NH4Cl and thus [OH–] diminishes to the extent that
[Mg2+] [OH–]2 < Ksp. Thus MgSO4 is not precipitated.
Exercise 18:
(i) CH3COOH (being weaker acid than oxalic acid) does not decompose CaC2O4; on the
other hand HCl being stronger acid forms Ca2+ and oxalate ion to pass CaC2O4 in
solution state.
2 4 2 2 2 4
CaC O HCl CaCl H C O
 
 
(ii) When AgCl is dissolved in AgNO3 solution then concentration of Ag+ ions increases
to such an extent that ionic product of Ag+ and Cl– ions exceeds the solubility product
of AgCl. While this is not so in pure water.
Exercise 19:
(i) If excess of H2SO4 is used, then the concentration of 2
4
SO 
ions increases to such an
extent that ionic product of Ba2+ and 2
4
SO 
ions exceeds the solubility product of
BaSO4. Thus complete precipitation of BaSO4 takes place.
(ii) CH3COOH checks up hydrolysis of CuSO4 by dissolving Cu(OH)2 formed during
dissolution.
Exercise 20:
(i) Sodium acetate being a strong electrolyte gives more CH3COO– ions which combine
with H+ ions to form CH3COOH (weak acid). Thus the concentration of S2– ions
increases to such an extent that the product of ionic concentration of Co2+ and
S2– ions exceeds the solubility product of CoS. Hence we get the precipitate of CoS.
(ii) On dissolution of Mohr salt [FeSO4(NH4)2.SO4.6H2O] to water, hydrolysis takes place
to give green turbidity of Fe(OH)2.H2SO4 added dissolves the Fe(OH)2 in it i.e. checks
the hydrolysis of Mohr’s salt and gives clear solution.

29. 29
MISCELLANEOUS EXERCISES
Exercise 1: A particular water sample has 131 ppm CaSO4. What fraction of the water must
be evaporated in a container before solid CaSO4 begins to deposit
Ksp of CaSO4 = 9.0  106?
Exercise 2: Calculate the solubility of AgCN in a buffer solution of pH = 3.
Given Ksp of AgCN = 1.2  1016 and Ka for HCN = 4.8  1010.
Exercise 3: Determine the number of mole of AgI which may be dissolved in 10 litre of
1.0 M CN
solution. Ksp for AgI and Kc for  

2
Ag CN are 1.2  1017 M2 and
7.1  1019 M2 respectively.
Exercise 4: 100.0 mL of a clear saturated solution of Ag2SO4 is added to 250.0 mL of a clear
saturated solution of PbCrO4. Will any precipitate form and if so what? Given,
Ksp values for Ag2SO4, Ag2CrO4, PbCrO4 are 1.4  105, 2.4  1012, 2.8  1013
and 1.6  108 respectively.
Exercise 5: What is the minimum volume of water required to dissolve 1 g of calcium
sulphate at 298 K. For calcium sulphate, Ksp = 9.1  106.
Exercise 6: A sample of hard water contains 0.005 mole of CaCl2 per litre. What is the
minimum concentration of Na2SO4 which must be added for removing Ca+2 ions
from this water sample? Ksp for CaSO4 is 2.4  105 at 25oC.
Exercise 7: Determine the solubility of barium sulphate in 0.005 M in a barium chloride
solution. The solubility product constant of barium sulphate is 1.1 1010 at 298 K.
Exercise 8: Calculate the amount of (NH4)2SO4 in g which must be added to 500 mL of
0.2 M NH3 to yield a solution of pH = 9.35. Kb for NH3 = 1.78  105.
Exercise 9: A solution contains 0.1 M H2S and 0.3 M HCl. Calculate the conc. of S2 and
HS
ions in solution. Given 1 2
a a
K and K for H2S are 107 and 1.3  1013
respectively.
Exercise 10: A solution contains a mixture of Ag+ (0.10 M) and 
2
2
Hg (0.10 M) which are to be
separated by selective precipitation. Calculate the maximum concentration of
iodide ion at which one of them gets precipitated almost completely. What % of
that metal ion is precipitated?
   
 
 
-
17 -
26
sp sp 2 2
K AgI = 8.5 ×10 ; K Hg I = 2.5 ×10

30. 30
ANSWER TO MISCELLANEOUS EXERCISES
Exercise 1: 0.68 V or 68%
Exercise 2: 1.58  105
mol litre1
Exercise 3: 2.91 mole
Exercise 4: 7.120  106
> Ksp (Ag2CrO4). Hence ppt. Is formed.
Exercise 5: 2.46 litre of water
Exercise 6: 4.8  103
mol litre1
Exercise 7: 2.2  109
Exercise 8: 5.248 g
Exercise 9: 3.3  108
M, 1.44  1020
M
Exercise 10: 5  1013
M, 99.83%

31. 31
SOLVED PROBLEMS
Subjective:
Board Type Questions
Prob 1. Explain the following:
Zn2+ is precipitated as ZnS by H2S in presence of NH4OH and not in presence of HCl.
Sol. In presence of HCl, the dissociation of H2S is suppressed appreciably due to
common ion effect and [Zn2+] [S2–] does not exceed higher values of Ksp of ZnS
(in comparison to II group sulphide). However, in presence of an alkali (say NH4OH)
the acid base reaction between H2S and alkali shows an increase in concentration of
S2– ions and [Zn2+] [S2–] exceeds over the Ksp of ZnS to show precipitation.
Prob 2. Electrical conductance of acetic acid increases on addition of NaOH in to it.
Sol. The dissociation of acetic acid increases in presence of NaOH which gives rise an
increase in 3 3 2
CH COOH NaOH CH COONa H O
strong
 
 
number of ions present in solution and thus conductance increases.
Prob 3. Why NH4Cl is added in III group precipitation before addition of NH4OH.
Sol. To prevent precipitation of IV group hydroxides (specially Mn) along with III group
hydroxides. NH4Cl decreases dissociation of NH4OH and thus limited [OH–] is
present in solution to precipitate III group only.
Prob 4. Silver nitrate solution does not give white ppt of AgCl with pure CHCl3 but gives white
ppt with NaCl.
Sol. Pure CHCl3 is covalent compound whereas NaCl is ionic. The formation of AgCl by
AgNO3 is an ionic reaction.
Prob 5. Explain the following:
(i) H2S is a weak acid, yet it precipitates CuS from a copper sulphate solution.
(ii) ZnS is precipitated by H2S from the solution of Zn(CH3COO)2 but not from a
solution of ZnCl2.
Sol. (i) H2S is a weak acid and gives few S2– ions on ionization. Even then the
concentration of S2– ions is sufficient to exceed the solubility product of CuS and
so CuS is precipitated out.
(ii) When H2S is passed in solution of Zn(CH3COO)2 and ZnCl2, CH3COOH and HCl
are formed. HCl being strong electrolyte suppresses the dissociation of H2S to
the extent that [Zn2+] [S2–] < Ksp and thus ZnS remains as unprecipitated mass.
On the other hand CH3COOH a weak acid does not suppress the dissociation of
H2S to an appreciable extent and [Zn2+] [S2–] > Ksp to show precipitation of ZnS.

32. 32
IIT Level Questions
Prob 6. Calculate the pH of the solution when 0.1 M CH3 COOH (50 ml) and 0.1 M NaOH
(50 ml) are mixed, [Ka (CH3COOH) = 10–5]
Sol. CH3 COOH CH3 COO
+ H+ …(I)
NaOH  Na+ + OH
H+ + OH– H2O …(II)
(I) + (II)
CH3COOH + OH– CH3COO– + H2O …(III)
0.05 – x 0.05 – x x
Keq of eq. (III) = a
w
K
K
a
w
K
K
=
 
 
 
 
 
-
3
-
3
CH COO
CH COOH OH
 conc. of H2O remains constant.
109=
 
2
x
0.05 x

because value of equilibrium constant is very high
there for x 0.05
let 0.05–x=a
109= 2
0.05
a
 a2=
2
9
5 10
10


a = 6
50 10

POH= 6–log 7.07
 POH= 6–0.85
PH= 14–6+0.85 =8.85
Prob 7. Given the solubility product of Pb3 (PO4)2 is 1.5 x 10–32.Determine the solubility in
gms/litre.
Sol. Solubility product of Pb3 (PO4)2 = 1.5  10–32
Pb3 (PO4)2 3Pb2+ + 3
4
2PO 
If x is the solubility of Pb3 (PO4)2
Then Ksp = (3x)3 (2x)2 = 108 x5
x =
Ksp
5
108
=
1
32 5
1.5 10
108

 

 
 
x = 1.692  10–7 moles/lit
Molecular mass of Pb3(PO4)2 = 811
x = 1.692  10–7  811 g/lit = 1.37  10–4 g/lit

33. 33
Prob 8. What is pH of 1M CH3COOH solution? To what volume must one litre of this solution
be diluted so that the pH of resulting solution will be twice the original value.
Given : Ka = 1.8  10–5
Sol. CH3COOH + H2O H3CCOO– + H3O+
t = 0 1 M 0 0
t = teq (1–x) M x x
Ka =
2
x
1 x


2
x
1
 x = a
K = 4.2  10–3 = [H3O+]
pH = – log [H3O+] = – log [4.2  10–3] = 3 – log 4.2 = 2.37
Now, let 1L of 1M AcOH solution be diluted to VL to double the pH and the conc. of
diluted solution be C.
H3CCOOH + H2O H3CCOO– + H3O+
t = 0 C 0 0
t=teq C– 1.8  10–5 1.8  10–5 1.8  10–5
New pH = 2  old pH
= 2  2.37 = 4.74
pH = – log [H3O+] = 4.74
 [H3O+] = 1.8  10–5
Ka = 3 3
3
[CH COO ] [H O ]
[CH COOH]
 

1.8  10–5 =
5 5
5
1.8 10 1.8 10
C 1.8 10
 

  
 
 C = 3.6  10–5 moles / litre
on dilution
M1V1 = M2V2
1M  1L = 3.6  10–5 moles/ litre  V2
 V2 = 2.78  104 litre
Prob 9. Find the concentration of H+, 3
HCO
and 3
CO
, in a 0.01M solution of carbonic acid if
the pH of this solution is 4.18
Ka1(H2CO3) = 4.45  10–7 and Ka2 = 4.69  10–11
Sol. pH = – log[H+]
4.18 = – log [H+]
[H+] = 6.61  10–5
H2CO3 H+ + 3
HCO
Ka = 3
2 3
[H ][HCO ]
[H CO ]
 
or, 4.45  10–7 =
5
3
6.61 10 [HCO ]
.01
 
 
 3
[HCO ]

= 6.73  10–5
again, 3
HCO
H+ + 3
CO
 Ka2 = 3
3
[H ][[CO ]
[HCO ]
 

= 4.69  10–11 =
5
3
5
6.61 10 [CO ]
6.73 10
 



 3
[CO ]

= 4.8  10–11

34. 34
Prob 10. Calculate the molar solubility of Mg(OH)2 in 1M NH4Cl
KspMg(OH)2 = 1.8  10–11, Kb(NH3) = 1.8  10–5
Sol. Mg(OH)2(s) Mg++ + 2OH– K1 = Ksp
4
2NH
+ 2OH– 2NH4OH K2 = 2
b
1
K
on addition Mg(OH)2(s) + 4
2NH
Mg++ + 2NH4OH K =
sp
2
b
K
K
1 – 2x x 2x at equilibrium
Let molar solubility of Mg(OH)2 in 1MNH4Cl be x
11
10
1.8 10
1.8 1.8 10



 
=
2
2
x (2x)
1

= 4x3 =
0.1
1.8
=
1
18
x = 0.24 M
Prob 11. An aqueous solution of metal bromide MBr2 (0.05M) is saturated with H2S. What is
the minimum pH at which MS will ppt.?
Ksp for (MS) = 6  10–21
Concentration of standard H2S = 0.1 M
Ka1(H2S) = 1  10–7, Ka2(H2S) = 1.3  10–13
Sol. In saturated solution of MS
MS(s) M++ + S2–
 [S– –] =
21
SP
K 6 10
0.05
[M ]



 = 1.2  10–19
The precipitate of MS will form only if [S2–] exceeds the concentration of
1.2  10–19
H2S H+ + HS– Ka1
HS– H+ + S2– Ka2
——————————————
H2S 2H+ + S2–K = 1.3  10–20
K =
2 2
2
[H ] [S ]
[H S]
 
 1.3  10–20 =
2 19
[H ] 1.2 10
0.1
 
 
[H+] = 0.109
pH = 0.96
Prob 12. How much AgBr could dissolve in 1.0 L of 0.4 M NH3? Assume that [Ag (NH3)2]+ is
the only complex formed given, Kf [Ag(NH3)2]+ =1.0108, Ksp (AgBr)= 5.010–13
Sol. AgBr Ag+ + Br–  Ag+ + 2NH3 [Ag (NH3)2]+
Let x= solubility,
Then x= [Br–] = [Ag+] = [Ag(NH3)2]+
 8
3 2
2
3
[Ag(NH ) ]
1 10
[Ag ] [NH ]


 
since, [NH3]>> [Br–],
8
3 2 3 2
2 7
[Ag(NH ) ] [Ag(NH ) ]
1.0 10 (Ag )
[Ag ] [0.4] 1.6 10
 


   

Since, the majority of silver is in the form of the complex, X=[Br–] = [Ag(NH3)2]+
 Ksp= [Ag+] [Br–]
=
2
13
3 2
7 7
[Ag(NH ) ] [Br ] x
5.0 10
1.6 10 1.6 10
 

  
 
x2=8.010–6  x=2.810–3 M

35. 35
Prob 13. What is the maximum pH of a 0.1 M Mg2+, solution from which Mg (OH)2 will not be
precipitated [KSP = 1.2  10–11)
Sol. Maximum pH means minimum H+ or maximum OH–. This means that I.P. = KSP
Therefore [Mg2+] [OH–]2 = 1.2  10–11
11
5
1.2 10
OH 1.09 10
0.1

 

    
 
pOH = 5 – log 1.09 = 4.960
Prob 14. The solubility product of Ca(OH)2 at 250C is 4.4210–5 mole3 lit–3. A 500 ml of
saturated solution of Ca(OH)2 is mixed with equal volume of 0.4 M NaOH solution.
How much Ca(OH)2 in milligrams is precipitated.
Sol. Let the solubility of Ca(OH)2 in pure water = S moles/litre
Ca(OH)2 Ca2+ + 2OH
S 2S
Then Ksp = [Ca2+] [OH
]2
4.42  105
= S  (2S)2  4.42  105
= 4S3
S = 2.224  102
= 0.0223 moles litre1
 No. of moles of Ca2+ ions in 500 ml of solution = 0.01115
Now when 500 ml of saturated solution is mixed with 500 ml of 0.4 M NaOH, the
resultant volume is 1000 ml. The molarity of OH
ions on the resultant solution would
therefore be 0.2 M.
 2
[Ca ]

=
 
5
sp
2 2
K 4.42 10
0.2
OH




 
 
= 0.001105 M
Thus, No. of moles of Ca2+ or Ca(OH)2 precipitated
= 0.01115  0.001105 = 0.010045
Mass of Ca(OH)2 precipitated = 0.010045  74 = 0.7433 g = 743.3 mg.
Prob 15. Two weak monobasic organic acids HA and HB have dissociation constants as
1.5  10–5 and 1.8  10–5 respectively at 25°C. If 500 ml of 1 M solutions of each of
these two acids are mixed to produce 1 litre of mixed solution, what is the pH of the
resulting solution?
Sol. Note that Ka of two acids is nearly same. In such cases, we have to consider H+ from
both HA & HB simultaneously. The concentration of HA & HB in the mixture = 0.5 M
[equal volumes are mixed ] = say ‘c’
HA  H+ + A–
HB  H+ + B–
Let x = [H+] from HA and Y = [H+] from HB
 [H+]final = x + y
KHA =
(x y)x
c

& KHB =
(x y)y
c

15 x
18 y
  6x = 5y
substitute for y =
6
5
x in KHA =
2
x xy
C

 1.5  10–5 =
2 2
5x 6x
2.5

x = 1.84  10–3; y = 2.21  10–3
[H+] = x + y = 4.055  10–3
pH = – log10 (4.055  10–3)  pH = 2.392

36. 36
Objective:
Prob 1. If the degree of ionization of water be 1.8  10–9 at 298K. Its ionization
constant will be
(A) 1.8  10–16 (B) 1 10–14
(C) 1  10–16 (D) 1.67  10–14
Sol. Ka =
14
w
2
K 10
[H O] 55.5

 = 1.8  10–16
(A)
Prob 2. When a solution of benzoic acid was titrated with NaOH the pH of the solution when
half the acid neutralized was 4.2. Dissociation constant of the acid is
(A) 6.31  10–5 (B) 3.2  10–5
(C) 8.7  10–8 (D) 6.42  10–4
Sol. C6H5COOH + NaOH C6H5COONa + H2O
After 0.5 0.5
neutralization
It is a buffer solution of weak acid and its salt
pH = pKa + log
[salt]
[acid]
pKa = 4.2
Ka = 6.31  10–5
(A)
Prob 3. 10–2 mole of NaOH was added to 10 litre of water. The pH will change by
(A) 4 (B) 3
(C) 11 (D) 7
Sol. Old pH = 7
New [OH–] = 10–2 
1
10
= 10–3
New pH = 11
Change in pH = 4
(A)
Prob 4. If an aqueous solution at 25°C has twice as many OH– ion as pure water its pOH
will be
(A) 6.699 (B) 7.307
(C) 7 (D) 6.98
Sol. [OH–] = 2  10–7
pOH = 14 – pH or – log [OH–]
(A)

37. 37
Prob 5. Solubility of AgCl in water, 0.01M CaCl2, 0.01M NaCl and 0.05M AgNO3 are S1, S2,
S3 and S4 respectively then.
(A) S1  S2  S3 < S4 (B) S1 > S3 > S2 > S4
(C) S1 > S2 = S3 > S4 (D) S1 > S3 > S4 < S2
Sol. AgCl Ag+ + Cl–
In CaCl2
CaCl2 Ca+2 + 2Cl–
0.01 0.01 20.01
In NaCl
NaCl Na+ + Cl–
0.01 0.01 0.01
In AgNO3
AgNO3 Ag+ + Cl–
0.05 0.05 0.05
common ion effect is maximum in AgNO3
So, S1 > S3 > S2 > S4
(B)
Prob 6. What would be the pH of an ammonia solution if that of an acetic acid solution of
equal strength is 3.2? Assume dissociation constant for NH3 & acetic acid are equal.
(A) 3.2 (B) 6.4
(C) 9.6 (D) 10.8
Sol. pH of CH3COOH = pOH of NH3 solution Ka = Kb
 pH of NH3 solution = 14  3.2 = 10.8
Prob 7. pH of Ba(OH)2 solution is 12. Its solubility product is
(A) 10–6M3 (B) 4  10–6M3
(C) 0.5  10–7M3 (D) 5  10–7M3
Sol. Since pH = 12 pOH = 14 – 12 = 2
[OH–] = 10–2M
We know Ba (OH)2 Ba++ + 2OH–
[Ba++] =
2
10
2

M
KSP = [Ba++] [OH–]2 =
2
10
2

 
 
 
 (10–2)2 = 5  10–7M3
(D)
Prob 8. The hydrolysis constant for ZnCl2 will be
(A) Kh =
Kw
K
b
(B)
2
Kw
K =
h K
b
(C) Kh =
2
Kw
2
K
b
(D)
K
b
K =
h 2
Kw
Where Kb is effective dissociation constant of base Zn++

38. 38
Sol. Zn++ + 2H2O Zn(OH)2 + 2H+
 Kh =
 
2
2
Zn OH H
Zn


   
 
 
 
 
… (1)
Zn(OH)2 Zn++ + 2OH–
Kb =
 
2
2
Zn OH
Zn OH
 
   
   
 
 
, Kw = [H+] [OH–]

2
w
b
K
K
= Kh
 (B)
Prob 9. M(OH)x has KSP 4  10–12 and solubility 10–4M. Then the value of x is
(A) 1 (B) 2
(C) 3 (D) –4
Sol. M(OH)x will ionize in the way
M(OH)x M+ x + x OH–
10–4 x  10–4
 Ksp = [M+x] [OH–]x  (10–4) (x  10–4)x = 4  10–12
by inspection we get this relation will hold good when x = 2
(B)
Prob 10. The pH of an aqueous solution of 0.1M solution of a weak monoprotic acid which is
1% ionised is
(A) 1 (B) 2
(C) 3 (D) 11
Sol. HA H+ + A–
0.1 0 0
0.1
1
1
100
 

 
 
1 0.1
100
 1 0.1
100

[H+] = 10–3 pH = 3
(C)
True or False
Prob 11. Silver chloride is more soluble in concentrated solution of NaCl then impure water.
Sol. False
Prob 12. Larger the value of a Ka, the stronger the acid is.
Sol. True

39. 39
Prob 13. Degree of hydrolysis of ammonia acetate does not depend upon its concentration.
Sol. True
Prob 14. A mixture of NaCl and NaOH can acts as a buffer.
Sol. False
Prob 15. When a strong acid is neutralized by an equal volume of a weak base, the resulting
solution will be acidic.
Sol. True
Fill in the Blanks
Prob 16. The solution of CuSO4 is …………. due to …………..
Sol. acidic, hydrolysis
Prob 17. The pH of 108 M NaOH solution is slightly ……. than 7.
Sol. higher
Prob 18. The solution that is made by mixing appropriate amount of weak acid ad its salt with
strong base is called …………
Sol. acidic buffer
Prob 19. The buffer capacity of a solution is maximum when the molar solution of the salt to
acid is ………
Sol. unity
Prob 20. Solution of equal pH and pOH are called ……….
Sol. neutral

40. 40
ASSIGNMENT PROBLEMS
Subjective:
Level – O
1. How is degree of dissociation of weak base related to its molarity?
2. Ka for a weak acid HA is 5
2 10
 . What is the value of Kb for A
?
3. Explain common ion effect and how does it effects degree of dissociation?
4. Derive and state Ostwald’s dilution law.
5. Explain the term ‘solubility product’. How does it differ from ionic product?
6. 10–2 mole of NaOH was added to 10 litres of water then calculate the change in pH value.
7. If an acidic indicator HIn ionises as HIn H+ + In–. To which maximum pH value its
solution has distinct colour characteristic of HIn.
8. pH of a mixture of 1M benzoic acid (pKa = 4.20) and 1M C6H5COONa is 4.5. Calculate the
volume of benzoic acid in 300 ml buffer.
9. Calculate the pH of an aqueous solution of 0.1M solution of a weak monoprotic acid which is
1% ionized.
10. Calculate the pH of a 10–10 M NaOH.
11. The solubility of CaF2 in water at 20C is 15.6 mg per dm3 of solution. Calculate the solubility
product of CaF2.
12. Explain the following:
(i) A white precipitate is obtained if HCl gas is passed into a concentrated solution of BaCl2.
(ii) The precipitation of IInd group sulphides in qualitative analysis is carried out with H2S gas
in the presence of HCl and not in HNO3.
13. Calculate the pH of solution obtained by mixing 10 ml of 0.1 M HCl and
40 ml of 0.2 M H2SO4.
14. The solubility product of PbBr2 is 810–5. If the salt is 80% dissociated in saturated solution,
find the solubility of the salt.
15. The solubility of Mg(OH)2 in pure water is 9.5710–3 g/l. Calculate its solubility in
0.02 M Mg(NO3)2 solution.

41. 41
Level – I
1. Given that KSP of CaF2 is 1.7  10–10, how many grams of it can be dissolved in
(a) 0.1M NaF, (b) 0.1M CaCl2.
2. Kw for H2O is 9.62  10–14 at 60°C. What is nature of the solution whose (a) pH = 6.7
(b) pH = 6.35.
3. Calculate (i) the hydrolysis constant of NH4Cl and (ii) the [OH–] in a 0.1 M NH4Cl solution.
Kb(NH4OH) = 1.75  10–5, (iii) what is its pH value ?
4. Calculate (i) the degree of hydrolysis of ammonium acetate. The dissociation constant for
NH4OH is 410–5 and that of CH3COOH is 1.810–5 (ii) also calculate its pH value.
5. Calculate the charge in pH after adding 0.1 moles of H+ to 1L of formic acid-formate buffer in
which the concentration of HCHO2 and CHO2
– are each 1M.
6. Determine the pOH of a solution after 0.1 moles of NaOH is added to 1L of a solution
containing 0.15 M CH3COOH and 0.2 M CH3COONa. Assume no change in volume.
(Ka of CH3COOH = 1.8  105
)
7. Calculate the pH of 0.1M, Kb(C6H5NH2) = 4.6  10–10
8. Kw for H2O is 9.62  1014
at 60°C. What is pH of water at 60°C.
9. Calculate the conc. of fluoroacetic acid which is required to get [H+] = 1.5  10–3.
Ka (acid) = 2.6  10–3.
10. Calculate the amount of NH3 and NH4Cl required to prepare a buffer solution of pH 9.0, when
total concentration of buffering reagents is 0.6  
b
K
1
3
mol . p for NH 4.7

 .
11. Calculate pH of 10–8 M HCl solution
12. Find the simultaneous solubility of AgCN and AgBr if their solubility products are
1.010–12 and 5  10–13 respectively.
13. Calculate the pH after the addition of 90 ml and 100ml respectively of 0.1N NaOH to
100 ml 0.1N CH3COOH (Given pKa for CH3COOH = 4.74)
14. Calculate the pH and % hydrolysis in a 1M NaCN solution. Ka = 410–10.
15. 0.2 M solution of Ba(OH)2 is found to be 90% ionized at 250C. Find the pH of the solution at
that temperature.

42. 42
Level – II
1. What is pH of 0.1M H2S solution? Given that;
Ka1 (H2S) = 1.0  10–7
Ka2 (H2S) = 1.3  10–14
2. 50 ml of 0.1M HCl solution is added into 100 ml of 0.1 M NaOH solution and then volume is
made upto 500 ml. What volume of resulting solution be required to neutralise
10 ml. of 10–2 M H2SO4 acid.
3. 50 ml of 0.1M KOH solution is added into 25 ml of 0.2 M HCl solution. The volume of the
resulting solution was made upto 500 ml. Now, 0.745 gm of solid KCl is added into solution.
What is concentration of Cl– ion in the final solution? What volume of
0.1M AgNO3 solution be required for complete precipitation of all the Cl
ions as silver
chloride.
4. 3ml of 0.1M NaOH solution is added into 100 ml of 0.1M CH3COOH solution. What is the pH
of resulting solution? What volume of 0.1M NaOH solution be further added so that pH of
resulting solution be 4.74 (Given than Ka(CH3COOH) = 1.8  10–5
5. Calculate the pH at the equivalence point of the titration between 0.1M CH3COOH(25 ml) with
0.05 M NaOH. Ka(CH3COOH) = 1.8  105
.
6. Equal volumes of 0.02 M AgNO3 and 0.02M HCN were mixed. Calculate [Ag+] at equilibrium
assuming no cyano complex formation. Ka(HCN) = 6.2  10–10 and Ksp (AgCN) = 2.2  10–16
7. Calculate the amount of (NH4)2SO4 in g which must be added to 500 mL of 0.2 M NH3 to yield
a solution of pH = 9.35 Kb for NH3 1.78  10–5.
8. Calculate [F–] in a solution saturated with respect to both MgF2 and SrF2 assuming no
hydrolysis of F–. [Ksp(MgF2) = 6.5  10–9, Ksp(SrF2) = 2.9  10–16]
9. A buffer solution with pH 9 is to be prepared by mixing NH4Cl and NH4OH. Calculate the
number of moles of NH4Cl that should be added to 1 liter of 1M NH4OH.
Kb (NH4OH) = 1.810–5.
10. Calculate (i) the degree of hydrolysis of a decinormal KCN solution at 250C. The dissociation
constant of HCN is 7.2 10–10 and ionic product of water is 10–14. Also find
(ii) the pH value of this solution.

43. 43
Objective:
Level – I
1. Which of the following, when mixed, will give a solution with pH greater then 7?
(A) 0.1M HCl + 0.2M NaCl
(B) 100 ml 0.2 M H2SO4 + 100ml. 0.3M NaOH
(C) 25 ml of 0.1M HNO3 + 25 ml 0.1M NH3
(D) 100 ml 0.1M CH3COOH + 100 ml 0.1M NaOH
2. 1cc. of 0.1N HCl is added to 999 cc. Solution of NaCl. The pH of resulting solution
will be
(A) 4 (B) 3
(C) 5 (D) 6
3. Which of the following substances when added into a solution of acetic acid will not supress
its degree of dissociation?
(A) dil HCl (B) CH3OONa
(C) water (D) dil. H2SO4
4. What is the pH of solution which have 1 ml NH4OH (conc. 0.1M) 1 ml and (NH4)2SO4
concentration 0.05M. Given that Kb(NH4OH) = 10–5
(A) 5 (B) 9
(C) 4.74 (D) 8.26
5. During salt analysis, the medium is made acidic in the precipitation of sulphides of group II
due to
(A) increase the S2– ion concentration (B) decrease the S2– ion concentration
(C) to dilute the solution (D) to make solution viscous
6. The pH of a soft drink is 3.82. Its hydrogen ion concentration in mole L–1 will be
(A) 1.96  10–2 (B) 1.96  10–3
(C) 1.5  10–4 (D) 1.96  10–1
7. Which of the following solution will have pH close to 1?
(A) 100 ml
N
10
HCl + 45 ml of
N
10
NaOH
(B) 55 ml of
M
10
HCl + 45 ml of
M
10
NaOH
(C) 10 ml of
N
10
HCl + 90 ml of
M
10
NaOH
(D) 75 ml of
M
5
HCl + 25 ml of
M
5
NaOH
8. The precipitate of CaF2 (Ksp = 1.7  10–10) is obtained when equal volume of following are
mixed
(A) 10–4 M Ca2+ + 10–4 M F– (B) 10–2 M Ca2+ + 10–3 MF–
(C) 10–5 M Ca2+ + 10–3 M F– (D) 10–3 M Ca2+ + 10–5 M F–
9. The dissociation constants of acids HA, HB, HC and HD are 2.6  103
, 5.3  109
, 1.1  102
,
7.5  105
respectively. The weakest acid among these acids is
(A) HA (B) HB
(C) HC (D) HD

44. 44
10. When HCl gas is passed through a saturated solution of common salt, pure NaCl is ppted
because
(A) HCl is highly ionised in solution
(B) HCl is highly soluble in water
(C) The solubility product of NaCl is lowered by HCl
(D) The ionic product of [Na+] exceeds the value of solubility product of NaCl
11. What is correct representation for the solubility product of SnS2?
(A) [Sn++] [S2
--] (B) [Sn++] [S--]2
(C) [Sn++] [S--] (D) [Sn++++] [S--]2
12. Which of the following will occur if a 0.1M solution of a weak acid is diluted to 0.01M at
constant temperature?
(A) [H+] will decrease to 0.01M (B) pH will decrease
(C) Ka will decrease (D) Percentage ionisation will increase
13. One litre of water contains 10–7 mole of H+ ions. Percentage degree of ionisation of
water is
(A) 1.8  10–7 (B) 0.8  10–9
(C) 3.6  10–7 (D) 3.6  10–9
14. The dissociation constant of a weak acid is 1.0  10–5, the equilibrium constant for its reaction
with strong base is
(A) 1.0  10–5 (B) 1  109
(C) 1.0  107 (D) 1.0  1014
15. Equilibrium constant for the reaction; CH3COOH + OH– CH3COO– + H2O; is
1.8  109. Hence the equilibrium constant for
H3COOH + H2O CH3COO– + H3O+ is
(A) 1.8  1023 (B) 1.8  105
(C) 1.8  10–5 (D) 5.55  10–10
16. pH of a mixture containing 0.10 M X– and 0.20 M HX is [pKb (X–) = 4]
(A) 4 + log 2 (B) 4 –log 2
(C) 10 + log 2 (D) 10 –log 2
17. H2O + H3PO4 H3O+ + 2 4
H PO
; 1
K
P = 2.15
H2O + 2 4
H PO
H3O+ + 4
HPO
; 2
K
P = 7.20. pH of 0.01M N NaH2PO4 is
(A) 9.350 (B) 4.675
(C) 2.675 (D) 7.350
18. To prepare a buffer of pH = 8.26, amount of (NH4)2SO4 to be added into
500 ml. of 0.01 M NH4OH solution [PKaNH4
+ = 9.26] is
(A) 0.05 mol (B) 0.025 mol
(C) 0.10 mol (D) 0.005 mol
19. Assuming 100% ionization, which will have maximum pH?
(A) 0.01M NH4Cl (B) 0.01M(NH4)2SO4
(C) 0.01M (NH4)3PO4 (D) equal
20. Which of the following not function as buffer solution?
(A) NaOH + NH4OH (B) NaH2PO4 + Na2HPO4
(C) Borax + boric acid (D) CH3COOH + CH3COONa

45. 45
Level – II
1. An acid type indicator, HIn, differs in colour form its conjugate base (In
). The human eye is
sensitive to colour difference only when the ratio [In
]/[HIn] is greater than 10 or smaller than
0.1.. What should be the minimum change in the pH of the solution to observe a complete
colour change (Kin = 1.0  105
)?
(A) 4 (B) 2
(C) 6 (D) 1
2. Which of the following mixture will be a buffer solution when dissolved in 500 ml. of water?
(A) 0.20 mole of aniline and 0.20 mole of HCl.
(B) 0.20 mole of aniline and 0.20 mole of NaOH
(C) 0.20 mole of NaCl and 0.20 mole of HCl
(D) 0.20 mole of aniline and 0.1 mole of HCl
3. pH of 0.01 (NH4)2SO4 and 0.02 M NH4OH buffer [pKa(NH4
+] = 9.26) is
(A) 9.26 (B) 9.26 + log2
(C) 4.74 (D) 4.74 + log2
4. The conjugate acid of NH2

is
(A) NH4
+ (B) NH3
(C) NH2OH (D) N2H4
5. Which of the following salts when dissolved in water will hydrolyse?
(A) NaCl (B) KCl
(C) NH4Cl (D) Na2SO4
6. The following reaction occurs in the body
CO2 + H2O H2CO3 H+ + HCO3
– . If CO2 escapes from the system
(A) pH will decreases
(B) hydrogen ion concentration will decrease
(C) H2CO3 concentration remain unaltered
(D) forward reaction will be promoted
7. The solubility product of BaSO4 is 1.5  109
. The precipitation in a 0.01 M Ba2+ ions solution
will start on adding H2SO4 of concentration
(A) 109
M (B) 108-
M
(C) 107
M (D) 106
M
8. There is no effect of dilution on pH of the following
(A) 0.01 M CH3COOH + 0.01M CH3COONa (B) 0.01M H3COONH4
(C) 0.01M NH4OH + 0.01 M NH4Cl (D) in all cases
9. The pKa of acetysalicylic acid (aspirin) is 3.5. The pH of gastric juice in human stomach is
about 2 – 3 and the pH in the small intestine is about8. Aspirin will be
(A) unionised in small intestine and in the stomach
(b) completely ionised in the small intestine and in the stomach
(C) ionised to the stomach and almost unionised in the small intestine
(D) ionised in the small intestine and almost unionised in the stomach
10. The molar solubility of calcium phosphate be S. What is its solubility product?
(A) S5 (B) 5
108S
(C)
5
S
108
(D) 5
108
S

46. 46
11. pH of Ba(OH)2 is 12. Its solubility product is
(A) 10–6 (B) 4  10–6
(C) 0.5  10–7 (D) 5  10–7
12. At 25°C, the pH of pure water is 7. It dissociates H2O(l) + H2O(l) H3O+ + OH–
H3O+ag. + OH–
aq. 2H2O(l) at 25°C;  H° = –13.7 K cal mol–1. pH of water of at 37°C is
expected to be
(A) greater than 7 (B) less than 7
(C) equal to 7 (D) none of true
13. The compound whose 0.1M solution is basic is
(A) ammonium acetate (B) ammonium chloride
(C) ammonium sulphate (D) sodium acetate
14. The dissociation constant of an acid HA is 1  10–5, the pH of 0.1M solution of acid will be
approximately
(A) 3 (B) 5
(C) 1 (D) 6
15. A 0.1 M solution of HCN is 0.01% ionised, the ionisation constant for HCN is
(A) 10–9 (B) 10–7
(C) 10–5 (D) 10–3
16. The solubility of sparingly soluble substance AgCl can be increased by the addition of
(A) aq. NH3 (B) aq. NaCN
(C) Both (D) None of these
17. The aqueous solution of NH4CN is slightly alkaline because
(A) CN– ion hydrolyses to a greater extent than NH4
+ ion.
(B) NH4
+ ion hydrolyses to a greater extent than CN– ion
(C) both hydrolyses to an equal extent
(D) it is a salt
18. If pKb of F– ion at 25C is 10.83, the ionisation constant of HF in water at this temperature is
(A) 1.75  10–5 (B) 3.52  10–3
(C) 6.75  10–4 (D) 5.38  10–2
19. The molar solubility of AgCl in 1.8 M AgNO3 solution is (Ksp AgCl = 1.8  10–10)
(A) 10–10 (B) 10–5
(C) 1.82  10–10 (D) None
20. To a solution having equal concentration of Cl–, Br– and I–, solid AgNO3 is slowly added.
Which one will precipitate out first? Ksp (AgCl}  Ksp (AgBr)  Ksp (AgI)
(A) AgCl (B) AgBr
(C) AgI (D) All precipitated simultaneously

47. 47
ANSWERS TO ASSIGNMENT PROBLEMS
Subjective:
Level – O
1.
K
; i.e. C ,
C
    
2. 5
a
K 2 10
  ,
14
10
w
b 5
a
K 10
K 5 10
K 2 10



   

3. Due to the presence of common ion in the solution, degree of dissociation of weak
electrolytes decreases
4. According to this law, on diluting in any weak electrolyte, degree of dissociation increases.
5. Solubility product is the product of concentration of ions in aqueous form with appropriate
power of their stoichiometric coefficients at the time of saturation. While ionic product is the
product of concentration at any time keeping temperature constant.
6. 4
7. pKin 1

8. 100 ml
9. 3
10. 7

11. 32  10–12
12. (i) When HCl gas is passed into a concentrated solution of BaCl2 the concentration of Cl–
ions increases. Thus product of ionic concentration of Ba2+ and [Cl–] ions exceeds the
solubility product of BaCl2 and a white precipitate of BaCl2 is formed.
(ii) The nitric acid being an oxidising agent and thus it oxidises H2S into sulphur. Therefore
HNO3 is not suitable reagent for precipitation of sulphides of the second group radicals.
13. 0.4685
14. 12.448 gL–1
15. 1.498  10–5

48. 48
Level – I
1. 1.326  107
g/ ; 1.606  103
g/ 2. (a) basic, (b) acidic
3. (i) Kh = 5.7  1010
(ii) [OH
] = 1.32  109
(iii) pH = 5.12
4. (i) 0.368%, (ii) pH = 7.175
5. pH = 0.097 6. pOH = 8.48
7. 8.83 8. pH = 6.51
9. 8.65  10–4M 10. NH3 = 3.4 g/ ; NH4Cl = 21.4 g/
11. 6.95 12. 8.166  10–7 M, 4.1  10–7
13. 5.695, 8.72 14. 11.7; 0.5%
15. 13.56
Level – II
1. 6.8617 2. 20 ML
3. 0.03M, 300 ml 4. 4.23; 47 ml
5. pH = 4.74 6.  1  10–5 M
7. 5.248g 8. 2.352  10–3
9. 1.82 moles 10. (i) 1.18 %; (ii) 11.07

49. 49
Objective:
Level - I
1. D 2. A 3. C
4. B 5. B 6. C
7. D 8. B 9. B
10. D 11. A 12. D
13. A 14. B 15. C
16. D 17. B 18. B
19. A 20. A
Level - II
1. B 2. D 3. A
4. B 5. C 6. B
7. D 8. D 9. D
10. B 11. D 12. B
13. D 14. A 15. A
16. C 17. A 18. C
19. A 20. C