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UNIT : 7 EQUILIBRIUM
Important Points
It is said that equilibrium is established when number of molecules moving from liquid state to
vapour state and number of molecules moving from vapour state to liquid state are same and it is
dynamic. Equilibrium is established in both physical and chemical types of reactions. At this point of
time the rates of forward and reverse reactions become equal. Equilibrium constant Kc is expressed
as the ratio of the multiplication of concentration of products to the multiplication of concentration of
reactants; concentration of each can be expressed as power of their stoichiometric coefficient.
For reaction aA + bB  cC + dD
Kc =
[C]c [D]d
a
b
[A] [B]
Equilibrium constant has constant value at constant temperature and at this stage macroscopic
properties like concentration, pressure, etc become constant. For gaseous reaction Kp is taken instead
of Kc and partital pressure of gaseous reactants and products are expressed instead of concentrations.
The relation between Kp and Kc is expressed as Kp = Kc(RT)ng. In which direction reaction will
occur (forward or reverse) can be expressed by reaction quotient Q c which is equal to Kc at
equilibrium. Le Chatelier's principle, mentions that if the equilibrium gets disturbed by change in factors
like concentration, temperature, pressure etc., then equilibrium will move in the direction whereby the
effect has been minimised or made negligible and the value of equilibrium constant will not change.
This can be used in industries to know how equilibrium can be obtained by study of changes in factors
like concentration, pressure, temperature, inert gas etc. In industries, we can change or control factors
accordingly so that reaction shifts from reactants to products (left to right). If catalyst is used, only
the rate of required reaction will increase but no change will occur in amounts of reactants or
products because the effect on forward and reverse reactions will be the same and so equilibrium
constant will not change.
The substances which allow the electric current to pass through their aqueous solutions are
called electrolytes. Acid, base and salt are electrolytes because their aqueous solutions conduct electric
current. The reason for the conduction of electric current in aqueous solution of electrolyte is the
formation of ions due to dissociation or ionisation which conducts electric current. While the weak
electrolytes are incompletely dissociated and so the equilibrium is established between its ions and
undissociated molecules. This is called ionic equilibrium.
According to Arrhenius ionisation theory, acid is called a substance which gives hydrogen ion
–
(H+) and base is called a substanec which gives hydroxyl ion (OH ) on ionisation. According to
Bronsted - Lowry theory, acid is defined as a proton donor and base is defined as proton acceptor.
Each acid has its conjugute base and each base has its conjugate acid. Hence, it is known as
conjugate acid - base or proton - transfer theory. Proton is tranferred between acid and base.
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Bronsted - Lowry is more general than Arrhenius definition. According to Lewis' definition, acid
means a substance which accepts a pair of electrons and base is a substance which donates a pair
of electrons. This definition can be applied to organic chemistry, complex compounds chemistry in
addition to acid-base. Hence, it is considered universally acceptable. Ionisation constant is also an
equilibrium constant. The ionisation constants of weak acid (Ka) and weak base (Kb) can be deter+
mined. Concentration of acid can be expressed as pH = –log10 [H3O ]. Hence, pH scale is determined
for acid - base. Similary, concentration of OH – can be expressed as pOH = –log10[OH –], ionisation
constant of water as pKw = –log10Kw [H3O ] and
[OH –] can be calculted by the use of
relation pKw = pH + pOH. If pH < 7 solution will be acidic, pH > 7 solution will be basic and pH
+
= 7 solution will be neutral.
Different salts can be obtained by neutralisation of strong or weak acid and strong or weak
base. In such salts, acidic, basic and neutral salts are included. When such salts react with water,
hydration (hydrolysis) reaction occurs and solution obtained can be acidic, basic or neutral. This is also
an equilibrium reaction and so corresponding equilibrium constant for it can be determined. Hydrolysis
constant is expressed as Kh. pH or pOH can be calculated from the values of Ka and Kb and the
value of Kh characteristic for the particular salt. Some solutions are such whose pH does not change
by addition of small amount of acid or base or in case they are being diluted. Such solutions are called
buffer solutions which can be acidic, basic or of neutral type. The control of pH is useful in the
control of biological reactions in our body and chemical reactions in analytical chemistry, industries etc.
Sparingly soluble salts (whose solubility is less than 0.01M in water) dissolve in water depending
on their solubility and equilibrium is established. Hence, equilibrium
constant for this can be
obtained which is known as solubility product constant or solubility product of the sparingly soluble salt.
The study of effect of common ion, acid, etc. on the solubility of sparingly soluble salt can be carried
out by application of Le Chatelier's principle. Generally, the solubility of sparingly soluble salt decreases due to effect of common ion. This is used in qualitative analysis. By mixing two solutions,
whether precipitates will be obtained or not, can be predicted by comparing concentration product Ip
with the solubility product Ksp. If Ip > Ksp precipitation will occur and if Ip < Ksp the precipitation
will not occur and if Ip = Ksp, the precipitation will not occur but solution will remain in saturated state.
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M.C.Q.
1.
Three gaseous equilibria have values of equilibrium constants as k1 , k 2 , k3. resp.
K1
(i) A + B 


 C
K2
(ii) B +C 


 P+Q
K3
(iii) A + 2B 


 P+ Q
What is the relation between k1 ,k2, k3. (?)
K
(a) K 3  1 K
2.
(b) K 3  K1  K 2
2
(c) K1  K 2  K 3
(d) K1  K 2  K 3  1
Three gaseous equilibria have value of equilibrium constants as K1 , K 2 , K 3 respectively..
(i) N 2  O 2  2 NO (K 1 )
(ii) N 2  2O 2  2 NO 2 (K 2 )
(iii) 2NO  O2  2 NO2 (K3 )
What is the relation between K1 , K 2 and K 3 (?)
(a) K 3  K1  K 2
3.
(b) K1  K 2  K 3
(c) K 2  K1  K 3
(d) K  K1 K2 K3
The equilibrium constant for the reaction,
N2(g)  O2(g)  2 NO(g)
4
4.
is 4.4  10 at 2000k temp
In presence of a catalyst, equilibrium is attained ten times faster. Therefore the equilibrium constant,
in presence of catalyst at 2000K is........
(a) 4.4  10 4
(b) 4.4  10 5
(c) 4.4  10 3
(d) difficult to compute
For the following reaction in gaseous phase
CO  1 2 O2  CO2
K C K P is
1
(a) RT  2
5.
(b) RT 
1
(c) RT
2
(d) RT 1
For the reaction equilibrium, N 2 O 4 (g )  2NO ( g ) the concentration of N 2O 4 and NO 2
at equilibrium are 4.8 102 and 1.2 102 molL1 respectively. The value of K C for the reaction
is :
6.
(a) 3.3  10 2 mol L1
(b) 3  101 mol L1
(c) 3  10 3 mol L1
Which one of the following statements is not true ?
(d) 3  103 mol L1
(a) The conjugate base of H 2 PO 4 is HPO 24
(b) P H  P OH  14 for all aqueous solutions at 298 k
(c) P H of 1 10 8 M HCl is 8
(d) NH 3 is a lewis base
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7.
What is the equilibrium expression for the reaction.
P4 (s)  5O 2 (g)  P4 O10(s)
P4O10 
(a) Kc  P O 5
4
8.
2
1
(b) Kc  O 5
P4O10 
(c) Kc  5P O 
(d) Kc  O 2 5
(c)
(d) 1.0
4
2
2
For the reaction
CO (g)  Cl2(g)  COCl2(g) then
Kp
Kc
is equal to
1
(b) RT
RT
The equilibrium constant for the reaction
(a)
9.
RT
N 2(g)  O 2(g)  2 NO (g)
at temperature. 300 k is 4  10 4 The value of Kc for the reaction
NO (g)  21 N 2 (g)  1 2 O 2(g) at the
same temperature is:
(a) 2.5 10 2
10.
11.
(b) 4  10 4
(c) 0.02
(d) 50
Hydrogen ion concentration in mol L in a solution of P H  5.4 will be,
(a) 3.98  10 6
(b) 3.68  10 6
(c) 3.88  10 6
(d) 3.98  108
The equilibrium constants Kp1 and Kp 2 for the reactions, X  2Y and Z  P  Q respectively..
are in the ratio of 1:9. If the degree of dissociation of x and z are equal then the ratio of total pressure
at these equlibria is.
(a) 1:36
(b) 1:1
(c) 1:3
(d) 1:9
12.
In a reaction, CO(g)  2H2(g)  CH3OH(g) ΔHO  92 KJ mol1 concentration of hydrogen,
13.
carbon monoxide and methanol become constant at equilibrium. what will happen if an inert gas is
added to the system.?
(a) reaction becomes fast
(b) reaction becomes slow
(c) equilibrium state disturbs
(d) equilibrium state remains undisturbed
At 473 k, equilibrium constant KC for the reaction
PCl5(s)  PCl3(g)  Cl2(g) is 8.3 103
14.
what is the value of KC for the reverse reaction at the same temperature ?
(a) 8.3 10 3
(b) 120.48
(c) 16.6  10 3
(d) 4.15  103
At 473 k, equilibrium constant KC for a reaction,
PCl5 s   PCl3(g)  Cl 2(g) is 8.3 10 3 (ΔH  124 KJmol-1 )
What would be the effect on KC if
(i) more PCls is added
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15.
(ii) the pressure is increased
(iii) the temp is increased
(a) KC remains unchanged, unchanged, increase.
(b) KC increases, increases, decreases.
(c) KC remains unchanged, increase, unchanged.
(d) KC decreases, increases, unchanged.
The equilibrium constant for the reaction,
N2(g)  O2(g)  2NO(g) is 4 104 at 2000 k
temperature. what is the value of kc for the reaction
16.
3 N  3 O  3 NO
(g)
2 2(g)
2 2(g)
(a) 4  10 4
(b) 8  106
For the reversible reaction
(c) 8  104
(d) 16  10 4
N 2(g)  3H 2(g)  2NH 3(g) at 500 c the value of
KP is 1.44 10 5 . what would be the value of KC for the same reaction
17.
(a) 1.44  10 5 / 0.082  500 2
(b) 1.44  10 5 / 8.314  7732
(c) 1.44  10 5 / 0.082  7732
The following equilibria are given
(d) 1.44  10 5 / 0.082  7732
N 2  3H 2  2NH3     K1
N 2  O 2  2NO     K 2
H 2  1 2 O 2  H 2O    K 3
2 NH3  5 2 O 2  2 NO  3H 2O The equilibrium constant of the reaction
in terms of k1 , k 2 and k 3 is
(a) k 1 k 2 / k 3
18.
19.
(b) k1 k 32 / k 2
(c) k 2 k 33 / k1
(d) k 1 k 2 k 3
The P Ka of a weak acid HA is 4.80 The pKb of a weak base BOH is 4.78 The P H of an aqueous
solution of the corrosponding salt BA will be
(a) 9.58
(b) 4.79
(c) 7.01
(d) 9.22
The equilibrium constant for the reaction
SO3(g)  SO2(g)  1 2 O 2(g) Kc  4.9  102
The value for the KC of the reaction
2SO2(g)  O2(g)  2SO3(g) will be
(a) 416
20.
(b) 2.40  10 3
(c) 9.8  10 2
(d) 4.9  10 2
P H of 0.1 M solution of weak acid is 3. The value of ionisation constant Ka of acid is
(a) 3  101
(b) 1 10 3
(c) 1 10 5
(d) 1  1 0  7
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21.
A vessel contains CO2 at 1000k temperature with a pressure 0.5 atm.some of CO2 is converted in
to CO on addition of graphite. If total pressure at equilibrium is 0.8 atm the value of KP is:
(a) 3 atm
(b) 0.3 atm
(c) 0.18 atm
(d) 1.8 atm
22.
Three reactions involving H 2 PO 4 are given below..
(i) H 3 PO 4  H 2 O  H 3 O   H 2 PO 4
(ii) H 2 PO -4  H 2 O  HPO 42  H 3 O
(iii) H 2 PO 4-  OH -  H 3 PO 4  O 2 
23.
In which of the above does H 2 PO-4 act as an acid.
(a) (i) only
(b) (ii) only
(c) (i) and (ii)
For the reaction
(d) (iii) only
2 NO2(g)  2 NO(g)  O2(g)
Kc  1.810 6 at 184c R  0.0831 KJ/mol.K
24.
25.
26.
When KP and KC are compared at 184 C it is found that :
(a) whether Kp is greater than less than or equal to Kc depends upon the total gas pressure
(b) Kp=Kc
(c) Kp<Kc
(d) Kp>Kc
Water is a
(a) Protophobic solvent
(b) Protophilic solvent
(c) Amphiprotic solvent
(d) Aprotic acid
Ammonium ion is
(a) Conjugate acid
(b) Conjugate base
(c) Neither an acidnor a base
(d) both an acid and a base.
Species acting both as bronsted acid and a base is
(a) HSO 4
27.
(b) Na 2 CO 3
30.
31.
(b) 4  10 10 M
(c) 2  10 5 M
(d) 9  10 4 M
(c) H 3PO 4
(d) HPO 24
The conjugate base of H 2 PO 4 is
(a) PO34
29.
(d) OH -
A solution of an acid has P H  4.70 find out the concentration of OH . pK w  14
(a) 5  10 10 M
28.
(c) NH 3
(b) HPO 4
What is the conjugate base of OH  ?
(a) O 2
(b) H 2 O
(c) O 
(d) O 2 
An example for lewis acid is
(a) Ammonia
(b) Aluminium chloride
(c) Pyridine
(d) Amines.
Which of the following molecule act as a lewis acid ?
(a) CH 3 2 O
(b) CH 3 3 P
(c) CH 3 3 N
(d) CH 3 3 B
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32.
33.
34.
35.
In a given system, water and ice are in equilibrium. If pressure is applied to the above system, then,
(a) More ice is formed
(b) Amount of ice and water will remain same
(c) More ice is melted
(d) Either (a) or (c)
In 2 HI  H 2  I 2  H  0 the forward reaction is affected by change in
(a) Catalyst
(b) Pressure
(c) Volume
(d) Temp
In which case KP is less than KC (?)
(a) PCl5(g)  PCl3(g)  Cl2(g)
(b) H 2(g)  Cl2(g)  2HCl(g)
(c) 2 SO2  O2  2SO3(g)
(d) all of these.
If K1 and K 2 are respective equlibrium constants for two reaction,
XeF6(g)  H2O(g)  XeOF4(g)  2HF(g)
XeF4(g)  XeF6  XeOF4(g)  XeO3 F2(g)
the equilibrium constant for the reaction
XeF4(g)  2HF(g)  XeO3F2(g)  H2O(g) will be
36.
37.
38.
(a) K1  K 2
(b) K 2  K1
For a homologous reaction,
(c) K 2 / K1
(d) K1 / K 2
4 NH 3  5O 2  4 NO  6H 2 O
the dimensions of equilibrium constant KC is
(a) conc.10
(b) conc.1
(c) conc.1
(d) It is dimensionless
For a reversible reaction, if the concetration of the reactants are doubled, the equilibrium constant
will be
(a) The same
(b) Halved
(c) Doubled
(d) One fourth
The molar solubility of a sparingly soluble salt AB4 is 's' mol lit The corrosponding solubility product
KSP is given in terms of KSP by the relation
K

(a) S   SP 128 


39.
41.
4
1
(b) S  128 K SP  4
(c) S   256 K SP 
1
5
(d) S   K SP / 256 
1
5
The solubility product of a salt having general formula MX2 in water is 4  10 12 The concentration
of M 2 ions in an aqueous solution of the salt is:
(a) 4.0  10 10 M
40.
1
(b) 1.6  104 M
(c) 1.0  104 M
(d) 2.0 10 6 M
Ka
P H of 0.005 M calcium acetate ( P of CH 3COOH  4.74 ) is
(a) 7.04
(b) 9.37
(c) 9.26
(d) 8.37
One of the following equilibria is not affected by change in volume of the flask.
(a) PCls(g)  PCl3(g)  Cl2(g)
(b) N2(g)  3H2(g)  2 NH3(g)
(c) N2(g)  O2(g)  2 NO(g)
(d) SO2 Cl2(g)  SO2(g)  Cl2(g)
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42.
Equal volmes of two solution of P H 3 and 4 are mixed The P H of the resulting solution will be
(a) 7
(b) 3.5
(c) 2.96
(d) 3.26
43.
P H of 10 8 M solution of Hcl in water is
(a) 8
(b) -8
(c) between 7 and 8
(d) between 6 and 7

44.
A certain buffer solution contains equal concentration of X and HX . Ka for HX is 10 8 . The P H
of buffer is
(a) 3
(b) 8
(c) 11
(d) 14
45.
The solubility product of AgCl is 4  10 10 at 298 k The solubility of AgCl in 0.04 M CaCl2 will be
46.
(a) 2  10 -5 M
(b) 1 10-4 M
(c) 5  10-9 M
(d) 2.2  10-4 M
Calculate concentration of sodium acetate which should be added to 0.1 M solution of


CH 3 COOH P K a  4.5 to give a solution of P H 5.5
47.
(a) 1.0 M
(b) 0.1 M
(c) 0.2 M
(d) 10.0 M
Which of the following is a base according to lowry-bronsted concept ?
48.
(a) I 
(b) H 3O 
(c) HCl
(d) NH 4
According to lowry-bronsted concept which one of the following is considered as an acid ?
(a) H 3O 
(b) BF3
(c) OH 
(d) Cl 
49.
The conjugate acid of NH 2  is
(c) N 2 H 4
(d) NH 2OH
50.
(a) NH 4
(b) NH 3
conjugate base of hydrazoic acid is
(a) HN 3
(c) N 3
(d) N 3
51.
(b) N 2
In which of the following reaction NH 3 acts as acid ?
(b) NH 3  H   NH 4
(a) NH 3  HCl  NH 4 Cl
52.
1
(c) NH 3  Na  NaNH 2  H 2
2
Consider the following reactions.
(d) NH 3 cannot act as acid.
(i) CO 32 H2O  HCO3  OH
(ii) CO 2  H 2O  H 2CO3
(iii) NH 3  H 2 O  NH 4OH.
53.
(iv) HCl  H 2 O  Cl   H 3O 
Which of the pairs of reaction proves that water is amphoteric in character ?
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iii)
One of the following is a bronsted acid but not a bronsted base :
(a) H 2S
(c) HCO 3
(b) H 2 O
(d) NH 3
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54.
The conjugate base in the following reaction
H 2SO 4  H 2 O  H 3 O   HSO 4 are
57.
(a) H 2 O, H 3O 
(b) HSO 4 , H 2 O
(c) H 3 O  , H 2SO 4
(d) H 2SO 4 , HSO 4
With increase in temperature, ionic product of water
(a) Decreases
(b) Increases
(c) Remains same
(d) May increase or decrease.
EDTA is a/an
(a) Arrhenius acid
(b) Bronsted base.
(c) Lewis base
(d) All of above
The units of lonic product of water (kw) are :
(c) mol 2 L1
58.
(a) mol 1 L1
(b) mol 2 L2
Whichof the following weakest ?
(a) C6 H 5 NH 2 : K b  3.8 1010
(b) NH 4 OH : K b  1.6  10 5
(c) C 2 H 5 NH 2 : K b  5.6  10 4
(d) C9 H 7 N : K b  6.3 1010
On adding ammonia to water,
(a) Ionic product will increase
(b) Ionic product will decrease
55.
56.
59.

(d) [H 3 O  ] will decrease
(c) [H 3 O ] will increase
60.
61.
 
According to lowry and bronsted system, the chloride ion Cl  in aqueous solution is a
(a) Weak base
(b) Strong base
(c) Weak acid
(d) Strong acid
H
The P of a buffer containing equal molar concentration of a weak base and its
chlorides is  K b for weak base 2  10 5   log 2  0.3010 
(a) 5
(b) 9
(c) 4.7
62.
63.
65.
(d) 9.3
Solution of 0.1 N NH 4OH and 0.1N NH 4Cl has P H 9.25.Then PK b is
(a) 9.25
(b) 4.75
(c) 3.75
(d) 8.25

9

The solubility of product barium sulphate is 1.5  10 at 18 C Its solubility in water at 18 C is
(a) 1.5  10 9 mol L1
64.
(d) mol 2 L2
(b) 1.5  10 5 mol L1
(c) 3.9  10 9 mol L1
(d) 3.9  10 5 mol L1
The least soluble compound (salt) of the following is
(a) ZnS  Ksp  .2 10 23 
(b) OH >Cl  CH3COO
(c) CsCl  Ksp  1  10 12 
(d) PbCl 2  Ksp  1.7  10 5 
What is the value of P H of 0.01 M glycine solution ? For glycine Ka 1  4.5  10 3 and
Ka 2  1.7  10 10 at 298k.
(a) 3.0
(b) 10.0
(c) 6.1
(d) 7.06
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66.
Solid Ba NO 3 2 is gradually dissolved in 1.0  10 4 M Na 2 CO 3 solution. At what concentration of
will precipitate ?
Ksp of BaCO3  5.110 9
(a) 4.110 5 M
67.
(b) 5.1 105 M
70.
71.
(c) 0.12 M
(d) 0.1 M
(b) 5.65  10 13
(c) 5.65  10 12
(d) 5.65  10 10
The dissociation consant of a substituted benzoic acid at 25 C is 1.0  10 4 The P H of 0.01 M
solution of its sodium salt is
(a) 0
(b) 1
(c) 7
(d) 8
H

Number of H ions present in 500 ml of lemon juice of P  3 is
(a) 1.506  10 22
(b) 3.012  10 20
(c) 3.102  10 22
(d) 1.506 1020
Equimolar solution of the following were prepared in water separately. Which one of the solutions
will record the highest P H (?)
(a) SrCl 2
72.
(b) 0.05 M
The ionisation constant of NH 4OH is 1.77  105 at 298 k. Hydrolysis constant of it is
(a) 6.5  10 12
69.
(d) 8.1107 M
What is the OH  in the final solution prepared by mixing of 20.0 ml of 0.05 M HCl with 30.0
ml of 0.1 M Ba OH 2 (?)
(a) 0.4 M
68.
(c) 8.1 10 5 M
(c) MgCl2
(b) BaCl2
(d) CaCl2
Solubility products constants (KSP) of the salt types MX, MX 2 and M 3X at temp T. are
4  10 8 ,3.2 1014 and 2.7 1015 respectively. Solubility of the salts at temp. T are in the order,,
73.
74.
(a) MX  M 3 X  MX 2
(b) MX 2  M 3 X  MX
(c) M 3 X  MX 2  MX
(d) MX  MX 2  M 3 X
When H  ion concentration of a solution increases
(a) PH increses
(b) PH decreases
(c) no change in PH
(d) POH decreases
At 25 C temp. the value of pk b for NH 3 in aqueous solution is 4.7. what is the value of P H of 0.1
M aqueous solution of NH 4Cl with 0.1 M NH 3 (?)
`(a) 8.3
(b) 9
(c) 9.5
(d) 10
75. The aqueous solution of HCOO Na, C 6 H5 NH 3 Cl, and KCN are respectively
(a) Acidic, acidic, basic
(b) Acidic, basic, neutral
(c) Basic, acidic, basic
(d) Basic, neutral, basic
76.
KSP of AgIO3 is 1 10-8 at a given temperature what is the mass of AgIO3 in 100 ml of its
saturated solution ?
(a) 1.0  10 4 gm
(b) 28.3  102 gm
(c) 2.83 103 gm
(d) 1.0  10 7 gm
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77.
78.
79.
80.
81.
82.
83.
84.
PH of a solution containing 50 mg of sodium hydroxide in 10 dm3 of the solution is
(a) 9
(b) 3.9031
(c) 10.0969
(d) 10
Which one of the following has the lowest P H value ?
(a) 0.1 M HCl
(b) 0.1 M KOH
(c) 0.01 M HCl
(d) 0.01 M KOH
Equal volumes of three basic solutions of 11,12 and13 are mixed in a vessel. what will be the H+
ion concentration in the miture ?
(a) 1.11  10 11 M
(b) 3.7  10 12 M
(c) 3.7  10 11 M
(d) 1.11 10 12 M
Columm I
Column II
(Buffer Solution)
PH
(A) O.1M CH3 COOH + 0.01M
(p) 3.8
CH3 COONa
(B) 0.0 1M CH3COOH+0.1M CH3COONa (q) 5.8
(C) 0.1M CH3COOH+0.1M CH3COONa
(v) 7.0
(D) 0.1M CH3COONH4
(s) 4.8
(Pka Of CH3COOH=Pkb of NH4OH=4.8)
(a) A-r, B-q, C-s, D-p
(b)A-q, B-p, C-s, D-r
(c)A-q, B-p, C-r, D-s
(d)A-r, B-q, C-s, D-p
PH of a soda water bottle is
(a) > 7
(b) = 7
(c) < 7
(d) unpredictable.
H
-8
Statement :1 P of 10 HCl solution is not equal to 8
Statement :2 HCI does not dissociate properly in very dilute solution.
(a) Statement -1 is true.
Statement -2 is true.
Statement-2 is a correct explanation for statement-1
(b) Statement- 1 is true.
Statement-2 is true.
Statement-2 is not a correct explanation of statement-1
(c) Statement-1 is true.
Statement-2 is false.
(d)Statement-1 is false.
Statement-2 is true.
For preparing a buffer solution of PH 6 by mixing sodium acetate and acetic acid , the ratio of
concentration of salt and acid should be, (Ka=10-5)
(a) 1:10
(b) 10:1
(c) 100:1
(d) 1:100
-32
The KSP of CuS, Ag2S and HgS are 10 , 4  10 45 and 10-54 respectively. The solubility of
these sulphides are in order of
(a) Ag 2S  HgS  CuS
(b) Ag 2S  CuS  HgS
(c) HgS  AgS2  CuS
(d) CuS > Ag2S > HgS
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85.
The PH of neutral water is 6.8 .Then the temperature of H 2 O
(a) is 250C
(c) is less than 250C
86.
On adding 0.1 M solution each of Ag  ,Ba 2 and Ca 2 ions. in a Na 2SO 4 solution. species first
precipitated is
K
SP
of BaSO4  1011 , KSP of CaSO4  106 , KSP of Ag 2 SO4  105
(a) BaSO4
87.
(b) is more than 250C
(d) can not be predicted.
(b) CaSO4
(c) Ag 2SO4

(d) all of these
The solubility of A 2 B3 is x mol L-1. It solubility product is
(a) 6 x 5
(b) 64 x 5
(c) 36 x 5
(d) 108 x 5
88.
How much volume of 0.1 M CH3COOH should be added to 50ml of 0.2 M CH3COONa If we
89.
want to prepare a buffer Solution of P H 4.91. given pKa for acetic acid is 4.76
(a) 80.92 ml
(b) 100 ml
(c) 70.92 ml
(d) 60.92 ml

4
The ionzation constant of formic acid is 7.8  10 . Calculate ratio of sodium formate & formic acid
90.
91.
92.
93.
94.
in a buffer of P H 4.25
(a) 9.63
(b) 3.24
(c) 6.48
(d) 3.97
The ionization constants of HF is 6.8  10 4 . Calculate ionization constant of corresponding conjugate
base.
(a) 1.9  1010
(b) 1.7 10 10
(c) 1.5 10 11
(d) 2.9  10 11
The ionization constant of formic acid is 1.8  10 4 around what P H will its mixture with sodium
formate give buffer solution of highest capacity
(a) 3.74
(b) 7.48
(c) 4.37
(d) 3.96
What is PH of our blood ? why does it remains constant inspite the variety of the foods and spices
we eat ?
(a) of blood is 5.4. It remains constant because.it is acidic.
(b) of blood is 7.4. It remains constant because.it is buffer.
(c) of blood is 10.8.It remains constant because.it is basic.
(d) of blood is 7.0. It remains constant because it is neutial.
The precipitate of CaF2 (K SP  1.7  10 10 ) is obtained when equal volumes of the following are
mixed.
(a) 10 4 M Ca 2  10 4 M F
(b) 10 2 M Ca 2  10 3 M F
(c) 10 5 M Ca 2  10 3 M F 
(d) 10 3 M Ca 2  10 5 M F 
A certain buffer solution contains equal concentrations of X and HX  The Ka for HX is10 8 The
P H of buffer is
(a) 8
(b) 3
(c) 7
(d) 4
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95.


The correct order of increasing H 3O  in the following aqueous solution is.
(a) 0.1 M H2S  0.1 M H2SO4  0.1 M NaCl  0.1 M NaNO2
(b) 0.1 M H 2SO4  0.1 M NaCl  0.1 M NaNO2  0.1 M H 2S
(c) 0.1 M NaNO2  0.1 M NaCl  0.1 M H 2S  0.1 M H 2SO4
96.
(d) 0.1 M NaNO2  0.1 M H2S  0.1 M H2SO4  0.1 M NaCl
What is the % hydrolysis of NaCN in N/80 solution when dissociation constant for HCN is
1.310 9& KW  11014
(a) 2.48
(b) 8.2
(c) 5.26
(d) 9.6
97.
The KSP of AgCl is 4.0  10 10 at 298 k.solubility of AgCl is 0.04 M CaCl2 will be
98.
(a) 2  10 5 M
(b) 1 10 4 M
(c) 5  10 9 M
(d) 2.2  10 4 M
How much sodium acetate should be added to 0.1 M solution of CH3COOH to give a solution of
99.
P H 5.5  pKa of CH 3CooH  4.5 
(a) 0.1 M
(b) 1.0 M
(c) 0.2 M
(d) 10.0 M
H
The P of solution obtained by mixing 50ml 0.4 N HCl & 50ml 0.2 N NaOH is
(a)  log 2
(b) 1.0
(c)  log 0.2
(d) 2.0
100. Ionisation constant of CH3COOH is 1.7  105 and concentration of H  ions is 3.4  10 4 The initial
concentration of CH 3COOH molecules is
(a) 3.4  10 4
(b) 3.4  10 3
(c) 6.8  10 3
(d) 1.7  10 3
101. In the reversible reaction A  B  C  D , the concentration of each C and D at equilibrium was
0.8 mollitre, then the equilibrium constant KC will be.
(a) 6.4
(b) 0.64
(c) 0.16
(d) 16.0
102. 4 moles of A are mixed with 4 moles of B.At equilibrium for the reaction A  B  C  D 2 moles
of C and D are formed . The equilibrium constant for reaction will be
1
1
(b) 4
(c)
(d) 1
4
2
103. A reversible chemical reaction having two reactants in equilibrium. If the concentration of the
reactants are doubled, then equilibrium constantwill
(a) become double
(b) become half
(c) become 4 times
(d) remains same
104. Two moles of PCls are heated in a closed vessel of 2L capacity. At equilibrium, 40 % of PCl5 is
dissociated in to PCl3 & Cl2. The value of equilibrium constant is,
(a) 0.266
(b) 0.53
(c) 2.66
(d) 5.3
(a)
105. The dissociation constant for acetic acid and HCN at 25 C are 1.5 105 and4.5 10 10 respectively..
the equilibrium constant for reaction CN   CH 3COOH  HCN  CH 3COO 
(a) 3 105
(b) 3  105
(c) 3  104
(d) 3 10 4
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106. A  B  C  D if finally the concentration of A and B are both equal but at equilibrium concentration
of  will be twice of that of A. Then what will be equilibrium constant of the reaction.
(b) 9 4
(a) 4 9
(c) 19
(d) 4
107. If in the reacton N2O4  2NO2 ,  is that part of N 2O 4 will dissociate, then the number of moles
at equilibrium will be,
(a) 3
(b) 1   
(c) 1   
(d) 1
108. 4.5 moles each of hydrogen and iodine heated in a sealed ten litre vessel. At equilibrium, 3 moles of
HI were found. The equilibrium constant for H 2(g)  I 2(g)  2HI(g) is
(a) 1
(b) 5
(c) 10
(d) 0.33
109. The rate constant for forward and backward reaction of hydrolysis of ester are 1.1  10 2 &1.5  10 3
per minute respectively. Equilibrium constant for reaction is,
CH 3 COOC 2 H 5  H 2 O  CH 3 COOH  C 2 H 5OH
(a) 4.33
(b) 5.33
(c) 6.33
(d) 7.33
110. Two moles of NH 3 ,when put in to a previously evacuated vessel(1 Litre), partially dissociate in to
.
N2 and H2 If at equlibrium one mole of NH 3 is present, the equilibrium constant is,
2
2
(a) 3 4 mol Litre
2
2
(b) 27 64 mole Litre
2
2
(c) 27 32 mole Litre
111. What is equilibrium expression for the reaction
2
2
(d) 27 16 mol Litre
P4  5O 2  P4O10
(b) KC   P4O10  / 5 A O2 
5
(a) K C   O 2 
5
5
(d) K C  1 / O 2 
(c) K C   P4O10  /  P4 O 2 
112. Partial pressure of O 2 in
2 Ag 2 O (g)  4 Ag(s)  O 2(g)is
(a) kp
(b)
(c) 3 kp
kp
(d) 2 kp
113. For reaction H 2(g)  CO2(g)  CO(g)  H 2 O(g) ,If the initial concentration of H 2   CO 2  and x
moles / litre of hydrogen is consumed at equilibrium, the correct expression of KP is
x2
(a)
1  x 2
(b)
1  x 2
1  x 2
x2
(c)
2  x 2
x2
(d)
1 x2
(c) mole litre 1
(d) Litre 2 mole 2


114. Consider the imaginary equlibrium
4A  5B  4x  6y
The equilibrium constant K C has unit
(a) mole 2 Litre 2
(b) Litre mole 1
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1
2
115. For reaction CO(g)  O2(g)  CO2(g)
(a) 1
Kp
is equivalent to
Kc
(b) RT
(c)
1
RT
(d) RT  1 2
116. For the reaction
CH4(g)  2O2(g)  CO2(g)  2H2O(L)
 H  170.8KJ mol1
Which of following statement is not true
(a) Adding of CH 4 ( g ) on O 2 ( g ) at equilibrium will cause a shift to the right
(b) Thee reaction is exothermic
(c) At equilibrium, concentrations of CO2(g) and H2O(L) are not equal.
(d) Equilibrium constant for the reaction is given by KP 
CO2 
CH4 O2 
117. The reaction Quetient (Q) for the reaction
N 2(g)  3H 2(g)  2NH 3
2
is given by
 NH3 
Q
3 The reaction will proceed from right to left is:
 N2 H2 
(a) Q  0
(b) Q  K C
(c) Q < KC
118. If concentration of reactants is increased by 'x' then k become
(a) ln( k / x )
(b) k/x
(c) k+x
119. Which of following is not favourable for formating SO3 formation
(d) Q  KC
(d) k
2SO2(g)  O2(g)  2SO3(g)  H  45.0 Kcal.
(a) High pressure
(c) High temperature
120. The most important buffer in blood consists of
(a) HCl andCl
(b) H 2 CO 3 and Cl
(b) Decreasing SO3 concentration
(d) Increasing reactants concentration
(c) H2CO3 and HCO3 (d) HCl and HCO3
121. Given that dissociation constant for H 2 O is K W  1 10 14 moles 2 / Litre2 What is P H of 0.001 M
NaOH
(a) 10-11
(b) 10 3
(c) 11
(d) 3
122. Select the pKa value of strongest acid from following
(a) 1.0
(b) 3.0
(c) 2.0
(d) 4.5
123. The degre of hydrolysis equilibrium
A   H 2 O  HA  OH 

at salt concentration of 0.001 M is Ka  1 10 5
(a) 1 10 3
(b) 1 10 4

(c) 6.75  10 4
(d) 5.38  10 2
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124. If pK b for fluoride ion at 250C is 10.83, the ionisation constant of hydrofluoric acid in water at this
temperature is
(a) 1.74  10 3
(b) 3.52  10 3
(c) 6.75  10 4
(d) 5.38  10 2
H
125. Henderson's equation is P  pka  log
Salt 
Acid  If acid gets half neutralised the value of
P H will
be [pka=4.30]
(a) 4.3
(b) 2.15
(c) 8.60
(d) 7
126. The P H of a 0.01 M solution of acetic acid having degree of dissociation 1.25% is
(a) 5.623
(b) 2.903
(c) 3.723
(d) 4.503
127. By adding 20 ml 0.1 N HCl to 20 ml 0.1 N KOH the P H of obtained solution will be :
(a) 0
(b) 7
(c) 2
(d) 9
128. If the Kb value in the hydrolysis reaction
B   H 2 O  BOH  H  is 1.0  10 6 then hydrolysis constant of salt would be
(a) 1.0  10 6
(b) 1.0  10 7
(c) 1.0  10 8
(d) 1.0  10 9
129. For a sparingly soluble salt ApBq relation ship of its solubility product(KSP) with its solubility(s) is.
(a) K SP  Sp  q P p q q
(b) K SP  Sp  q P q q p
(c) K SP  S pq P p q q
(d) KSP  Spq  pq 
p q
130. How many grams CaC2O 4 (mw  128) on dissolving in disttile water will give saturated
solution  K SP  CaC 2 O 4   2.5  10 9 mol 2 l 2 
(a) 0.0064gm
(b) 0.1280gm
(c) 0.0128gm
(d) 1.2800gm
131. If the concentration of CrO ion in a saturated solution of silver chromate is 2 104 solubility
product of silve chromate will be
(a) 4  10 8
(b) 8  10 12
(c) 12 10 12
(d) 32  10 12
132. According to bronsted- lowry concept. correct order of relative strength of bases follows the order
2–
4
(a) Cl   CH 3COO   OH 
(b) Cl  OH   CH 3COO
_
_
(d) OH  >Cl   CH 3COO 
(c) OH   CH 3COO  Cl 
133. HSO 4  OH   SO 24  H 2O Which is correct about conjugate acid base pair ?
(a) HSO 4 is Conjugate acid of base SO 42 
(b) HSO 4 is Conjugate base of acid SO 42 
(c) SO 42  is Conjugate base of acid HSO 4
(d) None of these
134. Which of following base is weakest
(a) NH 4 OH:K b  1.6  10 6
(b) C 6 H 5 NH 2 : K b  3.8  10 10
(c) C 2 H 5 NH 2 : K b  5.6  10 4
(d) C 2 H 7 N : K b  6.3 10 10
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135. HClO is a weak acid. concentration of H + ion in 0.1 M solution of HClO  K a  5  10 -8  will be
(a) 7.07  10 5 M
(b) 5  10 9 M
(c) 5  10 7 M
(d) 7  10 4 M


136. Upto what P H must a solution containing a precipitate of Cr  OH 3 be adjusted so that all
precipitate dissolves
(a) upto 4.4
(b) upto 4.1
(c) upto 4.2
(d) upto 4.0
137. NH 4Cl is acidic because
(a) On hydrolysis NH 4Cl give weak base NH 4OH and strong acid HCl
(b) Nitrogen donates a pair of e (c) It is a salt of weak acid and strong base
(d) On hyrdolysis NH 4Cl gives strong base and weak acid
138. 100 ml of 0.04 N HCl aqueous solution is mixed with 100 ml of 0.02 N NaOH solution. The P H of
resulting solution is
(a) 1.0
(b) 1.7
(c) 2.0
(d) 2.3
ANSWER KEY
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
B
C
A
A
C
C
B
A
D
A
A
D
B
A
B
D
C
C
A
C
B
B
D
C
A
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
A
A
D
D
B
D
C
D
C
C
B
A
D
C
D
C
D
D
B
C
A
A
A
B
C
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
C
C
A
B
B
D
D
A
D
A
D
B
D
B
D
B
D
D
D
B
B
A
B
A
C
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
C
C
A
B
B
C
C
B
B
B
A
D
C
B
C
A
B
B
A
C
A
C
B
B
C
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
D
D
D
A
D
D
B
A
D
B
D
A
A
C
C
D
D
D
C
C
C
A
A
C
A
167
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126
127
128
129
130
131
132
133
134
135
136
137
138
B
B
C
A
A
D
C
C
B
A
D
A
C
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Hint : Chemical equilibrium .
Hint :
1.
Addition of (1) and (2) gives (3) then
C   PQ
A B  BC
PQ   K

AB2 3
K1  K2 
 K 3  K1  K 2
when the addition of equilibria leads to another equilibria then the product of their equilibria constants
gives the equilibria constant of the resultant equilibrium.
3.
The value of equilibrium constant does not change in presence of catalyst.
4.
Formula :
Kp  Kc(RT)
Δn (g)
Δn (g)  n p - n r
6.
H 2 O is also present in 1 10-8 M HCl solution. So due to self ionisaion of H 2 O ,
H 2 O  H   OH 
1 10 7 M at 298 K
so conc. of H  ion in solution increase due to self ionisation of H 2 O Hence P H of HCl solution
decreases and its value is less than 7.
9.
N 2(g)  O 2(g)  2 NO (g)
P 2NO
Kp 
PN 2  PO2
For reaction,
1
N O (g )  1 O 2 (g )  N 2
2
2
1
K'P 
1
2
2
PO 2  PN 2
PNO
1
1
= Kp 1 2 
4  10  4


1
 50
2
10.
P H  5.4
P H  log  H  
5.4    log  H  
6.6  3.981  10 6   H  
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11
For a reaction X  2Y
initialmol 1
0
at equilibrium mol 1-x 2x
Total moles = 1-x+2x = 1+x
2
 2x 
P1 

2
PY
4x 2  p1
1 x 

 Kp1 


Px
1  x 1  x  ------(1)
1 x 

 p1
1 x 
For a reaction Z  P  Q
Initial 1 0 0
at eqm.
1-x x x
Total moles = 1+x
Kp 2 
x  P2  x  p 2
x2

 P2

1  x 1  x  -------(2)
2 1 x 
1  x   
P2
1 x 

Kp1 1

Kp 2 9

4 P1 1

P2
9
P1
1

 P1 : P2  1 : 36
P2 36
For salt of weak acid and weak base
1
P H    log Ka  log kw  log K b 
2
1
1
1
  pKa  pkw  pk b
2
2
2
1
1
1
  4.8   14   4.78
2
2
2
 7.01
Equilibrium constant for the reaction

18.
19.
SO 2(g)  1 2 O2(g)  SO3(g)
KC 
1
4.9  10  2
and for reaction 2SO2(g)  O2(g)  2SO3(g)
2
1


KC  
 416.49
2 
 4.9  10 
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20.


P H  3  Η 3 O  1 103
H O  110 
Ka 

3
2
3 2
C
21.
0.1
 110 5
CO2(g)  C(g)  2CO(g)
initial pressure 0.5 atm
0
final pressure(0.5-x)
2x
Total pressure = 0.5 -x+2x=0.5 +x = 0.8 atm
 x=0.3 atm.
2
KP 
22.
23.
 p CO   (0.6) 2 1.8
 PCO 2  0.2
atm.
According to lowry-bronsted acid base theory in (ii) reaction H 2 PO 4 donates H  ion to H 2 O so
it acts as an acid.
Kp  Kc(RT) n(g)
24.
n( g )  1
 Kc  0.0831 457 It means Kp > Kc
Water is an amphiprotic solvent as it can accept protons as well as give protons.
25.
Ammonium ion NH 4  is a conjugate acid of NH 3


NH3  H2O  NH4  OH
Bronsted base Conjugate acid
26.
HSO 4 can act as a bronsted acid as well as bronsted base.
HSO 4  H 2 O  SO 42   H 3 O
Acid
HSO 4  H 2 O  H 2SO 4  OH 
Base
27.
P H  4.7   H 3 O   1.995  10 5 pK w  14  K W  1  10 14
 log H3O NowKW  H3O OH 


 OH  
110 14
 5 10 10 M
5
1.995 10
28.
conjugate base of H 2 PO 4 is HPO 24
29.
H 2 PO 4  H 2 O  HPO 42   H 3O 
Acid
conjugate base of OH 

OH   H 2 O  O 2   H 3 O
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30.
Lewis acid always accepts a pair of e here AlCl 3 accepts a pair of e.
31.
32.
CH 3 3 B accept a pair of e so lewis acid
Pressure on equilibrium system increses ,so volume decreses. volume of ice is more than liquid
H 2 O so more ice is malted
33.
For endothermic H  O  reaction, change in temperature affects the equilibrium system and forward
reaction takes place . by increasing temp .
34.
K1 
and
XeOF4 HF2 K  XeOF4 XeO3F2 
2
XeF6 H 2 O
XeF4 XeF6 
XeO3 F2 H 2O
K 
is obtained by
XeF4 HF2
K2
K1
K  K2
36.
37.
38.
4
6

NO H 2 O
Kc 
NH3 4 O 2 5
K1
 (conc.)46(45)
 conc1
According to Le -chatelliers principle, if conc. of reactant become doubled, then forward reaction
takes place and concentration of product also increases. so equilibrium constant also remains same.
AB4(s)  A 4(aq)  4 B (aq)
_
K SP   A 4    B 
 
4
S(45)4  256S5
1
 K 5
S   SP 
 256 
39.
2
KSP  M2  X   4S3
 M   S   KSP 
4

2
40.
1
3
 4 1012 


4 

1
3
 1104
 CH3COO 2 Ca  Ca 2  2CH3COO 
0.005 2  0.005  0.01M.
PH   1  log KW  logKa  log C 
2
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1
  log K W  1 log K a  1 log C
2
2
2
1
  pkw  1  pka  1 log 1 10 -2
2
2
2
41.
 1 2  14  1 2  4.74  1 2 (2)
= 7  2.37  1  8.37
change in volume affects number of moks per unit voulme. In a reaction
N2(g)  O2(g)  2 NO(g)
no. of moles of reactants and product are equal so volume change does not affects the equilibrium.
42.
P H  3 means  H    103 M
P H  4 means  H    104 M
After mixing equal volume total
H   110
-3

 10 4 1
2

0.110-4  110 4
2
=
1.110 3
 5.5 10  4
2
 H    5.5 104 M
P H  log  H    log(5.5  10 4 )  4  0.7404  3.26
43.
P H of HCl should be less than 7. due to self ionisation of H 2 O
from acid  H    10  8 M from H 2 O  H    10  7 M
Total  H    10 8  10 -7  10 8 (1  10)  11  10 8 M
PH  log H   log(11108 )  (1.0414  8)
= 6.96
H
P  6.96
44.
P of buffer solution
H
 pK a  log
salt 
 Acid 
  log K a  log1
  log108  0   8
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45.
if x is the solubility of AgCl in 0.04 M cacl2 ,
then
 Ag    x mol L1
Cl  2  0.04  x  0.08  x  0.08 M
KSP of AgCl
  Ag    Cl 
4  1 0  10
46.
P H  pka  log
5.5  4.5  log
0.08
  A g    5  10  9 M
CH3COONa 
CH3COOH
 CH3COONa 
0.1
5.5  4.5  logCH 3CooNa   1
 logCH 3CooNa   0
 CH 3CooNa   1 M
47.
I  can accept protons and hence is a base.
50.
N3H  N3  H
53.
55.
H 2S can donate proton but can't accept proton.
With increase in temperature, ionic product increases. because self ionisation of is endothermic
process
EDTA is Arrhenius acid as it can give H  ions in aqueous solution, bronsted base. as it can accept
protons and lewis base because N and O in it can donate lone paris of electrons.
56.
Hydrazoic acid  N3 H
57.
KW   H  OH 
58.
 mol L1  mol L1  mol 2 L2
Smaller the Kb value, weaker is the base.
59.
On adding NH 3 to water, OH -  will increse,
Kw   H3O  OH   is constant.
Therefore H3O  will decrease
61.
 POH  P K b  log
Salt 
 Base
[Salt] = [Base]
=  log(2  10 5 )  log 1
=  0.3010  5.000  0  4.6990
P H  14  p OH  14  4.6990  9.3010
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62.
P OH  pK b  log
Salt 
 Base 
P OH  14  P H  4.75
4.75  p kb  log 1
 pk b  4.75
63.
 log 1  0
2

BaSO4 (s)  Ba (aq)
 So24(aq)
x mol/lit
x mol/lit
K SP   Ba 2   So 24  
K SP  x 2
1
64
65.
1
 x  (ksp) 2  (1.5 10 9 ) 2  3.9 10 5 molL1
Hint : As value of KSPis less , solubility is also less.
glycine (NH2CH2COOH) is more acidic than basic.
overall ionisation constant
K  K a1  K a 2  4.5 10 3  1.7 10 10
 7.65  10 13
H  

K  C  7.65 10 13  0.01
 0.87  10 7 M
P H  log(8.7  10 8 )  7.0605
66.
KSP of BaCo3   Ba 2  CO32 


 Ba 2 
67.
5.110 9
 5.110 5 M
110  4
Ba(OH)2  2HCl  BaCl2  2H2 O
2 mol HCl neutralize1 mole Ba(OH)2
1 mol HCl neutralize 0.5 mole Ba(OH)2
Ba(OH)2  Ba 2  2OH 
1
2
 no.of moles of Ba(OH)2  3.  1  2
 Ba(OH)2 left  3  0.5  2.5
Ba(OH) 2  

2.5
 0.05 M
50

or OH   2  0.05 M  0.1 M
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68.
69.
K W 1.0  1014
Kh 

 5.65  1010
5
K b 1.7  10
PH  
1 K
P W  PKa  log C
2


 1 2  14  1 2  4  1 2 log10 2
=7+2-1=8
70.
 
P H  3 means H   10 3 M
1000 ml juice contains 10 3 mole H  ions
 no.of H  ions  10 3  6.022  10 23 in 1000 ml
 500 ml juice contains H  ions 
71.
10 3  6.022 10 23  500
1000
= 3.011 10 20
All alkaline earth metal chlorides MCl 2  on hydrolysis will produce acidic solution
MCl 2  H 2 O  M(OH) 2  2HCl
because M(OH)2 is a weak base and HCl is a strong acid. but as we go down the group, basic
character of hydroxides increses. Hence acidic character decreses. So BaCl2 will have the highest P H .
72.
MX (s)  M   X 
K SP  S 2  S  (K SP )
1
2
 (4  10 8 )
MX 2(s)  M  2X K SP
2

1
2
 2  10 4 M.
K 
 4S  S   SP 
 4 
3
1
3
 2 105 M.
M3X (s)  3M  X 3
 KSP 
KSP = 27S S  

 27 
4
1
4
 2.7 1015 


27


1
4
 1104 M
 2  10 4  1  10 4  2  10 5
 MX  M 3X  MX 2 .
74.
P OH  pk b  log
Salt 
weak base 
 4.7  log
0. 1
 5.7
0.01
P H  14  5.7  8.3
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75.
HCOONa is a Salt of weak acid (HCOOH) and Strong base (NaOH) So it is basic. C 6 H 5 NH 3Cl
is a Salt of weak base C 6 H 5 NH 2  and strong acid (HCl) so it is acidic. KCN is a Salt of Strong
base (KOH) and weak acid (HCN) so it is basic.
76.

AgIO3(s)  Ag (aq)
 IO3 (aq)
K SP  S2
 S  (K SP )
1
2
 (1.0 108 )
1
2
 1104 mol
 S  110-4  283  283 10  4 gm
lit
lit
1000 ml contains 283  10 4 gm of AgIO 3
100 ml contains 28.3  10 4 gm of AgIO 3
77.
Molar concentration of NaOH 
O H
P
50  10 3 gm
-4
40 gm mol 1  10dm 3  1.25 10 M
  lo g (1 .2 5  1 0
4
)
 0.0969  4.0  3.9031
H
 P  14  3.9031  10.0969
78.
0.1 M HCl m eans  H    10  1  P H  1
0.1M KOH means  OH    10 1 P OH  1  P H  13
0.01 M HCl means  H    10 2  P H  2
0.01M KOH means OH   102  POH  2  PH  12
79.
 H  
ion concentrations are 10 11 , 10 12 and 1013 on mixing equal volumes,  H   in final solution
10 11 10 12 10 13 10 10 12  110 12  0.110 12

=
3
3

80.
H
(A) P  pka  log
11.110 12
 3.7 10 12 M.
3
Salt   4.8  log 0.1  4.8  1  5.8
acid 
0.01
H
(B) P  4.8  log
0.01
 4.8  log 10-1  4.8  1  3.8
0.1
H
(C) P  4.8  log
0.1
 4.8  0  4.8
0.1
1
1
H
(D) P  7  (pka  pkb)  7  (4.8  4.8) =7.0
2
2
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81.
Soda water contains weak acid H 2 CO 3 So its P H  7
83.
P H  p K a  log
6  5  log
 log
84.
Salt 
 acid 
Salt 
acid 
Salt   1
acid 
Salt   10.
acid 
or
Solubility of Cus  (ksp)
1
2
 (110
 ksp 
Solubility of Ag 2S  

 4 
)  11016 M
1
3
 4  10-45  3
15

  110 M
4


1
5

  110
At 25 C temp P of H O  7 H   10 M
P  6.8 means P  7  H is more than 10 M
Solubility of HgS   ksp 
85.
1
1
1
32 2

 4 10-54
2

H
27
M
7
2
H

H
7
Self ionisation of H 2 O is endothermic so by increasing temp  H   ion increases.
86.
ksp for BaSo 4   Ba 2   SO 42  
10  11  0.1   SO 24  
  SO 24    10  10 M
ksp for CaSO 4   Ca 2    SO 24  
10 6
0.1

 10 5 M  SO 24 

2
ksp for Ag 2So 4   Ag   SO 42  
10 5

(0.1)2

 10 3 M  SO 24


As SO 24  10 10 M in BaSO 4 (least value)
it can be precipitated first
87.
A2 B3(s)  2A3  3B2
2x
3x
  B 
ksp  A
3 2
2 3
 (2 x) 2  (3 x) 3
 4 x 2  27 x 3  108x 5
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88.
P H  P K a  log
Salt 
 Acid 
4.91  4.76  log
 log
Salt 
Acid 
Salt   0.15
Acid 

Salt   antilogof0 .15  1.41
Acid 
0.2
 50
1000

 1.41  V  70.92 ml
0.1
V
1000
89.
pka  log(1.8  10 4 )  3.74
log

90.
91.
Salt   P H  pka  4.25  3.74  0.51
Acid 
Salt   antilog of0.51  3.24
 Acid 
14
11
 10
 4  1.47  10
kc
6.8 10
Buffer Solution of highest capacity is formed at which
K b  kw
P H  pka  log(1.8  10 4 )  3.74
93.
Ionic Product of CaF2

   10  ksp
 10   10  ksp
 10   10  ksp
 10   10  ksp
in (i) IP  Ca 2  F
(ii) IP  10 2
(iii) IP  10 5
(iv) IP  10 3
94.
P H  pka  log
2
12
3 2
8
3 2
11
5 2
13
 ppt. obtain
Salt  Salt   Acid 
Acid 
= 8 + log 1 = 8
95.
H 2S  weak acid H 2SO 4  Strong acid
NaCl  neutral NaNO 2  basic
Hence H3 O  will be in the order of
NaNO2  NaCl  H 2S  H 2SO4
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96.
NaCN is Salt of weak acid (HCN) and Strong base (NaOH) Hence
h
97.
kw
10 14

 2.48  10 2
1
k aC
9
(1.3  10 ) 
80
 Percen tan ge Hydrolysis  (2.48 10 2 )100
= 2.48
If x is Solubility of AgCl in 0.04 M CaCl2 , then
 A g    x m ol L 1
 C l    (0.04  2)  x  0.08M
0.08(x)  4  10 10
x  5.0  10 9 M
98.
P H  pka 
log CH 3COONa 
1 CH 3COOH 
5.5  4.5  log
CH 3COONa 
0.7
 4.5 + log  CH3COONa 1
 log  CH 3 COONa   0
  CH 3 COONa   7 M
99.
50 ml of0.4 N HCl 
0.4
 50  0.02 g eq.
1000
0.2
 50  0.01g eq.
1000
0.01 g eq of NaOH will Neutalilise 0.01 g eq of HCl
 HCl left unneutralised = 0.01 g eq
vol of Sol. =50+50
=100ml
50 ml of 0.2 N NaOH 
 HCl 
0.01
 1000  0.1N
100
or  H    0.1M
 p H  log(0.1) = 1.0
100.
CH 3 COOH 
at eq (a  3.4  10 4 )
CH 3COO  
3.4  10 4
 3.4  10   3 .4  1 0   1 .7  1 0
 a  3.4  10 
4
H
3.4 10 4
4
4
5
(G iven)
 a  6.8  10 3
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101.
A  B C  D
initial 1 1 0 0
conc.
Ateqm(10.8) 0.8 0.8
conc.(10.8)
0.2 0.2
Kc 
CD  0.8  0.8  16.0
AB 0.2  0.2
Kc 
CD  2  2  1
AB 2  2
102.
A BCD
initial 4 4
conc.
0 0
Ateqm(4  2) 2 2
conc.(4  2)
2
2
103. KC remains same beacause KC is a characteristic constant.
PCl5  PCl3  Cl2
2
0
0
104.
2  60
100
2  40
100
moles  1.2
0.8
mol 1.2 0.8
conc. 
 2 2
lit
 Kc 
2  40
100
0.8
0.8
2
PCl 3 Cl 2   0.4  0.4  0.266
PCl 5 
0.6
105. Dissociation of CH3COOH
CH3COOH  H  CH3COO k a1  1.5 103
Dissociation of HCN: HCN  H   CN 
for a reaction
k a 2  4.5 10 3
CN   CH3COOH  CH3COO  HCN is
Ka 
Ka 1 1.5  10 3

 3.33  10 4
Ka 2 4.5  10 10
180
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106. A  B  C  D
x x
0 0
CD  2x  2x  4
AB x  x
kc 
2x 2x
107. N 2 O 4  2 NO 2
1
0
(1-  )
2
108.
H 2  I 2  2 HI
Total moles = 1-  +2  =1+ 
from equation 2x=3
Initial
4.5 4.5
0
conc.
at eqm. (4.5-x)(4.5-x) 2x
 x  3 2  1.5
H 2   4.5  1.5  3
Kc 
I 2   4.5  1.5  3
109.
K f  1.1 10 2
HI2  3  3  1
H 2 I 2  3  3
K b  1.5 10 3
kf 1.1 10 2
kc 

 7.33
k b 1.5  10 3
110.
initial
conc.
At eqm.
conc.
2 NH3  N 2  3H 2
2
0
0
1
1
3
2
3
KC 
111.
 N 2  H 2 
 NH 3 
2
 
1  3
 2 22
(1)
3
 27
64
P4(S)  5O2(g)  P4 O10(s)
KC 
 P4O10 
P4(s)  O2(g) 
5
we know that concentration of a solid component is always taken as a unity
 Kc 
1
O 2 5
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112. 2Ag 2O(s)  4 Ag (s)  O2(g )
for this reaction Kp  Po 2
H2(g)  CO2(g)  CO(g)  H2 O(g)
113.
initial
conc. 1
At eqm.
(1-x)
Kp 
129.
PC O  PH 2 O
PH 2  PC O

1
(1-x)
0
x
0
x
x2
(1  x ) 2
2
q
p
ApBq (s)  PA (aq)
 q B(aq)
Solubility is PS
qS
mol/lit
K SP  (PS) P  (qs) q
 S( p  q )  P P  q q
135.
Co  0.1 M Ka  5  10 8
H  

Ka  Co
 (5  10 8  0.1)
1
2
 (50  10 10 )
1
2
 7.07 10 5 M.
182
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