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CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 14 - MCQExams.com
CBSE
Class 11 Medical Chemistry
Equilibrium
Quiz 14
At
90
∘
C
, pure water has
[
H
3
O
+
]
=
10
−
6.7
m
o
l
/
l
i
t
. What is the value of
K
W
at
90
∘
C
:
−
Report Question
0%
10
−
6
0%
10
−
12
0%
10
−
14
0%
10
−
13.4
Which of the following will not function as buffer solution?
Report Question
0%
Na
H
2
P
O
4
+ N
a
2
HP
O
4
(
1:1
Molar
Ratio
)
0%
N
H
4
Cl + NaOH
(
2:1
Molar
Ratio
)
0%
HCOOH + NaOH
(
1:1
Molar
Ratio
)
0%
HCOOH + HCOONa
(
1:1
Molar
Ratio
)
In which of the following solution of electrolyte, ionic equilibrium is set up?
Report Question
0%
H
2
S
O
4
0%
H
C
l
0%
N
a
O
H
0%
H
C
N
Which of the following is a buffer solution?
Report Question
0%
500 mL of 0.1 N CH
3
COOH +500 mL of 0.4 N NaOH
0%
500 mL of 0.1 N CH
3
COOH +500 mL of 0.1 N HCl
0%
500 mL of 0.1 N CH
3
COOH +500 mL of 0.2 N NaOH
0%
500 mL of 0.1 N CH
3
COOH +500 mL of 0.3N NaOH
On adding NaCl solution to the saturated solution of
P
b
C
l
2
then the concentration of :
Report Question
0%
P
b
2
+
will increase
0%
C
l
−
will decrease
0%
P
b
2
+
will decrease
0%
Ions will remain unaltered
The ionic product of water at
25
o
C
is
10
−
14
The ionic product at
90
o
C
will be
Report Question
0%
1
×
10
−
20
0%
1
×
10
−
12
0%
1
×
10
−
14
0%
1
×
10
−
16
In a buffer solution the ratio of concentration of
N
H
4
Cl
and
N
H
4
OH
is
1:1
when it changes in 2:1 what will be the value of pH of buffer:-
Report Question
0%
Increase
0%
Decrease
0%
No effects
0%
N. O. T.
A hydrogen electrode placed in a buffer solution of NaCl and HCN in the ratio of X:Y and Y:X has an oxidation electrode potential values
E
1
a
n
d
E
2
volts respectively at
25
0
C
. The pKa value of HCN is
Report Question
0%
(
E
1
+
E
2
)
0.118
0%
−
(
E
1
+
E
2
)
0.118
0%
(
E
1
−
E
2
)
0.118
0%
(
E
2
−
E
1
)
0.118
If the degree of ionization of water is 1.8 x
10
−
9
at a given temperature. Its ionic product will be
Report Question
0%
1.8 x
10
−
16
0%
1 x
10
−
14
0%
1 x
10
−
16
0%
1.67 x
10
−
14
2
H
2
O
⇋
H
3
O
+
+
O
H
−
,
K
w
=
1
×
10
−
14
at
25
∘
C
hence
K
a
is :
Report Question
0%
1
×
10
−
14
0%
18
×
10
−
17
0%
5.55
×
10
−
3
0%
1.00
×
10
−
7
The dissociation constant of a weak acid is
1.0
×
10
−
5
, the equilibrium constant for its reaction with strong base is
Report Question
0%
1.0
×
10
−
5
0%
1.0
×
10
9
0%
1.0
×
10
7
0%
1.0
×
10
14
One litre of water contains
10
−
7
mole of
H
+
ions. Degree of ionisation of water is:
Report Question
0%
1.8
×
10
−
7
%
0%
0.8
×
10
−
9
%
0%
3.6
×
10
−
9
%
0%
3.6
×
10
−
7
%
Explanation
1
litre of water contains
1000
/
18
mole.
So degree of ionization
=
10
−
7
×
18
1000
=
1.8
×
10
−
7
%
Option A is correct.
Aqueous solution of NaCl is neutral because
Report Question
0%
N
a
+
undergoes hydrolysis
0%
Cl undergoes hydrolysis
0%
Both
N
a
+
and Cl undergo hydrolusis
0%
Does not undergo hydrolysis
Which of the following may be regarded as buffer solution?
Report Question
0%
50 ml of 0.1 M acetic acid + 50 ml of 0.05 M
N
a
O
H
0%
50 ml of 0.05 ammonia + 50 ml of 0.1 M
H
C
l
0%
25 ml of 0.1 M acetic acid + 50 ml of 0.1 M
N
a
O
H
0%
50 ml of 0.1 ammonia + 50 ml of 0.05 M
H
C
l
The total number of different kind of buffer obtained during the titration of
H
3
P
O
4
with NaOH are:-
Report Question
0%
3
0%
1
0%
2
0%
0
Select the correct acid-base equilibrium.
Report Question
0%
H
N
O
3
+
H
F
⇌
H
2
N
O
+
3
+
F
−
0%
H
N
O
3
+
H
F
⇌
H
2
F
+
+
N
O
−
3
0%
C
H
3
C
O
O
H
+
N
H
3
⇌
C
H
3
C
O
O
H
+
2
+
N
H
+
2
0%
H
F
+
H
2
O
⇌
H
2
F
+
+
O
H
−
H
C
O
O
H
and
C
H
3
C
O
O
H
solutions have equal
p
H
. If
K
1
K
2
(ratio of ionisation constants of acids) is 4, their molar concentration ratio will be:
Report Question
0%
2
0%
0.5
0%
4
0%
0.25
At
90
∘
C
,
pure water has
[
H
3
O
+
]
=
10
−
6
m
o
l
litre
−
1
.
The value of
K
w
at
90
∘
C
is:
Report Question
0%
10
−
6
0%
10
−
12
0%
10
−
14
0%
10
−
8
Hydrogen ion concentration of an aqueous solution is
1
×
10
−
4
M
. The solution is diluted with equal volume of water. Hydroxyl ion concentration of the resultant solution in terms of mol
d
m
−
3
is __________.
Report Question
0%
1
×
10
−
8
0%
1
×
10
−
6
0%
2
×
10
−
10
0%
0.5
×
10
−
10
Explanation
Concentration of aqueous solution=
10
−
4
M
If we add equal volume, let say V of water. The final volume of solution will be 2V. The new concentration is
⇒
M
1
V
1
=
M
2
V
2
⇒
10
−
4
M
×
V
=
M
2
×
2
V
⇒
M
2
=
5
×
10
−
5
M
⟶
New concentration of
[
H
+
]
.
p
H
=
−
l
o
g
[
H
+
]
=
−
l
o
g
[
5
×
10
−
5
]
⇒
p
H
=
4.3
⇒
p
O
H
=
14
−
p
H
=
9.69
⇒
−
l
o
g
[
O
H
−
]
=
9.69
⇒
[
O
H
−
]
=
10
−
9.69
=
2
×
10
−
10
Thus the concentration of
[
O
H
−
]
i
s
2
×
10
−
10
M
.
If
1
M
C
H
3
C
O
O
N
a
is added to
1
M
C
H
3
C
O
O
H
, then:
Report Question
0%
p
H
of the solution increases
0%
p
H
decreases
0%
p
H
does not change
0%
cannot be predicted
The dissociation constants for acetic acid and HCN at
25
0
C are
1
.
5
×
10
−
5
and
4
.
5
×
10
−
10
, respectively. The equilibrium constant for the equilibrium
C
N
−
+
C
H
3
C
O
O
H
⇋
H
C
N
+
C
H
3
C
O
O
−
would be :
Report Question
0%
3
×
10
4
0%
3
×
10
5
0%
3
×
10
−
5
0%
3
×
10
−
4
At
25
∘
C
the dissociation constant of a base , BOH, is
1.0
×
10
−
12
. The concentration of hydroxyl ions in 0.01 M aqueous solution of the base would be
Report Question
0%
1.0
×
10
−
6
m
o
l
L
−
1
0%
1.0
×
10
−
7
m
o
l
L
−
1
0%
2.0
×
10
−
6
m
o
l
L
−
1
0%
1.0
×
10
−
5
m
o
l
L
−
1
A solution is saturated with respect to
A
g
S
C
N
as well as
A
g
B
r
. The concentration of
A
g
+
in the solution would be:
(
K
s
p
for
A
g
S
C
N
=
1.2
×
10
−
12
and for
A
g
B
r
=
5
×
10
−
13
)
Report Question
0%
1.3
×
10
−
5
0%
1.3
×
10
−
12
0%
1.3
×
10
−
6
0%
1.3
×
10
−
8
Which of the following salt evolve(s)
N
2
gas on heating ?
Report Question
0%
(
N
H
4
)
2
C
r
O
4
0%
N
H
4
N
O
2
0%
N
H
4
N
O
3
0%
(
N
H
4
)
2
C
O
3
Which of following salt solution will impart greenish colour ?
Report Question
0%
K
2
C
r
2
O
7
0%
Z
n
C
l
2
0%
K
2
M
n
O
4
0%
K
2
C
r
O
4
10 ml of 0.2 M
A
g
N
O
3
is mixed with 10 ml of 0.1 M
N
a
C
l
. What is the concenctration of
C
l
−
in the resulting solution
[
K
s
p
(
A
g
C
l
)
=
10
−
10
M
2
]
Report Question
0%
10
−
5
M
0%
10
−
9
M
0%
2
×
10
−
9
M
0%
10
−
8
M
For preparing a buffer solution of pH = 7.0, which buffer system you will choose?
Report Question
0%
H
3
P
O
4
,
H
2
P
O
−
4
0%
H
2
P
O
−
4
,
H
P
O
2
−
4
0%
H
P
O
2
−
4
,
P
O
3
−
4
0%
H
3
P
O
4
,
P
O
3
−
4
The percentage of pyridine
(
C
5
H
5
N
)
that forms pyridinium ion
(
C
5
H
5
N
+
H
)
in a 0.10 m aqueous pyridine solution
(
K
b
f
o
r
C
5
H
5
N
=
1.7
×
10
−
9
)
is:
Report Question
0%
0.0060%
0%
0.013%
0%
0.77%
0%
1.6%
A weak acid react with strong base, ionisation constant of weak acid is
10
−
4
. Find out equilibrium constant for this reaction.
Report Question
0%
10
−
10
0%
10
10
0%
10
−
9
0%
10
9
Which statement is/are correct-
Report Question
0%
If
0.1
M
C
H
3
C
O
O
H
aqueous solution
(
K
a
=
1.8
×
10
−
5
)
is diluted at
2560
o
C
, then
[
H
+
]
will increase
0%
On heating a sample of pure water
[
H
+
]
and
[
O
H
−
]
increases but the sample still remains neutral
0%
10
−
4
M
H
C
l
solution is less acidic than
0.1
M
H
C
N
(
K
a
H
C
N
=
10
−
5
)
0%
10
−
8
M
H
C
l
aqueous solution will have pH exactly equal to 8
The self ionisation constant of
N
H
3
at
50
∘
C
is given by
K
N
H
3
=
[
N
H
+
4
]
[
N
H
−
2
]
=
10
−
30
. How many
N
H
−
2
ions are present per
c
m
3
of pure liquid
N
H
3
?
Report Question
0%
6
×
10
6
0%
6
×
10
5
0%
6
×
10
−
5
0%
6
×
10
−
6
The dissociation constant of an acid HA is
1
×
10
−
5
. The pH of 0.1 molar solution of the acid will be approximately:
Report Question
0%
3
0%
5
0%
1
0%
6
The silicate anion in the mineral kinetic is a chain of three
S
i
O
4
−
4
tetrahedral that shares conners with adjacent tetrahedral. The mineral also contains
C
a
icon
C
u
ions and water molecules in a 1:1:1 ratio mineral represented as:
Report Question
0%
C
a
C
u
2
S
i
3
O
10
.
H
2
O
0%
C
a
C
u
S
i
3
O
10
.
H
2
O
0%
C
a
2
C
u
2
S
i
3
O
10
.2
H
2
O
0%
None of these
Which of the following is ionic?
Report Question
0%
H
C
l
0%
C
H
C
l
3
0%
I
F
5
0%
K
I
For which salt the
p
H
of its solution does not change with dilution?
Report Question
0%
N
H
4
C
l
0%
C
H
3
C
O
O
N
H
4
0%
C
H
3
C
O
O
N
a
0%
None of these
Consider the following equilibrium.
Liquid
⇌
Vapour
Which of the following relation is correct?
Report Question
0%
d
l
n
P
d
T
=
Δ
H
v
R
T
2
0%
d
l
n
G
d
T
2
=
Δ
H
v
R
T
2
0%
d
l
n
P
d
T
=
−
Δ
H
v
R
T
0%
d
l
n
P
d
T
2
=
−
Δ
H
v
T
2
The
p
H
of an aqueous solution of sodium chloride at
60
∘
C
is
Report Question
0%
7.0
0%
>
7.0
0%
<
7.0
0%
0
Explanation
We know that at
25
∘
C
dissociation constant
K
w
of water is,
H
2
O
⇌
H
+
+
O
H
−
[
H
+
]
[
O
H
−
]
=
10
−
14
=
K
w
So,
[
H
+
]
=
[
O
H
−
]
=
10
−
7
M
Thus
p
H
=
7
for
H
2
O
at
25
∘
C
Adding
N
a
C
l
does not affect the
p
H
whatever be the temperature because sodium chloride is salt of strong acid and strong base. So, it is
H
+
ion of water which is affecting the
p
H
.
Dissociation of water is an endothermic process. With an increase in
temperature
dissociation
constant of water increases. So the dissociation constant is more i.e.,
K
w
>
10
−
14
.
Therefore,
[
H
+
]
>
10
−
7
and
p
H
is less than 7.
Hence the correct option is (C).
The EMF of cell:
H
2
(
g
)
|
Buffer
∥
Normal calmelelectrod, is 0.70 V at
25
∘
C
,
when barometric pressure is 760 mm. What is the pH of the buffer solution?
E
0
C
a
l
o
m
e
l
=
0.28
V
.
[
2.303
R
T
I
F
=
0.06
]
Report Question
0%
3.5
0%
7.0
0%
tending to zero
0%
tending to 14.0
The dissociation constant of formic acid is
0.00024
. The hydrogen ion concentration in
0.002
M
H
C
O
O
H
solution is nearly
Report Question
0%
6.93
×
10
−
4
M
0%
4.8
×
10
−
7
M
0%
5.8
×
10
−
4
M
0%
1.4
×
10
−
4
M
The degree of dissociation of pure water at
25
∘
C
is found to be
1.8
×
10
−
9
. The dissociation constant,
K
d
of water, at
25
∘
C
is
Report Question
0%
10
−
14
0%
1.8
×
10
−
16
0%
5.56
×
10
−
13
0%
1.8
×
10
−
14
Explanation
Given:
Degree of dissociation of pure water,
α
=
1.8
×
10
−
9
We know that,
Density of pure water =
1
g
c
m
−
3
=
1000
g
L
−
1
Molar mass of water =
18
g
m
o
l
−
1
C
o
n
c
e
n
t
r
a
t
i
o
n
=
n
u
m
b
e
r
o
f
m
o
l
e
s
V
o
l
u
m
e
=
1000
g
L
−
1
18
g
m
o
l
−
1
=
55.5
M
Dissociation of water,
H
2
O
⇌
H
+
+
O
H
−
Dissociation constant,
K
d
=
[
H
+
]
[
O
H
−
]
[
H
2
O
]
=
C
α
×
C
α
C
−
C
α
As
α
<<<
1
,
[
H
2
O
]
is not affected.
Therefore,
K
d
=
c
α
2
=
55.5
×
(
1.8
×
10
−
9
)
2
=
1.79
×
10
−
16
≈
1.8
×
10
−
16
Hence, the correct option is (B).
Which one of the following salts gives aqueous solution which is neutral?
Report Question
0%
N
a
H
C
O
3
0%
N
a
H
S
O
4
0%
N
a
C
l
0%
N
a
H
4
H
C
O
3
When ammonia is added to pure water, the concentration of which of the following species already present, will decrease?
Report Question
0%
O
H
−
0%
H
3
O
+
0%
N
H
+
4
0%
None of these
n-caproic acid,
C
5
H
11
C
O
O
H
, found in coconut and palm oil is used in making artificial flavors, has solubility in water equal to
11.6
g
/
L
. The saturated solution has
p
H
=
3.0
. The
K
a
of acid is
Report Question
0%
10
−
6
0%
10
−
5
0%
2
×
10
−
5
0%
2
×
10
−
6
What is
[
H
+
]
in mol/L of a solution that is 0.20 M
C
H
3
C
C
O
N
a
and 0.10 M in
C
H
3
C
O
O
H
?
(
K
a
f
o
r
C
H
3
C
O
O
H
=
1.8
×
10
−
5
)
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0%
3.5
×
10
−
4
0%
1.1
×
10
−
5
0%
1.8
×
10
−
5
0%
9.0
×
10
−
6
A solution has initially
0.1
M
−
H
C
O
O
H
and
0.2
M
−
H
C
N
.
K
a
of
H
C
O
O
H
=
2.56
×
10
−
4
,
K
a
of
H
C
N
=
9.6
×
10
−
10
. The only incorrect statement for the solution is
(
l
o
g
2
=
0.3
)
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0%
[
H
]
+
=
1.6
×
10
−
3
M
0%
[
H
C
O
O
]
−
=
1.6
×
10
−
3
M
0%
[
C
N
]
−
=
1.2
×
10
−
7
M
0%
p
O
H
=
2.8
Calculate the formation constant for the reaction of a tripositive metal ion with thiocyanate ions to form the monocomplex if the total metal concentration in the solution is
2
×
10
−
3
M
, the total
S
C
N
−
concentration is
1.51
×
10
−
3
M
and the free
S
C
N
−
concentration is
1.0
×
10
−
5
M
.
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0%
7.55
×
10
4
0%
3
×
10
5
0%
3.33
×
10
−
6
0%
1.5
×
10
5
Small amount of freshly precipitated magnesium hydroxides are stirred vigorously in a buffer solution containing
0.25
M
of
N
H
4
C
l
and
0.05
M
of
N
H
4
O
H
.
[
M
g
2
+
]
in the resulting solution is (
K
b
for
N
H
4
O
H
=
2.0
×
10
−
5
and
K
s
p
of
M
g
(
O
H
)
2
=
8.0
×
10
−
12
)
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0%
4
×
10
−
6
M
0%
2
×
10
−
6
M
0%
0.5
M
0%
2.0
M
A volume of
1.0
L
of solution which was in equilibrium with solid mixture of
A
g
C
l
and
A
g
2
C
r
O
4
was found to contain
1
×
10
−
4
moles of
A
g
+
ions,
1.0
×
10
−
6
moles of
C
I
−
ions and
8.0
×
10
−
4
moles of
C
r
O
2
−
4
ions.
A
g
+
ions are added slowly to the above mixture (keeping the volume constant) till
8.0
×
10
−
7
moles of
A
g
C
l
got precipitated. How many moles of
A
g
2
C
r
O
4
were precipitated simultaneously?
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0%
7.68
×
10
−
4
0%
4.8
×
10
−
4
0%
8.0
×
10
−
4
0%
7.68
×
10
−
5
Which one of the following is not an acid salt?
Report Question
0%
N
a
H
2
P
O
2
0%
N
a
H
2
P
O
3
0%
N
a
H
2
P
O
4
0%
N
a
2
H
P
O
4
Explanation
N
a
H
2
P
O
2
is not an acidic salt. It is a basic salt. It is a salt of weak acid
H
3
P
O
2
and a strong base
N
a
O
H
. Hence, the aqueous solution of
N
a
H
2
P
O
2
is alkaline in nature. In
N
a
H
2
P
O
2
,
there is no acidic hydrogen.
Hence, the correct option is (A).
The solubility of
A
g
C
N
in a buffer solution of
p
H
−
3.0
is (
K
s
p
of
A
g
C
N
=
1.2
×
10
−
16
;
K
a
of
H
C
N
=
4.8
×
10
−
10
)
Report Question
0%
1.58
×
10
−
5
M
0%
2.0
×
10
−
5
M
0%
1.58
×
10
−
4
M
0%
2.5
×
10
−
9
M
Explanation
Assume solubility of AgCN be a M in presence of a buffer of pH = 3
Given :-
K
s
p
of AgCN =
1.2
×
10
−
16
and
K
a
for HCN =
4.8
×
10
−
10
A
g
C
N
⇌
A
g
+
+
C
N
−
[
A
g
+
]
=
a
;
[
H
+
]
=
10
−
3
M
;
For weak HCN,
K
a
=
[
H
+
]
[
C
N
−
]
[
H
C
N
]
⇒
4.8
×
10
−
10
=
[
10
−
3
]
[
C
N
−
]
a
\therefore [CN^{-}] = 4.8 \times 10^{-7} \times a
For \ AgCN : K_{sp} = Ag^{+} \times CN^{-}
1.2 \times 10^{-16} = a \times 4.8 \times 10^{-7} \times a
a^{2} = \dfrac{1.2 \times 10^{-16}}{4.8 \times 10^{-7}}
a = 1.58 \times 10^{-5} M
0:0:1
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Practice Class 11 Medical Chemistry Quiz Questions and Answers
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