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CBSE Questions for Class 11 Medical Chemistry Equilibrium Quiz 13 - MCQExams.com
CBSE
Class 11 Medical Chemistry
Equilibrium
Quiz 13
When
0.1
m
o
l
arsenic acid,
H
2
A
s
O
4
is dissolved in
1
L
buffer solution of
p
H
=
4
, which of the following hold good?
K
1
=
2.5
×
10
−
4
,
K
2
=
5
×
10
4
,
K
3
=
2
×
10
−
23
for arsenic acid
[
′
<
<
′
sign denotes that the higher concentration is at least
100
times more than the lower one].
Report Question
0%
[
H
2
A
s
O
4
]
<<
[
H
2
A
s
O
−
4
]
0%
[
H
2
A
s
O
4
]
<<
[
H
A
s
O
2
−
4
]
0%
[
H
A
s
O
2
−
4
]
<<
[
H
A
s
O
−
4
]
0%
[
A
s
O
2
−
4
]
<<
[
H
A
s
O
2
−
4
]
Explanation
p
H
=
4
−
log
[
H
+
]
=
4
[
H
+
]
=
10
−
4
[
H
3
A
s
O
4
]
[
H
2
A
s
O
−
4
]
=
[
H
+
]
K
1
=
10
−
4
2.5
×
10
−
4
∴
1
2.5
⇒
[
H
3
A
s
O
4
]
<
[
H
2
A
s
O
−
4
]
[
H
2
A
S
O
−
4
]
[
H
A
s
O
2
−
4
]
=
[
H
+
]
K
2
=
10
−
4
5
×
10
4
=
10
−
4
−
4
5
=
1
5
×
10
8
⇒
[
H
2
A
s
O
−
4
]
<<
[
H
A
s
O
2
−
4
]
[
H
A
s
O
4
2
−
]
<<
[
H
2
A
s
O
−
4
]
not correct proof above.
[
A
s
O
2
−
4
]
[
H
A
s
O
2
−
4
]
=
K
3
[
H
+
]
=
2
×
10
−
23
10
−
4
=
2
10
×
10
18
=
1
5
×
10
18
⇒
[
A
s
O
4
2
−
]
<<
[
H
A
S
O
4
2
−
]
To prepare a buffer of
p
H
8.26
amount of
(
N
H
4
)
2
S
O
4
to be added to
500
m
L
of
0.01
M
N
H
4
O
H
solution is:
[
p
K
a
(
N
H
+
4
)
=
9.26
]
Report Question
0%
0.05
m
o
l
e
0%
0.025
m
o
l
e
0%
0.10
m
o
l
e
0%
0.005
m
o
l
e
Percentage ionisation of water at certain temperature is
3.6
×
10
−
7
%. Calculate
K
w
and
p
H
of water.
Report Question
0%
10
−
14
,
p
H
=
6
/
7
0%
4
×
10
−
14
,
p
H
=
6.7
0%
2
×
10
−
14
,
p
H
=
7
0%
10
−
14
,
p
H
=
7
Explanation
Given that
α
=
3.6
×
10
−
7
100
No. of ionized species
=
c
×
α
Molarity of water, C
=
1000
18
M
C
×
α
=
1000
18
×
3.6
×
10
−
7
100
=
2
×
10
−
7
M
[
H
+
]
[
O
H
−
]
=
2
×
10
−
7
×
2
×
10
−
7
=
4
×
10
−
14
∴
K
w
=
4
×
10
−
14
p
k
W
=
14
−
log
4
p
H
=
p
k
W
2
=
7
−
log
2
=
6.7
1
litre buffer solution
(
H
A
+
N
a
A
)
p
H
=
4
, is mixed with
2
litre buffer solution
(
H
A
+
N
a
A
)
p
H
=
4.3
.
p
H
of resulting solution if both solutions are initially
0.1
M
M
A
.
[Given
(
K
a
)
H
A
=
10
−
4
;
log
2
=
0.3
;
log
3
=
0.48
;
log
5
=
0.7
]
Report Question
0%
4.2
0%
4.22
0%
4.15
0%
6.4
What is the concentration of unionized acetic acid in the solution?
Report Question
0%
0
0%
4.9
×
10
−
10
M
0%
7
×
10
−
8
M
0%
2.45
×
10
−
10
M
The relationship between the molar solubility(z) of
M
g
F
2
in 0.10 M
M
g
(
N
O
3
)
2
and of
K
s
p
of
M
g
F
2
is:
Report Question
0%
(
K
s
p
/
0.10
)
1
/
2
0%
(
K
s
p
/
0.40
)
1
/
2
0%
(
K
s
p
/
4
)
1
/
2
0%
(
K
s
p
)
1
/
2
Which of the following mixtures will be a buffer solution when dissolved in
500.00
m
l
of water?
Report Question
0%
0.200
mol of aniline and
0.200
mol of
H
C
l
0%
0.200
mol of aniline and
0.400
mol of
N
a
O
H
0%
0.200
mol of
N
a
C
l
and
0.100
mol of
H
C
l
0%
0.200
mol of aniline and
0.100
mol of
H
C
l
An aqueous solution contains
0.01 M \ RNH_2 (K_b=2×10^{−6})
and
10^{−4} M\ NaOH.
The concentration of
OH^−
is nearly:
Report Question
0%
2.414\times 10^ {-4}M
0%
10^ {-4}M
0%
1.41\times 10^ {-4}M
0%
2\times 10^ {-4}M
One litre of water contains
1.0\times { 10 }^{ -7 }
moles of
{ H }^{ + }
ions. The degree of ionization of water is:
Report Question
0%
1.8\times { 10 }^{ -9}%
0%
0.8\times { 10 }^{ -9 }%
0%
3.6\times { 10 }^{ -9 }%
0%
3.6\times { 10 }^{ -7 }%
Explanation
Volume=
1L
Density of water=
1g/ml
Mass of water=
1000g
Moles of water =
\dfrac{1000}{18}
H_2O\rightleftharpoons H^{+}+OH^-
t=O
C
O
O
t.teq
c(1-\alpha)
e\alpha
e\alpha
c\alpha=10^{-7}
\Rightarrow \dfrac{1000}{8}\times \alpha=10^{-2}
\Rightarrow \alpha=1.8\times 10^{-9}
Solid
Ba{({NO}_{3})}_{2}
is gradually dissoved in a
1\times {10}^{-4}M
{Na}_{2}{CO}_{3}
solution. At what minimum conc. of
{Ba}^{-2}
will a precipitate of
Ba{CO}_{3}
begin to form? (
{K}_{sp}
for
Ba{CO}_{3}=5.1\times {10}^{-9}
)
Report Question
0%
4.1\times {10}^{-5}M
0%
8.1\times {10}^{-7}M
0%
5.1\times {10}^{-5}M
0%
8.1\times {10}^{-8}M
Explanation
Alc to the problem
\begin{aligned} \mathrm{Na}_{2} \mathrm{CO}_{3} \longrightarrow & 2 \mathrm{Na}^{+}+\mathrm{CO}_{3}^{2-} \\ \left[\mathrm{cO}_{3}^{2-}\right] &=1 \times 10^{-4} \mathrm{M} \end{aligned}
For,
\mathrm{BaCO}_{3} \rightleftharpoons\left[\mathrm{Ba}^{2+}\right]+\left[\mathrm{CO}_{3}^{2-}\right]
\mathrm{KG}_{p}=\left[\mathrm{Ba}^{2+}\right] \cdot\left[\mathrm{CO}_{3}^{2-}\right]
\begin{aligned} \Rightarrow \quad 5.1 \times 10^{-9} &=\left[B a^{2+}\right] \cdot\left[1 \times 10^{-4}\right] \\ &\left[\begin{array}{ll} {\left[B a^{2+}\right]=5.1 \times 10^{-9+4}} \\ {\left[B a^{2 +}\right]} =5.1 \times 10^{-5} \mathrm{M} \end{array}\right. \end{aligned}
\text { (c) is correct option }
Which of the following is a true statement:
Report Question
0%
The ionisation constant and ionic product of water are same.
0%
Water is a strong electrolyte.
0%
The value of ionic product of water is less than that of its ionisation constant
0%
At
298\ K
, the number of
{ H }^{ + }
ions in a litre of water is
6.023\times { 10 }^{ 16 }
.
Explanation
Let's check the options one by one:
(1) False. They are not the same.
Ionisation constant of water is
{ K }_{ a\left( { H }_{ 2 }O \right) }=\dfrac { \left[ { H }^{ + } \right] \left[ { OH }^{ - } \right] }{ \left[ { H }_{ 2 }O \right] }
Ionic product of water is
{ K }_{ w }={ K }_{ a\left( { H }_{ 2 }O \right) }\times \left[ { H }_{ 2 }O \right] \\ { K }_{ w }=\left[ { H }^{ + } \right] \left[ { OH }^{ - } \right]
(2) False. Water is a weak electrolyte. In fact, pure water is a non-electrolyte.
(3) False.
{ K }_{ w }={ K }_{ a\left( { H }_{ 2 }O \right) }\times \left[ { H }_{ 2 }O \right]
Clearly,
{ K }_{ w }>{ K }_{ a\left( { H }_{ 2 }O \right) }
value of ionic product is more than value of ionisation constant.
(4) At 298K, pH of water is 7
\therefore conc\quad of\quad \left[ { H }^{ + } \right] ={ 10 }^{ -7 }
moles per litre
1 mole contains
6.023\times { 10 }^{ 23 }
ions
\therefore conc\left[ { H }^{ + } \right] ={ 10 }^{ -7 }\times 6.023\times { 10 }^{ 23 }=6.023\times { 10 }^{ 16 }
The ionization constant of benzoic acid is
6.46 \times 10^{-5}
and
K_{sp}
for silver benzoate is
2.5\times 10^{-13}
. How many times is silver benzoate more soluble in a buffer of
pH = 3.19
compared to its solubility in pure water?
Report Question
0%
4
0%
3.32
0%
3.01
0%
2.5
Explanation
Since
pH=3.19
[H_3O^+]=6.46\times10^{-4}\ M
C_6H_5COOH+H_2O\leftrightharpoons C_6H_5COO^-+H_3O
K_a=\dfrac{[C_6H_5COO^-][H_3O^+]}{[C_6H_5COOH]}
\dfrac{[C_6H_5COOH]}{[C_6H_5COO^-]}=\dfrac{[H_3O^+]}{K_a}=\dfrac{6.46\times10^{-4}}{6.46\times10^{-5}}=10
Let the solubility of
C_6H_5COOAg
be
x\ mol/L
Then,
[Ag^+]=x
[C_6H_5COOH]+[C_6H_5COO^-]=x
10[C_6H_5COO^-]+[C_6H_5COO^-]=x
[C_6H_5COO^-]=\dfrac{x}{11}
K_{sp}[AG^+][C_6H_5COO^-]
2.5\times10^{-13}=x\left(\dfrac{x}{11}\right)
x=1.66\times10^{-6}\ mol/L
Thus, the solubility of silver benzoate in a pH 3.19 solution is
1.66\times10^{-6}\ mol/L
Now, let the solubility of
C_6H_5COOAg
be
x'\ M
Then,
[Ag^+]=x' \ M
and
[C_6H_5COO^-]=x'\ M
K_{sp}=[Ag^+][C_6H_5COO^-]
K_{sp}=(x')^2
x'=\sqrt{K_{sp}}=\sqrt{2.5\times10^{-13}}=5\times10^{-7}\ mol/L
\therefore \dfrac{x}{x'}=\dfrac{1.66\times10^{-6}}{5\times10^{-7}}=3.32
Hence,
C_6H_5COOAg
is approximately
3.32
times more soluble in low pH solution.
A sodium salt on treatment with
{MgCl}_{2}
gives white precipitate only on heating. The anion of the sodium salt is:
Report Question
0%
{ HCO }_{ 3 }^{ - }
0%
{ CO }_{ 3 }^{ 2- }
0%
{ NO }_{ 3 }^{ - }
0%
{ SO}_{ 4 }^{ 2- }
Explanation
The salt will be sodium bicarbonate and the anion is
HC{ O }_{ 3 }^{ \ominus }
When sodium bicarbonate react with
Mg{ Cl }_{ 2 }
it produce Magnesium bicarbonate which is also soluble in water
But, on heating
Mg{ \left( HC{ O }_{ 3 } \right) }_{ 2 }
it produce
Mg\left( C{ O }_{ 3 } \right)
which gives white precipitate
Mg{ \left( HC{ O }_{ 3 } \right) }_{ 2 }\rightarrow Mg\left( C{ O }_{ 3 } \right)+C{ O }_{ 2 }+{ H }_{ 2 }O
If the ionic product of water varies with temperature as follows and the density of water is nearly constant for this range of temperature.
The given equilibrium process is:
{ H }^{ + }+OH\rightleftharpoons { H }_{ 2 }O
Report Question
0%
Exothermic
0%
Endothermic
0%
Cant say
0%
Ionization
The
K_{sp}
of
Ag_{2}CrO_{4}, AgCl, AgBr
and
AgI
are respectively,
1.1\times 10^{-12}
,
1.8\times 10^{-10}
,
5.0\times 10^{-13}
and
8.3\times 10^{-17}
. Which of the following salts will precipitate last if
AgNO_{3}
solution is added to the solution containing equal moles of
NaCl, NaBr, NaI
and
Na_{2}CrO_{4}
?
Report Question
0%
Ag_{2}CrO_{4}
0%
AgI
0%
AgCl
0%
AgBr
Explanation
1.
Ag_2CrO_4\rightleftharpoons 2Ag^++{CrO_4}^{2-}
Ksp=(2s)^2\times s=4s^2
Ksp=(1.1\times 10^{-12})
S=3\sqrt {\cfrac {Ksp}{4}}=6.5\times 10^{-5}
M
2.
AgCl\rightleftharpoons Ag^++Cl^-
Ksp=S\times S
;
Ksp=1.8\times 10^{-10}
S=\sqrt {Ksp}=1.34\times 10^{-5}
M
3.
AgBr\rightleftharpoons Ag^++Br^-
Ksp=S\times S
;
Ksp=5\times 10^{-13}
S=\sqrt {Ksp}=0.71 \times 10^{-6}
M
4.
AgI\rightleftharpoons Ag^++I^-
Ksp=S\times S
;
Ksp=8.3\times 10^{-17}
S=\sqrt{Ksp}=0.9\times 10^{-8}
M
\therefore
Solubility of
Ag_2CrO_4
is highest, so it will precipitate last.
Which of the following statements is true?
Report Question
0%
{H}_{3}{PO}_{3}
is stronger acid than
{H}_{2}{SO}_{3}
0%
In aqueous medium
HF
is a stronger acid than
HCl
0%
HCl{O}_{4}
is a weaker acid than
HCl{O}_{3}
0%
H{NO}_{3}
is a stronger acid than
H{NO}_{2}
If
S_0, S_1, S_2
and
S_3
are the solubilities in water
of
AgCl
,
0.01 \,M \,CaCl_2, 0.01 \,M \,NaCl
and
0.5 \,M \,AgNO_3
solutions, respectively, then which of the following is true?
Report Question
0%
S_0 > S_2 > S_1 > S_3
0%
S_0 = S_2 = S_1 > S_3
0%
S_3 > S_1 > S_2 > S_0
0%
none of these
Explanation
The solubility of
AgCl
or its ion formation will depend inversely on the concentration
S_0=H_2O
S_1=0.01\ M\ CaCl_2
S_2=0.01\ M\ NaCl
S_3=0.05\ M\ AgNO_3
H_2O
is dilute solution and least concentrated hence have maximum solubility of
AgCl
in it. Out of
NaCl
and
CaCl_2
, the solubility of
NaCl
is high due to loess number of ions produced from
NaCl
as compare to
CaCl_2
. More the ion produced lesser is the solubility of coming salt.
So, the Correct order is
S_0 > S_2 > S_1 > S_3
We know that concentration of common ion
\alpha \dfrac{1}{solubility}
. The order of solubility of
AgCl : S_1 > S_3 > S_2 > S_4
As_2S_3
solution has negative charge, capacity to precipitate is highest in:
Report Question
0%
AlCl_3
0%
Na_3PO_4
0%
CaCl_2
0%
K_2SO_4
Explanation
Solution:- (A)
Al{Cl}_{3}
According to Hardy-Schulze rule, more is the valence of effective ion, greater is its coagulating power.
Hence
{As}_{2}{S}_{3}
precipitate the most in
Al{Cl}_{3}
.
The number of
{ H }_{ 3 }{ O }^{ + }
ions present in 10 ml of water at
{ 25 }^{ 0 }C
is
Report Question
0%
6.023\times { 10 }^{ -14 }
0%
6.023\times { 10 }^{ 14 }
0%
6.023\times { 10 }^{ -19 }
0%
6.023\times { 10 }^{ 19 }
The amount
{ CH }_{ 3 }{ NH }_{ 2 }
dissolved in
2\ L
of water that it produces concentration
{ OH }^{ - }
equal to
5\times { 10 }^{ -4 }\ M
, is
[Given
{ K }_{ b }
of
{ CH }_{ 3 }{ NH }_{ 2 }=2\times { 10 }^{ -6 }
]
Report Question
0%
5.6\ gm
0%
3.88\ gm
0%
7.75\ gm
0%
8.3\ gm
On adding ammonia to water,
Report Question
0%
ionic product will increase
0%
ionic product will decrease
0%
[H_{3}O^{+}]
will increase
0%
[H_{3}O^{+}]
will decrease
With increase in temperature, ionic production of water
Report Question
0%
decreases
0%
increases
0%
remains same
0%
may increases or decreases
Which buffer solution has maximum
pH?
Report Question
0%
mixture which is
0.1
M in
\mathrm { CH } _ { 3 } \mathrm { COOH }
and
0.1
M in
\mathrm { CH } _ { 3 } \mathrm { COONa } \left[ \mathrm { pK } _ { \mathrm { a } } \left( \mathrm { CH } _ { 3 } \mathrm { COOH } \right) = 4.74 \right]
0%
mixture which is
0.2
M
\mathrm { CH } _ { 3 } \mathrm { COOH }
and
0.2
M in
\mathrm { CH } _ { 3 } \mathrm { COONa }
0%
mixture which is
0.1
M in
\mathrm { NH } _ { 4 } \mathrm { Cl }
and
0.1
M in
\mathrm { NH } _ { 4 } \mathrm { OH } \left[ \mathrm { pK } _ { \mathrm { a } } \left( \mathrm { NH } _ { 4 } ^ { + } \right) = 9.26 \right]
0%
all the solution have equal
pH
which is
4.74
At certain temperature
{ K }_{ w }
for water
4\times { 10 }^{ -14 }
. Which of the following is incorrect for impure water at given temperature?
[Given:
log\ 2 = 0.3
]
Report Question
0%
pH=6.7 and water is acidic.
0%
pH=6.7 and water is neutral.
0%
pOH=6.7 and water is neutral.
0%
pH + pOH = 13.4
How many grams of
NaOH
is to be added to one litre of
1 M H_2CO_3
to get a
HCO^{-}_3/ CO^{-2}_3
buffer of maximum capacity ?
Report Question
0%
90 gm
0%
60 gm
0%
20 gm
0%
50 gm
Solubility of silver cyanide is maximum in
Report Question
0%
Acidic buffer solution
0%
Basic buffer solution
0%
In pure water
0%
Equal in all
A sample of water containing some dissolved table sugar and common salt is passed through organic ion exchange resins. The resulting water will be
Report Question
0%
Sweet
0%
Salty
0%
Tasteless
0%
None of these
At
100^\circ C
, value of
K_{w}
is
Report Question
0%
1.0\times 10^{-14}\quad m^{2}
0%
less than
1.0\times 10^{-14}\quad m^{2}
0%
greater than
1.0\times 10^{-14}\quad m^{2}
0%
Zero
Explanation
At higher temperature the value of
kw
increases.This is in according with le-chatelier principle.
At
100^o kw=51.3\times 106{-14}
C is the correct answer.
Calculate the degree of ionization of 0.04M HOCl solution having ionization constant
1.25\times { 10 }^{ -4 }
?
Report Question
0%
0.025
0%
0.25
0%
0.5
0%
0.055
When a solution of silver nitrate is added to pure carbon tetrachloride :
Report Question
0%
A light blue precipitate soluble in ammonia is obtained
0%
A curdy precipitate insoluble in ammonia is obtained
0%
A Pale yellow precipitate in ammonia is formed
0%
No precipitate is formed
Which of the following compound has the maximum degree of ionization?
Report Question
0%
1 M
NH_3
0%
0.001 M
NH_3
0%
0.1 M
NH_3
0%
0.0001 M
NH_3
In which of the following , the solubility of
AgCl
will be minimum ?
Report Question
0%
0.01 \ M \ Na_2{ SO_4}
0%
Pure water
0%
0.01 \ M \ CaCl_2
0%
0.01 \ M\ NaCl
Which shows weak ionisation in water?
Report Question
0%
H_2SO_4
0%
NaCl
0%
HNO_3
0%
NH_3
Explanation
Because it is a weak electrolyte. Since the weak electrolytes have fewer ions in the solution.
Option D is correct.
The percentage of pyridine
\left( {{C_5}{H_5}N} \right)
that forms pyridinum ion
\left( {{C_5}{H_5}{N^ + }H} \right)
in a
0.10M
aqueous pyridine solution
\left( Given - {{K_b}, for \ {C_5}{H_5}N = 1.7 \times {{10}^{ - 9}}} \right)
is?
Report Question
0%
0.0060\%
0%
0.013\%
0%
0.77\%
0%
1.6\%
At any temperature, the proton concentration of water is
Report Question
0%
{10}^{-7}M
0%
<
{10}^{-7}M
0%
>
{10}^{-7}M
0%
\sqrt { { K }_{ w } }
Amongst the following hydroxides,the one which has the lowest value of
{ K }_{ SP }
is?
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0%
{ Mg(OH) }_{ 2 }\quad \quad \quad
0%
Ca(OH)_{ 2 }
0%
Ba(OH)_{ 2 }
0%
Be(OH)_{ 2 }
Degree of dissociation of 0.1 M HCN solution is 0.01%. Its ionisation constant would be
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0%
10^{-3}
0%
10^{-5}
0%
10^{-7}
0%
10^{-9}
If the ionic product of water is
1.96 \times 10 ^ { - 14 }
at
35 ^ { \circ } \mathrm { C } .
What is its value at
10 ^ { \circ } \mathrm { C }
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0%
2.95 \times 10 ^ { - 14 }
0%
1.96 \times 10 ^ { - 7 }
0%
2.95 \times 10 ^ { - 15 }
0%
3.9 \times 10 ^ { - 12 }
Explanation
Given:
Ionic product of water =
1.96 \times 10^{-14}
Initial temperature=
35^{0}C
Solution:
Ionic product follows a direct relationship with the temperature, i.e., the ionic product is directly proportional to temperature. As the temperature decreases, the ionic product also decreases.
According to question, as the temperature is decreasing
(10^{0}C)
so, ionic product of water also decreases.
From the given options it can be observed that
2.95 \times 10^{-15}
is the only value less than the present value.
Hence, the correct option is (C).
Which of the following solution mixture will show resistance to the small addition of acids?
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0%
500ml\ of\ 0.1N\ CH_{3}COOH+500ml\ of\ 0.1N\ NaOH
0%
500ml\ of\ 0.1N\ CH_{3}COOH+500ml\ of\ 0.1N\ HCl
0%
500ml\ of\ 0.1N\ CH_{3}COOH+500ml\ of\ 0.2N\ NaOH
0%
500ml\ of\ 0.2N\ CH_{3}COOH+500ml\ of\ 0.1N\ NaOH
Which of the following salt solution will act as a buffer?
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0%
C{H_3}COON{H_4}\left( {aq.} \right)
0%
N{H_4}Cl\left( {aq.} \right)
0%
C{H_3}COONa\left( {aq.} \right)
0%
NaCl\left( {aq.} \right)
\frac { N } { 10 }
acetic acid was titrated with
\frac { N } { 10 }
NaOH.When
25 \% , 50 \%
and
75
\%
of titration is over then the pH of the solution will be
: \left[ \mathrm { K } _ { a } = 10 ^ { - 5 } \right]
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0%
5 + \log 1 / 3,5,5 + \log 3
0%
5 + \log 3,4,5 + \log 1 / 3
0%
5 - \log 1 / 3,5,5 - \log 3
0%
5 - \log 1 / 3,4,5 + \log 1 / 3
Explanation
t CH_3COOH NaOH CH_3COO^-Na^+
0 0.1 0.1
25% 0.1-0.025 0.1-0.025 0.025
50% 0.1-0.050 0.1-0.050 0.050
75% 0.1-0.075 0.1-0.075 0.075
pH=- \log K_a+\log \dfrac{[salt]}{[acid]}
t= 25%
pH=5+\log \dfrac{0.025} {0.075}
pH=5+\log \dfrac{1} {3}
t= 50%
pH=5+\log \dfrac{0.050} {0.050}
pH=5+\log 1=5
t= 75%
pH=5+\log \dfrac{0.075} {0.025}
pH=5+\log 3
In decimolar solution, CH
_{3}
COOH is ionised to the extent of 1.3 %. If
log 1.3 = 0.11
, what is the pH value of the solution?
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0%
3.89
0%
4.89
0%
2.89
0%
Unpredictable
On addition of sodium acetate, the ionization of acetic acid:
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0%
cannot be predicted
0%
decreases
0%
increases
0%
remains unaffected
The concentration of hydronium
(H_{3}O^+)
ion in water is
Report Question
0%
Zero
0%
1\times 10^{-14}gm\ ion/litre
0%
1\times 10^{7}gm\ ion/litre
0%
1\times 10^{-7}gm\ ion/litre
Explanation
as,
H_{3}O^{+}\rightleftharpoons OH^{-}+H_{2}
and,
pOH+pH=14
\Rightarrow [H^{+}]+[OH^{-}]=10^{-14}
\Rightarrow 10^{-7}+10^{-7}=10^{-14}
\Rightarrow [H_3O^+]=[OH^{-}]=10^{-7}\ gm\ ion/litre
.
Which may be added to one litre water to act as a buffer.
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0%
1 mole of
HC_{2}H_{3}O_{2}
and 1 mole of Hcl
0%
1 mole of
NH_{4}OH
and 0.5 mole of
NaOH
0%
1 mole of
NH_{4}Cl
and 1 mole of Hcl
0%
1 mole of
HC_{2}H_{3}O_{2}
and 0.5 mole of
NaOH
At certain temperature, the
{ H }^{ + }
ion concentration of water is
4\times { 10 }^{ -7 }
M then the value of
{ K }_{ w }
at the same temperature is
Report Question
0%
{ 10 }^{ -14 }{ M }^{ 2 }
0%
4\times { 10 }^{ -14 }{ M }^{ 2 }
0%
1.6\times { 10 }^{ -13 }{ M }^{ 2 }
0%
4\times { 10 }^{ -7 }{ M }^{ 2 }
The mass in grams of potassium permanganate (MM = 158) crystals required to oxidize
750\,\,c{m^3}
of 0.1 M Mohr's salt solution in acidic medium is about?
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0%
11.8
0%
5.8
0%
6.0
0%
2.4
Degree of hydrolysis for a salt of weak acid and strong base can be changed by
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0%
Increasing concentration of salt
0%
Increasing temperature
0%
Adding base
0%
All of these
Which of the following mixture can form buffer solution ?
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0%
C{H_3}COOH + C{H_3}COONa
0%
NaCl + NaOH
0%
HCl + N{H_4}Cl
0%
C{H_3}COOH + HCl
If pH of a saturated solution of
Mg(OH)_2
is 12.
Find its
K_{sp}
Report Question
0%
2\times 10^{-4}
0%
5\times 10^{-5}
0%
4\times 10^{-6}
0%
5\times 10^{-7}
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Practice Class 11 Medical Chemistry Quiz Questions and Answers
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