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Standard XII
Chemistry
Salt of Strong Acid and Weak Base
Question
1.2
mol of
C
H
3
N
H
2
(
p
K
b
=
3.3
)
is added to
0.8
moles of
H
C
l
and the solution is diluted to one litre, resulting
p
H
of solution is:
10.7
10.4
3.6
11.3
A
3.6
B
10.4
C
10.7
D
11.3
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Solution
Verified by Toppr
As we know,
p
O
H
=
p
K
b
+
l
o
g
s
a
l
t
b
a
s
e
p
O
H
=
3.3
+
l
o
g
0.8
0.4
=
3.6
p
H
=
14
−
p
O
H
=
10.4
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Similar Questions
Q1
1.2
mol of
C
H
3
N
H
2
(
p
K
b
=
3.3
)
is added to
0.8
moles of
H
C
l
and the solution is diluted to one litre, resulting
p
H
of solution is:
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Q2
C
H
3
N
H
2
(0.1 mole
K
b
= 5
×
10
−
4
)
is added to 0.08 mole of HCl and the solution is diluted to one litre, resulting hydrogen ion concentration is
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Q3
C
H
3
N
H
2
(
0.1
mole,
K
b
=
5
×
10
−
4
)
is added to
0.08
moles of
H
C
l
and the solution is diluted to one litre. The resulting hydrogen ion concentration is :
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Q4
A solution of
H
C
l
has a
p
H
=
5
. If
1
m
L
of it is diluted to
1
litre,
p
H
of resulting solution will be:
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Q5
C
H
3
N
H
2
(0.1 mole
K
b
= 5
×
10
−
4
)
is added to 0.08 mole of HCl and the solution is diluted to one litre, resulting hydrogen ion concentration is
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