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Byju's Answer
Standard VI
Chemistry
Saturated Solution
If 6.539× 1...
Question
If
6.539
×
10
−
2
g of metallic zinc is added to
100
ml saturated solution of AgCl. Find the value of
l
o
g
10
[
Z
n
2
+
]
[
A
g
+
]
2
.
Open in App
Solution
According to the problem:-
2
A
g
+
+
2
e
→
2
A
g
;
(
E
∘
=
0.80
v
)
S
i
n
c
e
,
(
65.39
×
10
−
2
65.39
=
10
−
3
m
o
l
e
s
o
f
z
n
h
a
s
b
e
e
n
a
d
d
e
d
)
Z
n
→
Z
n
2
+
+
2
e
;
(
E
∘
=
0.80
v
)
Hence,
2
A
g
+
(
a
q
)
+
Z
n
(
s
)
→
Z
n
2
+
(
e
q
)
+
2
A
(
s
)
;
E
∘
=
1.57
v
log
10
K
(
e
q
)
=
52.8
Therefore, if the reaction will move in the forward direction completely. Hence moles of
A
g
formed will be
10
−
6
(At Equilibrium
E
c
e
11
=
0
)
E
∘
c
e
11
=
+
0.0591
2
log
10
[
Z
n
+
2
]
[
A
g
+
]
2
Hence,
1.56
×
2
0.0591
=
log
[
Z
n
+
2
]
[
A
g
+
2
]
2
=
52.8
Hence the answer is
52.8
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0
Similar questions
Q.
(I) For the reaction,
A
g
+
(
a
q
)
+
C
l
−
(
a
q
)
→
A
g
C
l
(
s
)
Write the cell representation of above reaction and calculate
E
o
at 298 K.
Given:
Species
Δ
G
f
(
k
J
/
m
o
l
)
A
g
+
(
a
q
)
+77
C
l
−
(
a
q
)
-129
A
g
C
l
(
s
)
-109
(II) Calculate the
log
10
K
s
p
(
A
g
C
l
)
at 298 K.
(III) If
6.539
×
10
−
2
g
of metallic zinc is added to 100 ml saturated solution of
A
g
C
l
, find the value of
log
10
[
Z
n
2
+
]
[
A
g
+
]
2
.
(IV)How many moles of Ag will be precipitated in the above reaction?
Given that-
A
g
+
+
e
−
→
Ag;
E
o
=
0.80
V
Z
n
2
+
+
2
e
−
→
Zn;
E
o
=
−
0.76
V
(It was given that Atomic mass of Zn
=
65.39
)
Q.
100 mL of a clear saturated solution of
A
g
2
S
O
4
is added to 250 mL of a clear saturated solution of
P
b
C
r
O
4
. What is the precipitate formed? Given,
K
s
p
values for
A
g
2
S
O
4
,
A
g
2
C
r
O
4
,
P
b
C
r
O
4
and
P
b
S
O
4
are
1.4
×
10
−
5
,
2.4
×
10
−
12
,
2.8
×
10
−
13
and
1.6
×
10
−
5
respectively.
Q.
In
1
L
saturated solution of
A
g
C
l
[
K
s
p
(
A
g
C
l
)
=
1.6
×
10
−
10
]
,
0.1
m
o
l
of
C
u
C
l
[
K
s
p
(
C
u
C
l
)
=
1
×
10
−
6
]
is added. The resultant concentration of
A
g
+
in the solution is
1.6
×
10
−
x
. The value of
x
is:
Q.
In
1
L
saturated solution of
A
g
C
l
(
K
s
p
(
A
g
C
l
)
=
1.6
×
10
−
10
]
,
0.1
mol of
C
u
C
l
[
K
s
p
(
C
u
C
l
)
=
1.0
×
10
−
6
]
is added. The resultant concentration of
A
g
+
in the solution is
1.6
×
10
−
x
. The value of
′
x
′
is:
Q.
In 1 L saturated solution of
A
g
C
l
[
K
s
p
(
A
g
C
l
)
=
1.6
×
10
10
]
, 0.1 mol of
C
u
C
l
[
K
s
p
(
C
u
C
l
)
=
1.0
×
10
−
6
]
is added. The resultant concertraion of
A
g
+
in the solution is
1.6
×
10
−
x
. The value of
x
is:
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