0.5 mole of AgNO3 is mixed with 0.5 mole of NaCl in 1 litre of water. A white ppt is formed. The ratio of mole percent of each element in the formed ppt is x: 1. The value of x is ??
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Answer:
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Explanation:
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are The K
sp
values of Ag
2
CrO
4
and AgIO
3
reveals that CrO
4
2−
and IO
3
−
will be precipitated on addition of AgNO
3
as:
[Ag
+
][IO
3
−
]=10
−13
[Ag
+
]
needed
=
(0.005)
10
−13
=2×10
−11
M
[Ag
+
]
2
[CrO
4
2−
]=10
−8
∴ [Ag
+
]
needed
=
0.1
10
−8
=3.16×10
−4
M
Thus AgIO
3
will be precipiated first.
Now in order to precipitate AgIO
3
, one can slow.
0.010.005
AgNO
3
+
0.0050
NaIO
3
⟶
00.005
AgIO
3
+
00.005
NaNO
3
The left mole of AgNO
3
are now used to precipitate Ag
2
CrO
4
0.0050
2AgNO
3
+
0.10.0975
Na
2
CrO
4
⟶
00.0025
AgIO
3
+
00.005
2NaNO
3
Thus [CrO
4
2−
] left in solution =0.0975
Now solution has precipitates of AgIO
3
(s) (0.005mole)+Ag
2
CrO
4
(s)(0.0025mole) and CrO
4
2−
ions (0.0975)
Thus,[Ag
+
]
left
=
[CrO
4
2−
]
K
sp
Ag
2
CrO
4
=
0.0975
10
−8
=3.2×10
−4
M
[IO
3
−
]
left
=
[Ag
+
]
K
sp
AgIO
3
=
3.2×10
−4
10
−13
=3.1×10
−10
M