The solubility product of Ag2CrO4 at 298K is 4 x 10^-12 M^3. Find out its solubility at this temperature
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1
Answer:
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Answered by
9
Answer:
ANSWER
E
cell
=
1
0.059
log
[Ag
+
]
LHS
[Ag
+
]
RHS
0.164=
1
0.059
log
[Ag
+
]
LHS
0.1
or [Ag
+
]
LHS
=1.66×10
−4
M
So, [CrO
4
2−
]=
2
1.66×10
−4
K
sp(Ag
2
CrO
4
)
=[Ag
+
]
2
[CrO
4
2−
]
=(1.66×10
−4
)
2
(
2
1.66×10
−4
)
=2.287×10
−12
mol
3
L
−3
.
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