. The conductivity of saturated solution of AgCl
is found to be 1.86 x 10-6 ohm cm' and that
of water is 6 108 ohm'cm'. If 2° AgCl is
138 ohm cm² eq' the solubility product of
AgCl is
1) 1.3 x 10-5
2) 1.69 X 10-10
3) 2 x 10-10
4) 2.7 x 10-10
Answers
Answered by
1
Answer:
Correct option is
B
1.69×10
−10
Conductivity of distilled water= 6×10
−8
ohm
−1
cm
−1
Conductivity of AgCl solution= 1.86×10
−6
ohm
−1
cm
−1
Thus corrected conductivity of AgCl=K
AgCl
−K
H
2
O
=(1.86×10
−6
−6×10
−8
)ohm
−1
cm
−1
R
AgCl
=1.8×10
−6
ohm
−1
cm
−1
as concentration=
λ
o
conductivity
⇒C=
138ohm
−1
cm
2
eq
−1
1.8×10
−6
ohm
−1
cm
−1
⇒C=1.3×10
−8
cm
−3
eq
For AgCl 1eq= 1 moles
and 1cm
3
=10
−3
dm
3
or 1cm
−3
=10
3
dm
−3
⇒C=1.3×20
−5
mole dm
−3
⇒AgCl⇌[Ag
+
]
aq
+Cl
aq
−
⇒K
sp
=[Ag
+
][Cl
−
]=C
2
⇒K
sp
=(1.3×10
−5
)
2
mol
2
dm
−6
⇒K
sp
=1.69×10
−10
mol
2
dm
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