Calculate the solubility of AgCl(s) in pure water. Solubility product of AgCl is given as
2.8 X 10^(-10).
Answers
Answered by
0
Answer:
AgCl⇌Ag
+
+Cl
−
a 0 0
a-S S S+0.1
k
sp
=1.6×10
−10
=[Ag
+
][Cl
−
]
=S(0.1+S)
k
sp
is value seems to be very small
S value can be ignored, with respect to 0.1 M
1.6×10
−10
=S×0.1
S=1.6×10
−9
M
Hence option D is correct.
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