The solubility of AgCl is 0.005g/L. Calculate it's solubility product in mol/L.
(molecular weight of AgCl =143.5)
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Answer:
Correct option is C)
BaCl
⇌
Ba
2+
+2Cl
−1
K
sp
=[Ba
2+
][Cl
−
]
2
=x×(2x)
2
=4x
3
4x
3
=3.2×10
−9
⇒x=9.28×10
−4
=0.928×10
−3
=1×10
−3
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