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If the solubility product of AB2 is 3.20×10−11M3, then the solubility of AB2 in pure water is.......×10−4 mol L−1 [Assuming that neither kind of ion reacts with water]

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answer is 0.0002.

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Detailed Solution

AB2⇌A+2s+2B−2sKsp =s×(2s)2=4s34s3=3.2×10−11s3=8×10−12⇒s=8×10−123=2×10−4M

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