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QJEEM20Ch3773: JEE Main 2020 Chemistry - Class 11 JEEM20Ch3773 Solution: If the solubility product of AB 2 is 3.20 10 –11 M 3 , then the …

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

Step-by-Step Verified Solution

AB2 ⇌ A2+(aq) + 2B-(aq)
s 2s


Ksp = 4s3 = 3.2 × 10–11

⇒\Rightarrow s3 = 8 × 10–12

⇒\Rightarrow s = 2 × 10–4

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