1.

If the solubility product of lead iodide is 3.2 times 10^-8, its solubility will be……..

Answer»

`2 times 10^-3 M`
`4 times 10^-4 M`
`1.6 times 10^-5 M`
`1.8 times 10^-5 M`

Solution :`Pbl_s(s) leftrightarrow Pb^(2+) (AQ)+21^(-) (aq)`
`K_(SP)=(s) (2s)^2`
`3. 2 times 10^-8=4s^3`
`s=((3.2 times 10^-8)/4)^(1//3)=(8 times 10^-9)^(1//3)=2 times 10^-3 M`


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