1.

If the solubility product of lead iodide (Pbl_(2)) is 3.2 xx 10^(-8), then its solubility in moles/litre will be

Answer»

`2 xx 10^(-3)`
`4 xx 10^(-4)`
`1.6 xx 10^(-5)`
`1.8 xx 10^(-5)`

Solution :`K_(SP) = 4S^(3)`
`4S^(3) = 3.2 xx 10^(-8), S = 2 xx 10^(-3) M`.


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