1.

If the solubility of Pbbr_(2) is S g-mole per litre, its solubility product, considering it to be 80% ionized, is

Answer»

`2.048 S^(2)`
`20.48 S^(3)`
`2.048 S^(3)`
`2.048 S^(4)`

Solution :`PbBr_(2) hArr Pb^(2+) + 2Br^(-)`
`S(S xx 80)/(100) (2S xx 80)/(100)`
[`because PbBr_(2)` ionises to 80%]
`K_(SP) = [(S xx 80)/(100)][(2S xx 80)/(100)]^(2)`
`= 0.8 S xx 2.56 S^(2) = 2.048 S^(3)`


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