Question

If the solubility of `Pbbr_(2)` is S g-mole per litre, its solubility product, considering it to be 80% ionized, is
A. `2.048 S^(2)`
B. `20.48 S^(3)`
C. `2.048 S^(3)`
D. `2.048 S^(4)`

Answer 1

Correct Answer - C
`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)`