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

Options

A. `2.048 S^(2)`
B. `20.48 S^(3)`
C. `2.048 S^(3)`
D. `2.048 S^(4)`

Answer

C

Explanation

`PbBr_(2) hArrPb^(2+) +2Br^(-)`
`S (S xx 80)/100 (2S xx 80)/100`
`[ :. PbBr_(2) " ionises to " 80 %] `
`K_(sp) = [(Sxx80)/100] [(2S xx 80)/100]^(2)`
` = 0.8 S xx 2.56 S^(2) = 2.048 S^(3)`