`K_("sp")` of `PbBr_(2)` (Molar mass `=367`) is `3.2xx10^(-5)`. If the salt is `80%` dissociated in solution, caculate the solubility of salt in `g` per `litre`.

Answer

`00009.175`

Explanation

`00009.175`
Let solubility of `PbBr_(2)` is `S` mol `"litre"^(-1)`
`PbBr_(2)(s)+"Aq" hArr PbBr_(2)("aq")hArr underset ((Sxx80)/100)(Pb^(2))+underset((2Sxx80)/100)(2Br^(-)`
Since, `PbBr_(2)` ionises to `80%` only.
Now `K_("sp")=[Pb^(2+)][Br^(-)]^(2)`
or `3.2xx10^(-5)=[(Sxx80)/100][(2Sxx80)/100]^(2)`
`S=0.025 "mol litre"^(-1)`
`=0.025xx367g "litre"^(-1)`
`=19.175g "litre"^(-1)`