Calcualte the solubility of `M_(2)X_(3)` in pure water, assuming that neither kind of ion reacts with `H_(2)O`. The solubility product of `M_(2)X_(3), K_(sp) = 1.1 xx 10^(-23)`.

Answer

`1.0xx10^(-5) "mol" L^(-1)`

Explanation

`{:(A_(2)X_(3),hArr,2A^(3+),+,3X^(2-),,,),(s,,s,,s,,,):}`
`:. K_(sp)=(2s)^(2)(3s)^(3) = 108 s^(5)`
`s=(K_(ap)/108)^(1//5)=(1.1xx10^(-23)//108)^(1//5)`
`log s = (1)/(5) (log 1.1 + log 10^(-23)-log 108 ) = (1)/(5) (0.0414-23-2.0334)`
or `log s = - 4.9984 = bar(5) .0016 :. s= 10^(-5) "mol" L^(-1)`