The solubility product of `AI(OH)_(3)` is `2.7 xx 10^(-11)`. Calculate its solubility in `mgL^(-1)`. (Atomic mass of `AI = 27 u)`.

Explanation

Let the solubility of `Al(OH)_(3)` in water =S.
`Al(OH)_(3)overset((aq))hArrAl^(3+)(aq)+ +3OH^(-)(aq)`
Concentration of species at time (t) `= 0" " 1" "0" "0`
Concentration of species at equilibrium point. `(1-S)" " S" "3S"`
`K_(sp)=[Al^(3+)][OH^(-)]^(3)=[S][3S]^(3)=27S^(4)`
`S^(4)=(K_(sp))/(27)=(2.7xx10^(-11))/(27)=1xx10^(-12)`
`S^(4)=(1xx10^(-12))1//4=1xx10^(-3)mol L^(-1)`
Solubility of `Al(OH)_(3) " in gram "//" litres " =(1xx10^(-3)mol L^(-1))xx[Molar Mass of Al (OH)_(3)]`
`=(1xx10^(-3) mol L^(-1))xx(78g mol^(-1))`
`78xx10^(-3)g L^(-1)=7.8xx10^(-2)gL^(-1)`
Concentration of `(OH^(-)" in solution "=3S=3xx(1xx10^(-3)" mol " L^(-1))`
`P_(OH)=-log [OH^(-)]`
`=-log [3xx10^(-3)]=[3-log3]`
`=(3-0.4771)=2.5229`
`ph = 14 -P_(OH)= 14-2.5229=11.4771`