The solubility product of `Al (OH)_3` is `2.7 xx 10^(-11)`. Calculate the solubility in g `L^(-1)` and also find pH of this solution. (Atomic mass of Al = 27 u).

Explanation

Let S be the solubility of `Al(OH)_3`
`{:(Al(OH)_3 hArr, Al_((aq))^(3+) + , 3OH_((aq))^(-)),(1,0,0),(1-s,s,3s):}`
Concentration to species at t=0
Concentration of various species at equilibrium
`K_(sp)=[Al^(3+)] [OH^-]^3 =(S) (3s)^3 = 27S^4`
`S^4=K_(sp)/27=(2.7+10^(-11))/27=1xx10^(-12)`
`S=1xx10^(-3) "mol L"^(-1)`
(i) Solubility of `Al(OH)_3`
Molar mass of `Al(OH)_3` is 78 g. Therefore ,
Solubility of `Al(OH)_3` in g `L^(-1)`
`=1xx10^(-3) xx 78 g L^(-1) 78xx10^(-3) g L^(-1)`
`=7.8xx10^(-2) g L^(-1)`
(ii) pH of the solution
`S=1xx10^(-3) "mol L"^(-1)`
`[OH^-]=3S=3xx1xx10^(-3)=3xx10^(-3)`
pOH=3-log 3
pH=14-pOH =11+log 3 = 11.4771