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).

The solubility product of Al (OH)_3 is 2.7 xx 10^(-11). Calculate the

1.	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).
Answer» SOLUTION :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 , Solubilityof `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