The solubility product of `Al(OH)_3` is `2.7xx10^(-11)`. Calculate its solubility is `gL^(-1)` and also find out pH of this solution .

Explanation

Suppose the solubility is S mol `L^(-1)`. Then
`{:(Al(OH)_(3),hArr,Al^(3+),+,3OH^(-),,),(,,S,,3 S,,):}`
`K_(sp)= S xx (3 s)^(3) = 27 S^(4)`
`:. 27S^(4)=2.7xx10^(-11) or S^(4) = 10^(-12) or S=10^(-3) ` mol `L^(-1)`
Molar mass of `Al(OH)_(3) ` in g `L^(-1) = 10^(-3)xx78=7.8xx10^(-2) g L^(-1)`
`[OH^(-)]=3 S = 3 xx 10^(-3) ` mol `L^(-1)`
`:. pOH = - log (3xx10^(-3))=3-0.4771=3.5229`
pH = 14 - 2.5229 = 11.4771.