The molar solubility of `Al(OH)_(3)` in 0.2 M NaOH solution is `x xx 10^(-22)"mol/L"`. Given that, solubility product of `Al(OH)_(3)=2.4xx10^(-24)`. What is numerical value of x?

Options

A. `12xx10^(-23)`
B. `12xx10^(-21)`
C. `3xx10^(-19)`
D. `3xx10^(-22)`

Answer

D

Explanation

`Al(OH)_3 hArr Al^(3+) + 3OH^(-)`
If s is the solubility
`[Al^(3+)]=s, [OH^-]` =3s + 0.2 `approx` 0.2
`K_(sp)=[Al^(3+)][OH^-]^3`
`2.4xx10^(-24)=(s)(0.2)^3`
`s=(2.4xx10^(-24))/(0.2)^3=3xx10^(-22)` M