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The molar solubility of Al(OH)(3) in 0.2...

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?

A

`3xx10^(-22)`

B

`12xx10^(-22)`

C

`12xx10^(-21)`

D

`3xx10^(-19)`

Text Solution

Verified by Experts

The correct Answer is:
A
`K_(sp)=[s][0.2]^(3)`
`[s]=24xx10^(-24)=3xx10^(-22)//[0.2]^(3)`
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