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The solubility product of a rare earth m...

The solubility product of a rare earth metal hydroxide `M(OH)_(3)` at room temperature is `4.32 xx 10^(-14)`, its Solubililty is

A

`1.25xx10^(-10)M`

B

`2.0xx10^(-6)M`

C

`2.0xx10^(-4)M`

D

`1.25xx10^(-7)M`

Text Solution

Verified by Experts

The correct Answer is:
C
For `AB_(3)` type salt
`K_(sp)=27S^(4)`
`S=4 sqrt((K_(SP))/(27))`
`(16xx10^(-16))^(1//4)=2x10^(-4)`
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