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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

`1.25 xx 10^(-10) M`

B

`2.0 xx 10^(-6) M`

C

`2.0 xx 10^(-4) M`

D

`1.25 xx 10^(-7) M`

Text Solution

Verified by Experts

The correct Answer is:
C
For `AB_(3)` type salt
`K_(sp) = 27S^(-4)`
`S = 4sqrt((K_(sp))/(27))`
`(16 xx 10^(-16))^(1//4)`
`2x10^(-4)`
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