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The solubility product of a rare earth ...

The solubility product of a rare earth metal hydroxide `M(OH)_3` at room temperature is ` 4.32 xx 10^(-14) ` its solubility is

A

`2xx 10^(-3) M`

B

` 2.0 xx10^(-4) M`

C

` 2xx10^(-5)M `

D

` 2.0 xx 10^(-6) M`

Text Solution

Verified by Experts

The correct Answer is:
B
` M( OH)_3 hArr underset(S)M^(+3) = 3 underset( 3S) O H^(-) `
`KsP = (s) (3s) ^(3) = 274 ^(4) =4.32 xx 10 ^(-14) , S =2 xx 10 ^(-4) `
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