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The K(sp) of M(OH)(2)" is "5 times 10^(-...

The `K_(sp)` of `M(OH)_(2)" is "5 times 10^(-10)M^(3)`. The molar solubility of `M(OH)_(2)` in a 0.1 M NaOH solution is

A

`5xx10^(-12)M`

B

`5xx10^(-8)M`

C

`5xx10^(-10)M`

D

`5xx10^(-9)M`

Text Solution

Verified by Experts

The correct Answer is:
C
Ionic product of `M(OH)_(2)=5xx10^(-10)`
or `[M^(2+)][OH^(-)]^(2)=5xx10^(-10)`
Given `[OH^(-)]=10^(-1)"mol/L"`
So, `[M^(2+)]=(5xx10^(-10))/([10^(-1)]^(2))`
`=(5xx10^(-10))/(10^(-2))=5xx10^(-8)M`
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