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The solubility of Mg (OH)(2) in pure wat...

The solubility of `Mg (OH)_(2)` in pure water is `9.57 xx 10^(-3) gL^(-1)`. Calculate its solubility (in `gL^(-1))` in `0.02M Mg(NO_(3))_(2)` solution.

A

`1.5xx10^(-4)`

B

`8.69xx10^(-4)`

C

`0.5xx10^(-3)`

D

`0.5xx10^(-5)`

Text Solution

Verified by Experts

The correct Answer is:
B
solubility of `Mg(OH)_(2)`
`=(9.57xx10^(-3))/(58)=1.65x10^*-4) "mol lits"^(-1)`
`K_(SP)=4s^(3)=4x(1.65x10^(-4))=17.96x10^(-12)`
but in presence of `Mg(NO_(3))_(2)`
`K_(s)=(x+c)(2s)^(2)`.
`K_(s)=17.96x10^(-12)=(s+0.02)(2s)^(2)`
`s=14.98x10^(-6)"molts"^(-1)`
`s=14.98x10^(-6) gr "lit"^(-1)=8.69x10^(-4) "gr lits"^(-1)`
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