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pH of saturated solution of Ba(OH)(2) is...

`pH` of saturated solution of `Ba(OH)_(2)` is `12`. The value of solubility product `(K_(sp))` of `Ba(OH)_(2)` is

A

`5.0xx10^(-7)M^(3)`

B

`0.6xx10^(-12)M^(3)`

C

`4.0xx10^(-8)M^(3)`

D

`5.0xx10^(-9)M^(3)`

Text Solution

Verified by Experts

The correct Answer is:
A
Solubility expressed in terms of molarity `[OH^(-)]=10^(-2)`
`[Ba^(2+)]=0.5xx10^(-2)`
`K_(SP)=(0.5 x10^(-2))(10^(-2))^(2)`
`K_(SP)=5x10^(-7)`
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