The maximum `pH` of a solution which is having `0.10M` in `Mg^(2+)` and from which `Mg(OH)_(2)` is not precipated is: (Given `K_(sp)Mg)(OH)_(2)=4xx10^(11)M^(3)){log2=0.30}`
A
`10.3`
B
`3.7`
C
`7.5`
D
`9.3`
Text Solution
Verified by Experts
The correct Answer is:
D
The concentration of `OH^(-)` required to precipitate `Mg(OH)_(2)` is `[OH]=((K_(sp))/([Mg^(2+)]))^(1//2)=((4xx10^(-11)M^(3))/(0.10M))^(1//2)` `pOH=log(2xx10^(5))=4.7` and `pH=4.7=9.3`
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