1.

Which of the following will have lowest value of K_(sp) at room temperature?

Answer»

`Be(OH)_(2)`
`MG(OH)_(2)`
`Ca(OH)_(2)`
`Ba(OH)_(2)`

Solution :`Be(OH)_(2)` is least soluble in water hence, it will have lowest value of `K_(SP)`.
`Be(OH)_(2)HARR Be^(2+)+2OH^(-)`
`K_(sp)=[Be^(2+)][OH^(-)]^(2)`


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