1.

Which of the following will have lowest value of `K_(sp)` at room temperature?A. `Be(OH)_(2)`B. `Mg(OH)_(2)`C. `Ca(OH)_(2)`D. `Ba(OH)_(2)`

Answer» Correct Answer - A
`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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