If `K_(sp)`of `M(OH)_(3)` is `1 xx 10^(-12)` then `0.001 M.M^(2+)` is precipitate in a `pH gt 9`

If `K_(sp)`of `M(OH)_(3)` is `1 xx 10^(-12)` then `0.001 M.M^(2+)` is

1.	If `K_(sp)`of `M(OH)_(3)` is `1 xx 10^(-12)` then `0.001 M.M^(2+)` is precipitate in a `pH gt 9`
Answer» Correct Answer - T `M(OH)_(3) `is precipitate `[M^(3+)][OH^(Theta)]^(3) gt K_(sp) [OH^(Theta)]^(3)gt 1 xx 10^(-9)` `[OH^(Theta)] gt 1 xx 10^(-3)` Maximum `pOH = 3` Minimum `pH = 9` Hence true.

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If `K_(sp)`of `M(OH)_(3)` is `1 xx 10^(-12)` then `0.001 M.M^(2+)` is precipitate in a `pH gt 9`

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