1.

The K_(sp) of Mg(OH)_(2) is 1 xx 10^(-12), 0.01 M Mg(OH)_(2) will precipitate at the limiting pH

Answer»

3
9
5
8

Solution :`Mg(OH)_(2) Mg^(2+) + 2OH^(-)`
`K_(sp) = [Mg^(2+)] [OH^(-)]^(2)`
`1 XX 10^(-12) = 0.01 [OH^(-)]^(2)`
`[OH^(-)]^(2) = 1 xx 10^(-10) rArr [OH^(-)] = 10^(-5)`
`[H^(+) = 10^(-14)//10^(-5) = 10^(-9)`
`PH = -LOG[H^(+)] = -log[10^(-9)] = 9`.


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