1.

At 25^(@)C, the solubility product of Mg(OH)_(2) is 1.0 xx 10^(-11). At which pH, will Mg^(2+) ions start precipitating in the form of Mg(OH)_(2) from a solution of 0.001 M Mg^(2+) ions

Answer»

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11

Solution :`K_(SP) = [Mg^(+2)][OH^(-)]^(2)`
`1 xx 10^(-11) = 10^(-3)xx [OH^(-)]^(2)`
`[OH^(-)]^(2) =10^(-8)`
`OH^(-) = 10^(-4)`
`POH = 4[pH + pOH + pOH = 14]`
pH = 10.


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