At 25^(@)C, the solubility product of Mg(OH)_(2) is 1.0xx10^(-11). At which pH will Mg^(2+) ions start precipitating in the form of Mg(OH)_(2) from a solution of 0.0001 M Mg^(2+) ions ?

At 25^(@)C, the solubility product of Mg(OH)_(2) is 1.0xx10^(-11). At

1.	At 25^(@)C, the solubility product of Mg(OH)_(2) is 1.0xx10^(-11). At which pH will Mg^(2+) ions start precipitating in the form of Mg(OH)_(2) from a solution of 0.0001 M Mg^(2+) ions ?
Answer» 11 8 9 10 Solution :`K_(SP) `for `Mg(OH)_(2)=[Mg^(2+)][OH^(-)]^(2)` `:. 1.0xx10^(-11)=(0.001)[OH^(-)]^(2)` or `[OH^(-)]^(2)=10^(8) or [OH^(-)]=10^(-4)` i.e, `pOH=4 or pH = 10`