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 0.001 M Mg^(2+) ions ?

Answer»

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Solution :`Mg(OH)_(2)`DISSOCIATES as :
`Mg(OH)_(2)HARR Mg^(2+)+2OH^(-)`
`K_(SP)=[Mg^(2+)][OH^(-)]^(2)`
`[OH^(-)]=sqrt((K_(sp))/([Mg^(2+)]))`
`[OH^(-)]=sqrt((1xx10^(-11))/(0.001))=1xx10^(-4)`
`THEREFORE pOH = 4`andpH = 10.


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