1.

The pH of an aqueous solution of Mg(OH)_(2) is 9.0. If the solubility product of Mg(OH)_(2) is 1xx10^(-11), what is [Mg^(2+)] ?

Answer»

`1XX10^(-5)`
`1.0xx10^(-4)`
`1xx10^(-2)`
`0.1`

SOLUTION :`pH=9, "" therefore [H^(+)]=10^(-9)`
`[OH^(-)]=(1xx10^(-14))/(10^(-9))=10^(-5)`
`Mg(OH)_(2)hArr Mg^(2+)+2OH^(-)`
`K_(sp)=[Mg^(2+)][OH^(-)]^(2)`
`1xx10^(-11)=[Mg^(2+)][10^(-5)]^(2)`
`[Mg^(2+)]=(1xx10^(-11))/((10^(-5))^(2))=10^(-1)=0.1`


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