1.

The solubility product of a sparingly soluble metal hydroxide, M(OH)_(2) at 298 K is 5xx10^(-16)mol^(3)dm^(-9). The ph value of its aqueous and saturated solution is :

Answer»

5
9
`11.5`
`2.5`

SOLUTION :`M(OH)_(2)hArr underset(s)(M^(2+))+underset(2s)(2OH^(-))`
`K_(sp)=(s)(2s)^(2)=4s^(3)=5XX10^(-16)`
`s^(3)=1.25xx10^(-16)=125xx10^(-18)`
`s=(125xx10^(-18))^((1)/(3))=5xx10^(-6)`
`[OH^(-)]=2s=2xx5xx10^(-6)=10^(-5)`
`pOH=-log[OH^(-)]=-LOG10^(-5)=5`
`pH=14-pOH=14-5=9`.


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