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 :

Options

A. 5
B. 9
C. `11.5`
D. `2.5`

Answer

B

Explanation

`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`.