If the solubility product of `MOH` is `1xx10^(-10) mol^(2) dm^(-6)` then `pH` of its aqueous solution will be

Options

A. 11
B. 9
C. 13
D. 5

Answer

B

Explanation

`K_(SP)` of `MOH=1xx10^(-10)M^(2)`
`P^(H)=?`
MOH is AB type of salt
`:.K_(SP)=S^(2)`
`[OH^(-)]=S`
`S=sqrt(K_(SP))=sqrt(1xx10^(-10))`
`S=10^(-5)`
`[OH^(-)]=10^(-5)`
`P^(OH)=-log_(10)[OH^(-)]=-log_(10)10^(-5)=5`
`P^(H)+P^(OH)=14`
`:.P^(H)=14-5=9`