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

Options

A. 3
B. 6
C. 9
D. 12

Answer

C

Explanation

`K_(sp)` of MOH =`1xx10^(-10)`
`[M^+][OH^-]=1xx10^(-10)`
Now, `[M^+]=[OH^-]`
`therefore [OH^-]^2 =1xx10^(-10)` or `[OH^-]=10^(-5)`
`[H_3O^+]=10^(-14)/10^(-5)=10^(-9)`
pH=-log `(10^(-9))`=9