Calculate the `pOH` of solution at `25^(@)C` that contains `1xx10^(-10) M` of hydronium ions, i.e., `H_(3)O^(+)`

Options

A. `7.00`
B. `4.00`
C. `9.00`
D. `1.00`

Answer

B

Explanation

`[H_(3)O^(+)]=[H^(+)]=10^(-10)`
`pH+pOH=14` …..(i)
and `pH=-log[H^(+)]`
`pH=-log[10^(-10)]` …(ii)
pH = 10
from eq. (i) and (ii), we get
`pOH+10=14`
`pOH=14-10=4`