A monoprotic acid `(HA)` is `1%` ionised in its aqueous solution of `0.1M` strength. Its `pOH` will be

Options

A. `11`
B. `3`
C. `10`
D. `2`

Answer

A

Explanation

`{:(,HA+,H_(2)OhArr,H_(3)O^(o+)+,A^(Theta)),("Initial",1,0,0,0),("Final",C(1-alpha),,Calpha,Calpha):}`
`[H_(3)O^(o+)] = Calpha = 0.1 xx 0.01 = 10^(-3)`.
`pH = 3, pOH = 11`.