Calculate the concentration of `OH^(-)` in a fruit juice which contains `2xx10^(-3)M, H_(3)O^(+)` ion. Identify the nature of the solution.

Explanation

`"Given that "H_(3)O^(+)=2xx10^(-3)M`
`K_(w)=[H_(3)O^(+)][OH^(-)]`
`therefore [OH^(-)]=(K_(w))/([H_(3)O^(+)])=(1xx10^(-14))/(2xx10^(-3))=0.5xx10^(-11)M`
`2xx10^(-3) gt gt 0.5 xx 10^(-11)`
i.e., `[H_(3)O^(+)] gt gt [OH^(-)]`, hence the juice is acidic in nature