Calculate the `pH` at which an acid indicator with `K_(a) = 1.0 xx 10^(-5)` changes colour when the indicator is `1.00 xx 10^(-3)M`.

Explanation

The midpoint of the colour change range of an idicator is the point at which its acid and conjugate base forms are present in equal concentration, hence
`K_(ind) = ([H_(3)O^(o+)][Ind^( Θ)])/([HIn]) =[H_(3)O^(o+)] = 1.0 xx 10^(-5)` and `pH = 5.00`