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`.

Answer

5

Explanation

The dissociation equilibrium for acid indicator say H In is
`{:(H InhArr,In^(-),+H^(+)),("Colour A","ColourB",):}`
`K_(In)= ([In^(-)][H^(+)])/([H In])`
The middle point of the colour range of an indicator H In is the point at which
`[H In]=[In^(-)]=1`
`:. K_(In)= [H^(+)]`
or `[H^(+)]=10^(-5) or pH=5`