At $\mathrm{pH}=\mathrm{p}{K}_{\mathrm{Hin}}^{\circ }+1$, an indicator is present in

Here $\log \left(\left[{\mathrm{In}}^{-}\right]/\left[\mathrm{HIn}\right]\right)=1$ 

or $\left[{\mathrm{In}}^{-}\right]/\left[\mathrm{HIn}\right]=10/1.$

Percentage ionization$=[10/(10+1\left)\right]\times 100\simeq 91\mathrm{\%}$

The percentage of a chemical that has undergone ionization is said to be ionized. Therefore, we compare the concentration of the ion in solution with the initial concentration of the neutral species.