Question

Calculate the $pH$ at which an acid indicator $H{I}_n$ with concentration $0.1M$ changes its colour.
($K_a$ for $H{I}_n=1\times {10}^{-5}$)

Answer 1

This indicator $\left(H{I}_n\right)$ is a weak acid.

${HI}_n+H_{2}O\rightleftharpoons H_3{O}^{+}+I_n^{-}$

$\therefore$ $K_s=\frac{\left[H_3{O}^{+}\right]\left[I_n^{-}\right]}{\left[{HI}_n\right]}$

$=\left[H_3{O}^{+}\right]$ [Since at equivalence point the concentration of indicator and it's conjugate base are equal.]

$\therefore$ ${pK}_s=pH$

$\therefore$ $pH={pK}_{I_n}=5$