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Q.
Calculate $pH$ at which an acid indicator HIn with concentration $0.1 \,M$ changes its colour $\left( K _{ a }\right.$ for HIn $=\left.1 \times 10^{-5}\right)$
Equilibrium
Solution:
Indicator equation is: $pH = pK _{\text{In }}+\log \frac{\left[\text{In} ^{-}\right]}{[ \text{H In }]}$
Colour change takes place when $\left[ \text{In} ^{-}\right]=[\text{H In }]$
i.e., when $pH = pK _{\text{In }}=5$