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(A) For an acid-base indicator, the color change occurs when the concentration of the ionized form equals the concentration of the unionized form, i.e

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$5$

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(A) For an acid-base indicator, the color change occurs when the concentration of the ionized form equals the concentration of the unionized form, i.e., $[In^-] = [HIn]$.Using the Henderson-Hasselbalch equation for an indicator: $pH = pK_{in} + \log \frac{[In^-]}{[HIn]}$.Since $[In^-] = [HIn]$, the ratio $\frac{[In^-]}{[HIn]} = 1$.Therefore, $pH = pK_{in} + \log(1) = pK_{in}$.Given $K_{in} = 1.0 	imes 10^{-5}$, then $pK_{in} = -\log(1.0 	imes 10^{-5}) = 5$.Thus, $pH = 5$.

For an acid indicator with $K_{in} = 1.0 \times 10^{-5}$, the color change occurs when the concentration of the indicator is $1 \times 10^{-3} \ M$. Find the $pH$ at which the color change occurs.

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