Two drops of indicator \(\operatorname{HIn}\left(K_{\mathrm{a}}=1.0 \times 10^{-9}\right)\), where \(\mathrm{HIn}\) is yellow and \(\mathrm{In}^{-}\) is blue, are placed in \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{HCl}\). a. What color is the solution initially? b. The solution is titrated with \(0.10 \mathrm{M} \mathrm{NaOH}\). At what \(\mathrm{pH}\) will the color change (yellow to greenish yellow) occur? c. What color will the solution be after \(200.0 \mathrm{~mL} \mathrm{NaOH}\) has been added?

Since we have \(\mathrm{HIn}\) in \(0.10 \mathrm{M} \mathrm{HCl}\) solution, the strong acid will protonate the indicator, making being in the acid form. The acid form of the indicator (HIn) is yellow. Therefore, the initial color of the solution will be yellow. Answer (a): The initial color of the solution is yellow.

The color change occurs when the weak acid and its conjugate base are in equal proportion. The Henderson-Hasselbalch equation relates the \(\mathrm{pH}\), \(\mathrm{p}K_{\mathrm{a}}\), and the ratio of concentrations of the conjugate base to weak acid as follows: \[\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log \frac{[\mathrm{In}^{-}]}{[\mathrm{HIn}]}\] We need to find the \(\mathrm{pH}\) value when \([\mathrm{In}^{-}] = [\mathrm{HIn}]\). In this case: \[\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log \frac{[\mathrm{In}^{-}]}{[\mathrm{HIn}]} = \mathrm{p}K_{\mathrm{a}} + \log(1) = \mathrm{p}K_{\mathrm{a}}\] Now, calculate the \(\mathrm{p}K_{\mathrm{a}}\) value: \[\mathrm{p}K_{\mathrm{a}} = -\log(K_{\mathrm{a}}) = -\log(1.0 \times 10^{-9}) = 9\] Answer (b): The color change will occur at \(\mathrm{pH} = 9\).

To find the final color, we can calculate the final pH of the solution after adding 200.0 mL of 0.10 M \(\mathrm{NaOH}\) to 100.0 mL of 0.10 M \(\mathrm{HCl}\). Using the stoichiometry of the reaction, moles of \(\mathrm{HCl}\) can be calculated as: moles of \(\mathrm{HCl}\) = molarity × volume = 0.10 \(\mathrm{mol/L} \) × 0.100 \(\mathrm{L} \) = 0.010 \(\mathrm{mol} \) moles of \(\mathrm{NaOH}\) = 0.10 \(\mathrm{mol/L} \) × 0.200 \(\mathrm{L} \) = 0.020 \(\mathrm{mol} \) Since the moles of \(\mathrm{NaOH}\) are greater than the moles of \(\mathrm{HCl}\), the solution is now a basic solution (the excess OH- ions will remain in the solution). After the reaction between \(\mathrm{NaOH}\) and \(\mathrm{HCl}\), there will still be some \(\mathrm{NaOH}\) left: moles of excess \(\mathrm{NaOH}\) = moles of \(\mathrm{NaOH}\) - moles of \(\mathrm{HCl}\) = 0.020 \(\mathrm{mol} \) - 0.010 \(\mathrm{mol} \) = 0.010 \(\mathrm{mol} \) Now, calculate the final concentration of excess \(\mathrm{NaOH}\): \[ [\mathrm{OH}^-] = \frac{0.010 \, \mathrm{mol}}{0.100 \, \mathrm{L}+0.200 \, \mathrm{L}} = 0.0333 \, \mathrm{M} \] Now, calculate the \(\mathrm{pOH}\) and \(\mathrm{pH}\): \[\mathrm{pOH} = -\log({[\mathrm{OH}^-]}) = -\log(0.0333) = 1.48\] \[\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 1.48 = 12.52\] Since the final pH is greater than 9 (the pH at which the color change occurs), the indicator will be present predominantly in its conjugate base form (In-), which is blue. Answer (c): The final color of the solution will be blue.