Two drops of indicator HIn \(\left(K_{2}=1.0 \times 10^{-9}\right),\) where HIn is yellow and In - is blue, are placed in 100.0 \(\mathrm{mL}\) of 0.10 \(\mathrm{MCl}\) . a. What color is the solution initially? b. The solution is titrated with 0.10\(M\) NaOH. At what pH will the color change (yellow to greenish yellow) occur? c. What color will the solution be after 200.0 \(\mathrm{mL}\) NaOH has been added?

When the indicator HIn is mixed with 0.10 M HCl, it will react with the acidic solution as follows: HIn + HCl → H₂In⁺ + Cl⁻ Since HCl is a strong acid, it will completely dissociate, moving the equilibrium to the right and protonating the HIn to form H₂In⁺. Since HIn is yellow and In⁻ is blue, the initial color will be yellow, as the equilibrium heavily lies in the protonated form of the indicator. Answer (a): The initial color of the solution is yellow.

Now, the equilibrium expression for the dissociation of HIn is as follows: \[\frac{[In⁻]}{[HIn]} = \frac{1.0 \times 10^{-9}}{[H⁺]}\] We can rearrange this expression: \([H⁺] = \frac{[HIn]}{[In⁻]} \times (1.0 \times 10^{-9})\) Since the solution is titrated with NaOH, the OH⁻ ions created from NaOH will react with H₂In⁺ to form HIn and water. When the molar concentration of In⁻ equals that of HIn, the system will display a greenish-yellow color. Thus, if we want to find out the pH at which the color change occurs, we can set \([HIn] = [In⁻]\) and calculate the H⁺ concentration: \([H⁺] = \frac{1 \times (1.0 \times 10^{-9})}{1}\) \([H⁺] = 1.0 \times 10^{-9} \: M\) Now, we can calculate the pH: \(pH = -\log{[H⁺]}\) \(pH = -\log{(1.0 \times 10^{-9})}\) \(pH = 9\) Answer (b): The color change from yellow to greenish-yellow occurs at a pH of 9.

The initial volume is 100.0 mL of 0.10 M HCl, which means there are 0.01 mol of HCl. When we add the 200.0 mL of 0.10 M NaOH, it results in 0.02 mol of NaOH. By stoichiometry of the reaction between HCl and NaOH, one mole of each reactant yields one mole of each product: HCl + NaOH → NaCl + H₂O Since there is more NaOH than HCl, there will be an excess of 0.01 mol NaOH present in the solution. The pOH of the final solution can be calculated by dividing the excess moles of NaOH by the total volume of the solution (100 mL + 200 mL): \(pOH = -\log{\frac{0.01\: mol\: NaOH}{(100 + 200)\: mL\: solution}}\) \(pOH = -\log{\frac{0.01}{0.3}}\) \(pOH = 1.48\) Now let's calculate the pH: \(pH = 14 - pOH = 14 - 1.48 = 12.52\) Since the pH is above 9, which was the color-change transition value, the solution will have a blue color due to the high concentration of In⁻ ions in the solution. Answer (c): The color of the solution after adding 200.0 mL of NaOH will be blue.