Question

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?

Short answer

The initial color of the solution is yellow. The color change from yellow to greenish-yellow occurs at a pH of 9. The color of the solution after adding 200.0 mL of NaOH will be blue.

Step-by-step solution

a. Initial color of the solution

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.

b. pH at which the color change occurs

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.

c. Color after adding 200.0 mL of NaOH

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.

Key concepts

Equilibrium Constant

The equilibrium constant ( K ) is a fundamental concept in chemistry that describes the ratio of product concentrations to reactant concentrations at equilibrium for a reversible reaction. It is defined for a given reaction at a certain temperature. In this exercise, the equilibrium constant pertains to the acid-base indicator HIn:

• For the reaction HIn ⇌ H⁺ + In^-, the expression is K = [In^-][H⁺]/[HIn].
• In the context of acids and bases, K is crucial for understanding at what point the indicator changes color.

The equilibrium constant is particularly small in this case (1.0 × 10^-9), implying that, initially, HIn predominantly exists in its undissociated form (yellow HIn) – explaining the initial color of the solution in an acidic environment.
Understanding how K affects the equilibrium position helps predict when the color change of the indicator occurs during a titration.

Color Change

Color change in chemical solutions serves as a visual cue for various chemical processes. With indicators, a color change points to a shift in the equilibrium due to a change in pH . Indicators like HIn are specially chosen for their ability to show different colors based on pH changes.

• HIn is yellow, and upon deprotonation, In^- becomes blue.
• The transition between these colors helps identify a specific range of pH where the molecule shifts between the two forms.

In this exercise, the change from yellow to greenish-yellow (a mixture of undissociated HIn and deprotonated In^-) occurs when the pH is 9. This is when concentrations of HIn and In^- are equal, indicating a balance between the acidic and its conjugate base form.

Titration

Titration is an analytical technique used to determine the concentration of a solution, where you add a titrant to a solution until a reaction completes. In this problem, we titrate 0.10 M HCl with 0.10 M NaOH:

• The HCl is a strong acid, and NaOH is a strong base, they react completely to form water and salt.
• As NaOH is added, the pH of the solution gradually changes, leading to a color change of the indicator.

Titration is vital to see the endpoint of the reaction, which is detected by the color change of the indicator. It shows the point at which the amounts of titrant added and the analyte solution have reacted stoichiometrically, further confirmed by reaching a pH where the indicator changes color.

pH Calculation

Understanding how to calculate pH is essential in chemistry, indicating the acidity or alkalinity of a solution. It is inversely related to hydrogen ion concentration:

• \[ pH = -\log{[H^+]} \]
• On the pH scale, values range from 0 (highly acidic) to 14 (highly basic).

During this exercise, we saw two important calculations:

• At the pH where the color change occurs (pH = 9), [H⁺] = 10^-9 M, showing when HIn equals In^- concentration.
• After adding excess NaOH, calculating the remaining NaOH's concentration after reaction provides the pH of 12.52, indicating a basic solution.

These calculations help you understand the interplay between pH, the extent of neutralization in titration, and the visual indicator response.