Question

For the indicator phenolphthalein (In \(^{-}\) ), HIn is colorless and \(\mathrm{In}^{-}\) is red; \(\mathrm{K}_{\mathrm{diss}}\) is \(10^{-9} .\) If one has a colorless solution of 50 \(\mathrm{ml}\) phenolphthalein, how much \(0.10 \mathrm{M} \mathrm{NaOH}\) solution would one have to add to make it red: \(\mathrm{K}_{\mathrm{W}}=10^{-14}\).

Short answer

To make the phenolphthalein solution turn red, one would need to add approximately \(0.5\) mL of \(0.10\) M NaOH solution.

Step-by-step solution

Determine the pH at which phenolphthalein changes color

First, we need to find the pH at which phenolphthalein changes color. This occurs when the colorless HIn form is equal to the red In⁻ form. We can use the dissociation constant, Kdiss, to determine this point. The dissociation constant is defined as: \[ \mathrm{K_diss} = \frac{[\mathrm{In}^-]}{[\mathrm{HIn}]}\] Since the [HIn] and [In⁻] will be equal at the point of color change, we can denote this as x: \[10^{-9} = \frac{x}{x}\] The equation simplifies to x^2 = 10^{-9}. We can solve for x to find the concentration of HIn and In⁻ at the color change point.

Calculate x, the common concentration of HIn and In⁻ at the color change point

We can now solve for x, the common concentration of phenolphthalein's form: \[x^2 = 10^{-9}\] \[\Rightarrow x = \sqrt{10^{-9}}\] \[\Rightarrow x = 10^{-4.5}\] So at the point of color change, the concentration of HIn and In⁻ is 10^{-4.5} mol/L.

Use the concentration of HIn to calculate the pH at the color change point

Since HIn is the acidic form of phenolphthalein, we can treat it as a weak acid and find the pH of a solution containing HIn in equilibrium. Knowing that: \[10^{-4.5} = [\mathrm{HIn}^-]=[\mathrm{H}^+]\] We can convert the concentration of H+ ions to pH using the formula: \[ \mathrm{pH} = -\log [\mathrm{H}^+]\] Calculate the pH: \[\mathrm{pH} = -\log{10^{-4.5}}\] \[\mathrm{pH} = 4.5\] Thus, the color change occurs at pH = 4.5.

Determine the volume of NaOH needed to reach the color change point

Now that we know the pH at the color change point, we can calculate the volume of 0.1 M NaOH solution required to reach this pH. When NaOH is added to an acidic solution, it will react with the H+ ions, increasing the pH of the mixture: \[\mathrm{OH}^- + \mathrm{H}^+ \rightarrow \mathrm{H_2O}\] Let the volume of NaOH solution to be added be denoted as V. The moles of OH⁻ in the NaOH solution are given by the product of volume, V (in liters), and concentration, 0.10 mol/L: \[\mathrm{OH}^- = (0.10 \mathrm{M}) \times V\] Initially, there are moles of HIn in the solution: \[\mathrm{HIn} = (10^{-4.5}) \times 0.050 \mathrm{L}\] At the color change point, the moles of HIn and OH⁻ are equal. Therefore, we can write the following equation: \[(0.10 \mathrm{M}) \times V = (10^{-4.5}) \times 0.050 \mathrm{L}\]

Calculate the volume of NaOH needed

Now, we can solve for the volume of NaOH needed: \[V = \frac{(10^{-4.5}) \times 0.050 \mathrm{L}}{(0.10\mathrm{M})}\] \[V = 5 \times 10^{-4}\mathrm{L}\] Converting the volume to milliliters: \[V = 5 \times 10^{-4}\mathrm{L} \times \frac{1000 \mathrm{mL}}{1 \mathrm{L}}\] \[V \approx 0.5 \mathrm{mL}\] Thus, to make the phenolphthalein solution turn red, one would need to add approximately 0.5 mL of 0.10 M NaOH solution.

Key concepts

Phenolphthalein Indicator

Phenolphthalein is a common acid-base indicator used to determine the pH of a solution. It is initially colorless in acidic environments but turns pink or red as the pH becomes slightly basic. This color change occurs because phenolphthalein exists in two different forms: the acidic form (HIn), which is colorless, and the basic form (\(\mathrm{In}^{-}\)), which is colored. When you add a base to a phenolphthalein solution and the pH increases, the equilibrium between these two forms shifts, leading to a visible color change.
For phenolphthalein, this color transition typically happens around a pH of 8.3 to 10. However, in the given exercise, we figured out the specific conditions where the concentration of the colored and colorless forms are equal, leading to a pH of 4.5. This is quite different from the usual transition range, indicating particular experimental circumstances or assumptions.
To determine the exact pH for a color change, scientists utilize the dissociation constant (as explained in the next section) and equal concentrations of the forms in equilibrium.

Dissociation Constant

The dissociation constant (denoted as \(K_\text{diss}\)) is a crucial value in understanding the behavior of weak acids or bases, such as phenolphthalein. It gives us insight into how a compound dissociates into ions in a solution. In our exercise, \(K_\text{diss}\) was provided as \(10^{-9}\). This value tells us the ratio of the concentration of the dissociated ions to the undissociated form at equilibrium.
The dissociation constant can be given by the formula:

• \(K_\text{diss} = \frac{[\mathrm{In}^-]}{[\mathrm{HIn}]}\)

Here, \(\mathrm{In}^{-}\) is the ion that makes the solution turn red, while \(\mathrm{HIn}\) is the colorless form. When both forms are present in equal concentrations, the value of this expression is equal to \(K_\text{diss}\).
For phenolphthalein to change color, these two concentrations must become equal, which is how the pH at the color change is determined. Knowing \(K_\text{diss}\), you can calculate the necessary concentrations and subsequently, the pH, by using the relation of concentration with \(10^{-9}\).

pH Calculation

Calculating the pH of a solution during an acid-base reaction is a fundamental skill in chemistry and involves understanding how hydrogen ion concentration converts into a pH value. In this exercise, once the dissociation constant and concentration of ions are determined at the color change point, the pH is calculated as follows:

• Find the concentration of hydrogen ions \([\mathrm{H}^+]\)
• Use the formula \( \mathrm{pH} = -\log [\mathrm{H}^+]\) to find the pH

In our specific example, we calculated the hydrogen ion concentration at \(10^{-4.5}\) mol/L.
From this concentration, the pH is found using the formula: \( \mathrm{pH} = -\log(10^{-4.5})\), providing us with a pH of 4.5.
This calculation represents the precise moment of phenolphthalein's color-changing point under the conditions of the experiment, highlighting the importance of accurately determining \([\mathrm{H}^+]\) for pH-dependent reactions.

NaOH Solution Volume

To arrive at a specific pH, particularly to observe a color shift in a titration experiment, it's necessary to add an accurate amount of a base, such as NaOH. In this exercise, we focused on calculating the volume of a 0.10 M NaOH solution needed to reach the pH where phenolphthalein turns red.
NaOH is a strong base, and when dissolved, it provides hydroxide ions \((\mathrm{OH}^-)\) that react with hydrogen ions \((\mathrm{H}^+)\) from the acidic solution. This reaction can be expressed as:

• \(\mathrm{OH}^- + \mathrm{H}^+ \rightarrow \mathrm{H_2O}\)

We calculated the moles of NaOH by multiplying the volume of NaOH (in liters) by its concentration (0.10 M). These moles must equal the moles of hydrogen ions required to correct the pH:

• The formula used was \((0.10 \ \text{mol/L}) \times V = (10^{-4.5}) \times 0.050 \ \text{L}\).

After rearranging and solving, we determined \(V\) to be 5 mL or 0.5 mL when adjusted to more practical units. This calculation is critical in identifying just how much base is necessary to reach the target pH.