Problem 17 - Biochemistry: A Short Course

Question

A dye that is an acid and that appears as different colors in its protonated and deprotonated forms can be used as a pH indicator. Suppose that you have a \(0.001 \mathrm{M}\) solution of a dye with a \(\mathrm{p} K_{\mathrm{a}}\) of \(7.2 .\) From the color, the concentration of the protonated form is found to be \(0.0002 \mathrm{M} .\) Assume that the remainder of the dye is in the deprotonated form. What is the \(\mathrm{pH}\) of the solution? \( {6}\)

Step-by-Step Solution

Answer

The pH of the solution is approximately 7.802.

Step 1: Understand the Equilibrium Equation

The dye exists in equilibrium between its protonated form \( D_{H} \) and its deprotonated form \( D^{-} \). The equilibrium can be represented by the equation: \( D_{H} \rightleftharpoons H^{+} + D^{-} \). The equilibrium constant for this reaction is the acid dissociation constant \( K_a \), which relates to \( pK_a \) by \( pK_a = -\log K_a \).

Step 2: Use the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is given by: \( \text{pH} = pK_a + \log \left( \frac{[A^-]}{[HA]} \right) \). Here, \([A^-]\) is the concentration of the deprotonated form \( D^{-} \), and \([HA]\) is the concentration of the protonated form \( D_H \).

Step 3: Calculate the Concentration of the Deprotonated Form

Since the solution is \(0.001 \text{ M}\) and \(0.0002 \text{ M}\) is protonated, the concentration of the deprotonated form \([D^-]\) is \(0.001 \text{ M} - 0.0002 \text{ M} = 0.0008 \text{ M}\).

Step 4: Plug Values into the Henderson-Hasselbalch Equation

Substitute \( pK_a = 7.2 \), \([A^-] = 0.0008 \text{ M}\), and \([HA] = 0.0002 \text{ M}\) into the equation: \[ \text{pH} = 7.2 + \log \left( \frac{0.0008}{0.0002} \right) \]

Step 5: Calculate the Logarithmic Fraction

Calculate the logarithm: \( \frac{0.0008}{0.0002} = 4 \), thus \( \log(4) \approx 0.602 \).

Step 6: Determine the pH

Add \(0.602\) to \(7.2\): \( \text{pH} = 7.2 + 0.602 = 7.802 \).

Key Concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is an essential tool for calculating the pH of a solution containing a weak acid and its conjugate base. This equation provides a straightforward method to determine the pH by using the concentration of the protonated and deprotonated forms of an acid. It is particularly useful in buffer solutions and for pH indicators, where the acid and its conjugate base exist in equilibrium.

The equation itself is expressed as:

• \( \text{pH} = \text{p}K_a + \log \left( \frac{[A^-]}{[HA]} \right) \)

Where:

• \([A^-]\) is the concentration of the deprotonated form (or conjugate base).
• \([HA]\) is the concentration of the protonated form (or acid).
• \(\text{p}K_a\) is the negative log of the acid dissociation constant.

In the context of the exercise, the equation allows you to find the pH of a dye solution by using the known \(\text{p}K_a\) value and the concentrations of the dye in its protonated and deprotonated forms. This approach highlights how the balance between these forms influences the solution's pH.

acid-base equilibrium

Acid-base equilibrium describes the balance between acids and bases in a solution, particularly when they can donate or accept protons. This equilibrium is central in the study of reactions involving acids and bases, such as pH indicators and buffer systems.

For instance, in the exercise, the equilibrium exists between the protonated dye form \( D_H \) and the deprotonated form \( D^- \). The reaction can be written as:

• \( D_H \rightleftharpoons H^+ + D^- \)

This balanced reaction implies that at any given time, the forward reaction of donating protons (forming \(H^+\) and \(D^-\)) and the reverse reaction (where \(D^-\) gains protons to become \(D_H\)) occur at the same rate. This dynamic balance is what maintains the constant concentrations of the protonated and deprotonated dye under stable conditions.

The equilibrium constant \(K_a\) for this reaction helps quantify the strength of the acid, or how much it dissociates in solution. For dyes and pH indicators, this equilibrium is crucial, as it determines the color change that occurs at a specific pH value.

protonation and deprotonation

Protonation and deprotonation refer to the processes of gaining and losing a proton (\( H^+ \)), respectively. These processes are essential in many chemical reactions and play a crucial role in maintaining acid-base equilibrium.

For an acid like the dye mentioned in the exercise, protonation occurs when the dye molecule gains a proton, forming the protonated species \(D_H\). Deprotonation, on the other hand, occurs when the dye loses a proton, resulting in the deprotonated form \(D^-\). Both processes are reversible and occur readily in solution.

The understanding of protonation and deprotonation is pivotal in explaining the function of pH indicators. These indicators exist in two forms, and their color depends on the concentration ratio of these forms at a given pH. In our exercise, when you calculated the solution's pH, you effectively determined the balance point where the visible color change in the dye corresponds to the specific ratio of its protonated and deprotonated forms.